uVA :: 908 :: Re-connecting Computer Sites

Problem : Please find the problem here.

Explanation :The previously chosen path is already given so the first answer is just the sum of the cost of these paths. For the second cost, combine the original edges with the newly given edges and find the weight of the MST associated to that set of edges.

Code : Used Kruskal's Algorithm to find the MST.

#include <bits/stdc++.h>
using namespace std;

const int mxN = 1e6;
int n, t, m, k;
int parent[mxN], ranks[mxN];

struct Edge{
    int src, dst, cost;
    
    Edge(int _src, int _dst, int _cost){
        src = _src;
        dst = _dst;
        cost = _cost;
    }

    bool operator<(Edge const &other){
        return cost < other.cost;
    }
};

void make_set(int v){
    parent[v] = v;
    ranks[v] = 0;
}

int find_set(int v){
    if(v == parent[v]){
        return v;
    }
    return parent[v] = find_set(parent[v]);
}

void union_sets(int a, int b){
    a = find_set(a);
    b = find_set(b);
    if(a != b){
        if(ranks[a] < ranks[b]){
            swap(a, b);
        }
        parent[b] = a;
        if(ranks[a] == ranks[b]){
            ranks[a]++;
        }
    }
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    ofstream fout("out");
    bool first = 1;

    while(cin >> n){
        int src, dst, cost;
        vector<Edge> new_edges;
        int ans1 = 0, ans2 = 0;
        for(int i = 0; i < n-1; i++){
            cin >> src >> dst >> cost;
            //chosen_edges.emplace_back(Edge(src, dst, cost));
            ans1 += cost;
        }
        cin >> k;
        for(int i = 0; i < k; i++){
            cin >> src >> dst >> cost;
            new_edges.push_back(Edge(src, dst, cost));
        }
        cin >> m;
        for(int i = 0; i < m; i++){
            cin >> src >> dst >> cost;
            new_edges.emplace_back(Edge(src, dst, cost));
        }

        for(int i = 0; i < (int)new_edges.size(); i++){
            make_set(i);
        }

        sort(new_edges.begin(), new_edges.end());
        for(Edge e : new_edges){
            if(find_set(e.src) != find_set(e.dst)){
                ans2 += e.cost;
                union_sets(e.src, e.dst);
            }
        }
        if(!first){
            cout << '\n';
        }else{
            first = 0;
        }
        cout << ans1 << '\n';
        cout << ans2 << '\n';
    }
    return 0;
}

Uva :: 11631 :: Dark roads

Problem : Please find the problem here.

Explanation : The answer is the total weight of the graph minus the weight of the minimum spanning tree(MST).

Code : Used Kruskal's Algorithm to find the MST.

#include <bits/stdc++.h>
using namespace std;

const int mxN = 2e5;
int n, m;
int parent[mxN], ranks[mxN];

struct Edge{
    int src, dst, weight;
    bool operator<(Edge const& other){
        return weight < other.weight;
    }
    Edge(int _src, int _dst, int _weight){
        src = _src;
        dst = _dst;
        weight = _weight;
    }
};
void make_set(int v){
    parent[v] = v;
    ranks[v] = 0;

}

int find_set(int v){
    if(v == parent[v]){
        return v;
    }
    return parent[v] = find_set(parent[v]);
}

void union_sets(int a, int b){
    a = find_set(a);
    b = find_set(b);
    if(a != b){
        if(ranks[a] < ranks[b]){
            swap(a, b);
        }
        parent[b] = a;
        if(ranks[a] == ranks[b]){
            ranks[a]++;
        }
    }
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    while(cin >> n >> m && (n || m)){
        
        int src, dst, weight;
        int total_weight = 0;
        vector<Edge> edges, result;
        int cost = 0;
        // make disjoint sets for each of the vertex in the graph;
        for(int i = 0; i < n; i++){
            make_set(i);
        }

        // add edges;
        for(int i = 0; i < m; i++){
            cin >> src >> dst >> weight;
            total_weight += weight;
            //cout << src << ' ' << dst << ' ' << weight << '\n';
            edges.push_back(Edge(src, dst, weight));
        }

        // one can sort this edges or just store the edges in the priority queue with required operator..;
        sort(edges.begin(), edges.end());
        // now the unification process.
        // pick all the edges from first to last and if the ends of current edge belong to different set,
        // these sets are combined  and the edge is added to the result.
        for(Edge e : edges){
            if(find_set(e.src) != find_set(e.dst)){
                cost += e.weight;
                result.push_back(e);
                union_sets(e.src, e.dst);
            }
        }
        // After iterating through all the edges, all the vertices will belong to the same subtree.
        cout << total_weight-cost << '\n';
        // The answer is the total_weight of the graph minus the weight of the MST;
    }
    return 0;
}

UVa :: 10986 :: Sending email

Problem : Please find the problem here.

Explanation : Given an undirected weighted graph, find the minimum distance between two given nodes. The problem is naive and clearly suggested to use the algorithm suited to calculate the SSSP.

Code : Used Dijkstra's Algorithm.

#include <bits/stdc++.h>
using namespace std;

const int mxN = 2e5;
int n, m, s, t, d[mxN];
vector<array<int, 2>> adj[mxN];
bool vis[mxN];


void init(){
    memset(vis, false, sizeof(vis));
    for(int i = 0; i < n; i++){
        adj[i].clear();
        d[i] = INT_MAX;
    }
}

void solve(){
    cin >> n >> m >> s >> t;
    init();
    for(int i = 0, u, v, w; i < m; i++){
        cin >> u >> v >> w;
        adj[u].push_back({v, w});
        adj[v].push_back({u, w});
    }
    priority_queue<array<int,2>, vector<array<int, 2>>, greater<array<int, 2>>> pq;
    d[s] = 0;
    pq.push({0, s}); 
    while(pq.size()){
        array<int, 2> uu = pq.top();
        pq.pop();
        int u = uu[1];
        if(vis[u]) continue;
        vis[u] = 1;
        for(array<int, 2> vv : adj[u]){
            int v = vv[0], w = vv[1];
            if(d[v] > d[u]+w){
                d[v] = d[u]+w;
                pq.push({d[v], v});
            }
        }
    }
    if(d[t] == INT_MAX) cout << "unreachable\n";
    else cout << d[t] << '\n';
    
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int test_case;
    cin >> test_case;
    for(int i = 1; i <= test_case; i++){
        cout << "Case #" << i << ": ";
        solve();
    }
    return 0;
}

UVa :: 11709 :: Trust groups

Problem : Please find the problem here.

Explanation : Find the number of strongly connected components.  This can be done using Kosaraju's algorithm or Tarjan's Algorithm. I used the first one. Basically just do topological sort on the graph and then count the connected components with the DFS or BFS.

Code : Used Kosaraju's Algorithm.

#include <bits/stdc++.h>
using namespace std;

const int mxN = 1e3;
int n, m;
string in;
map<string, int> node_id;
vector<int> adj[mxN], adjr[mxN], ts;
bool vis[mxN];

void dfs1(int u){
    vis[u] = 1;
    for(int v : adj[u]){
        if(!vis[v]){
	    dfs1(v);
	}
    }
    ts.push_back(u);
}

void dfs2(int u){
    vis[u] = 1;
    for(int v : adjr[u]){
	if(!vis[v]){
	    dfs2(v);
	}
    }
}

// Kosaraju's Algorithm

int SCC_count(){
	
    for(int i = 0; i < n; i++){
	if(!vis[i]){
	    dfs1(i);
	}
    }
    reverse(ts.begin(), ts.end());
    memset(vis, false, sizeof(vis));
    int cnt = 0;
    for(int i = 0; i < n; i++){
        if(!vis[ts[i]]){
	    dfs2(ts[i]), cnt++;
	}
    }
    return cnt;
}

void init(){
    memset(vis, false, sizeof(vis));
    ts.clear();
    node_id.clear();
    for(int i = 0; i < n; i++){
        adj[i].clear();
	adjr[i].clear();
    }
}

int main()
{
    while(cin >> n >> m && (n || m)){
        init();
	cin.ignore();
	int id = 0;
	for(int i = 0; i < n; i++){
	    getline(cin, in);
	    if(node_id.find(in) == node_id.end()){
		node_id[in] = id++;
	    }
	}
	for(int i = 0, u, v; i < m; i++){
	    getline(cin, in);
	    u = node_id[in];
	    getline(cin, in);
	    v = node_id[in];
	    adj[u].emplace_back(v);
	    adjr[v].emplace_back(u);
        }
	cout << SCC_count() << '\n';
    }	
    return 0;
}

UVa : 11770 :: Lighting Away

Problem : Please find the problem here.

Explanation : Find the strongly connected components.  Basically just do topological sort on the graph and then count the connected components with the DFS or BFS.

Code : Used DFS.

#include <bits/stdc++.h>
using namespace std;

const int mxN = 1e4;
int n, m;
vector<int> adj[mxN], ts;
bool vis[mxN];

void dfs(int u){
    vis[u] = 1;

    for(int v : adj[u]){
	if(!vis[v]){
	    dfs(v);
	}
    }
    ts.push_back(u);
}

void init(){
    memset(vis, false, sizeof(vis));
    ts.clear();
    for(int i = 0; i < n; i++){
	adj[i].clear();
    }
}

int main()
{
    int t;
    cin >> t;
    for(int i = 1; i <= t; i++){
	cin >> n >> m;
	init();
	for(int j = 0, u, v; j < m; j++){
	    cin >> u >> v, u--, v--;
	    adj[u].emplace_back(v);
	}

        // doing topological sort;
        for(int j = 0; j < n; j++){
	    if(!vis[j]) dfs(j);
        }
        reverse(ts.begin(), ts.end());

        memset(vis, false, sizeof(vis));	
        int cnt = 0;
        for(int j = 0; j < n; j++){
            if(!vis[ts[j]]) dfs(ts[j]), cnt++;
        }
        cout <<"Case " << i << ": " << cnt << '\n';

    }	
    return 0;
}

Uva : 10199 :: Tourist Guide

Problem : Please find the problem here.

Explanation : Count the number of bridges in the graph

Code : Used DFS.

#include <bits/stdc++.h>
using namespace std;

const int mxN = 100;
int n, m, tin[mxN], low[mxN], timer;
string in, in2;
vector<int> adj[mxN];
map<string, int> node_id;
map<int, string> node_name;
set<string> ans;
bool vis[mxN];

void dfs(int u, int pu = -1){
    vis[u] = 1;
    tin[u] = low[u] = timer++;
    int children = 0;
    for(int v : adj[u]){
	if(v == pu) continue;
	if(vis[v]){
	    low[u] = min(low[u], tin[v]);
	}else{
	    dfs(v,u);
	    low[u] = min(low[u], low[v]);
	    if(low[v] >= tin[u] && pu != -1){
	        ans.insert(node_name[u]);
	    }
	    children++;
	}	
    }
    if(pu == -1 && children > 1){
	ans.insert(node_name[u]);
    }
}

void init(){
    memset(vis, false, sizeof(vis));
    memset(tin, -1, sizeof(tin));
    memset(low, -1, sizeof(low));
    for(int i = 0; i < n; i++){
	adj[i].clear();
    }
    node_id.clear();
    node_name.clear();
    ans.clear();
    timer = 0;
}

int main()
{
    ofstream fout("out");
    int __map = 1;
    cin >> n;
    while(1){
        if(n == 0){
	    break;
	}
	init();
	int id = 0;
	for(int i = 0; i < n; i++){
	    cin >> in;
	    node_id[in] = id;
	    node_name[id] = in;
	    id++;
	}
	cin >> m;
	for(int i = 0; i < m; i++){
            cin >> in >> in2;
	    adj[node_id[in]].emplace_back(node_id[in2]);
	    adj[node_id[in2]].emplace_back(node_id[in]);
	}
	for(int i = 0; i < n; i++){
	    if(!vis[i]) dfs(i);
	}
	printf("City map #%d: %d camera(s) found\n", __map, (int)ans.size());
	fout << "City map #" << __map++ <<": " << (int)ans.size() << " camera(s) found\n";
	for(string a : ans){
	    cout << a << '\n';
	    fout << a<< '\n';
	}
	cin >> n;
	if(n == 0){
	    break;
	}
	cout << '\n';
	fout << '\n';
    }
    return 0;
}

UVa : 796 :: Critical Links

Problem : Please find the problem here.

Explanation : Count the number of bridges in the graph. 

A bridge in a graph is an edge which when removed along with its associated edges disconnects the graph. 

Code : Used DFS.

#include <bits/stdc++.h>
using namespace std;

const int mxN = 1e3;
int n, m, tin[mxN], low[mxN], timer;
vector<int> adj[mxN];
string in;
set<array<int, 2>> ans;
bool vis[mxN];

void dfs(int u,  int pu = -1){

    vis[u] = 1;
    tin[u] = low[u] = timer++;
    for(int v : adj[u]){
	if (v == pu) continue;
	if (vis[v]){
	    low[u] = min(low[u], tin[v]);
	}else{
	    dfs(v, u);
	    low[u] = min(low[u], low[v]);
	    if(low[v] > tin[u]){
                array<int, 2> tmp={u, v};
		sort(tmp.begin(), tmp.end());
		ans.insert(tmp);
	    }
	}
    }
}

void init(){
    ans.clear();
    timer = 0;
    for(int i = 0; i < n; i++)
	adj[i].clear();

    memset(vis, false, sizeof(vis));
    memset(tin, -1, sizeof(tin));
    memset(low, -1, sizeof(low));
}

int main()
{
    ofstream fout("out");
    while(cin >> n){
	init();
	int u, v;
	for(int i = 0; i < n; i++){
	    cin >> u;
	    cin.ignore();
	    cin.ignore();
	    cin >> m;
	    cin.ignore();
	    while(m--){
		cin >> v;
		adj[u].emplace_back(v);
		adj[v].emplace_back(u);
	    }
	}
	for(int i = 0; i < n; i++){
	    if(!vis[i]) dfs(i);
	}
	printf("%d critical links\n", (int)ans.size());
	for(array<int, 2> a : ans){
	    cout << a[0] << " - " << a[1] << '\n';
	}
	cout << "\n";
    }
    return 0;
}

UVa : 315 :: Network

Problem : Please find the problem here.

Explanation : The problem basically asks to find the number of articulation points in the graph which can be done in using DFS with O(N+M) time complexity.

An Articulation Point is the verticex in the graph which when removed along with its associated edges, disconnects the graph.  

Code : Used DFS.

#include <bits/stdc++.h>
using namespace std;

const int mxN = 100;
int n, m, tin[mxN], low[mxN], timer, cnt;
bool vis[mxN];
string in;
vector<int> adj[mxN];
set<int> ans;

void dfs(int u, int pu = -1){

    vis[u] = 1;
    tin[u] = low[u] = timer++;
    int children = 0;
    for(int v : adj[u]){
	if(v == pu) continue;
	if(vis[v]){
	    low[u] = min(low[u], tin[v]);
	}else{
	    dfs(v, u);
	    low[u] = min(low[u], low[v]);
	    if(low[v] >= tin[u] && pu != -1){
		ans.insert(u);
		cnt++;
	    }
	    ++children;
	}
    }
    if(pu == -1 && children > 1){
	ans.insert(u);
	cnt++;
    }
}

void init(){
    timer = 0;
    cnt = 0;
    ans.clear();
    memset(vis, false, sizeof(vis));
    memset(tin, -1, sizeof(tin));
    memset(low, -1, sizeof(low));
    for(int i = 0; i < n; i++){

        adj[i].clear();
    }
}

int main()
{
    while(1){
	cin >> n;
	if(n == 0){
	    break;
	}
	init();
	cin.ignore();
	while(1){
	    getline(cin, in);
	    stringstream ss(in);
	    int u, v;
	    ss >> u;
	    if(u == 0){
		break;
	    }
	    while(ss >> v){
	        adj[u-1].emplace_back(v-1);
		adj[v-1].emplace_back(u-1);
	    }
	}
	for(int i = 0; i < n; i++){
	    if(!vis[i]) dfs(i);
	}
	cout << ans.size() << '\n';
    }	
    return 0;
}

UVa : 336 :: A Node Too Far

Problem : Please find the problem here.

Explanation : The problem basically asked and suggest to traverse the graph with BFS from a given node upto a given level. Remember that, BFS visit vertices that are direct neighbors of the source vertex (first layer), neighbors of direct neighbors (second layer), and so on, layer by layer. To identify which layer the given vertex belongs to, we can push the vertex with its level from the starting node in the queue. One more thing, the number of vertices are not given, so I made a edge-list first, mapped all the elements to the integers starting from zero, later created an adjacency-list with the given edge-list.

Code : Used BFS.

#include <bits/stdc++.h>
//#define debug
using namespace std;

const int mxN = 1e3;
int n, m;
vector<int> adj[mxN];
map<int, int> node_id;
vector<array<int, 2>> edges;
bool vis[mxN];

int bfs(int node, int time){

    memset(vis, false, sizeof(vis));

    queue<array<int, 2>> qu;
    qu.push({node, time});
    vis[node] = 1;
    while(qu.size()){
	array<int, 2> u = qu.front();
	qu.pop();
	if(u[1]){
	    for(int v : adj[u[0]]){
		if(!vis[v]){
		    qu.push({v, u[1]-1});
		    vis[v] = 1;
		}
	    }
	}

    }
    // now counting the NOT_VISITED nodes;
    int cnt = 0;
    for(int i = 0; i < n; i++){
        if(!vis[i]){
	    cnt++;
	}
    }
    return cnt;
}

void init(){

    edges.clear();

    node_id.clear();

    memset(vis, false, sizeof(vis));
    for(int i = 0; i < n; i++){
	adj[i].clear();
    }
}

void set_adj(){
    for(auto it : edges){
	adj[it[0]].emplace_back(it[1]);
	adj[it[1]].emplace_back(it[0]);
    }
}

int main()
{
    int __case = 1;
    while(cin >> m){
	int id = 0;
	init();
	for(int i = 0, u, v; i < m; i++){
	    cin >> u >> v;
	    if(node_id.find(u) == node_id.end()){
#ifdef debug
	        cout << u << " is mapped to " << id << '\n';
#endif
	        node_id[u] = id++;
	    }
	    if(node_id.find(v) == node_id.end()){
#ifdef debug

                cout << v << " v is mapped to " << id << '\n';
#endif
		node_id[v] = id++;
	    }
	    u = node_id[u], v = node_id[v];
	    edges.push_back({u, v});
	}
	n = node_id.size();
	set_adj();
#ifdef debug
	cout << "printing adjacency list\n";
	for(int i = 0; i < n; i++){
	    cout << i << "->";
	    for(int u : adj[i]){
		cout << u << ' ';
	    }
	    cout << '\n';
	}
#endif
	int node, time;
	while(cin >> node >> time && (node || time)){
	    printf("Case %d: %d nodes not reachable from node %d with TTL = %d.\n",__case++, bfs(node_id[node], time), node, time);
	}
    }	
    return 0;
}

UVa : 260 :: II Gioco dell'X

Problem : Please find the problem here.

Explanation : Check, for black, if there's path from first row to last row OR for white, if there's path from first column to last column. Notice that as stated in the problem statement that both black and white can never win, its either black or white, that means you dont need to check for both white and black, as if one wins, other loses ans vice-versa. 

Code : Used DFS.

#include <bits/stdc++.h>
using namespace std;

const int mxN = 200, di[6]={-1, -1, 0, 0, 1, 1}, dj[6]={-1, 0, -1, 1, 0, 1};
int n;
string s[mxN];
bool vis[mxN][mxN];

bool isok(int i, int j){
	return i>=0 && i<n && j>=0 && j< n && !vis[i][j];
}

void dfs(int i, int j){
	vis[i][j] = 1; 
	for(int k = 0; k < 6; k++){
		int ni = i+di[k], nj = j+dj[k];
		if(isok(ni, nj) && s[ni][nj] == s[i][j]){
			dfs(ni, nj);
		}
	}
}

int main()
{	
	ofstream fout("out");
	int t = 1;
	while(cin >> n  && n){
		for(int i = 0; i < n; i++){
			cin >> s[i];
		}
		bool ok = 0;
		memset(vis, false, sizeof(vis));
		for(int i = 0; i < n; i++){
			if(s[i][0] == 'w'){
				dfs(i, 0);
			}
		}
		for(int i = 0; i < n; i++){
			if(vis[i][n-1]){
				ok = 1;
				break;
			}
		}
		cout << t++ << ' ' << (ok?'W':'B') << '\n';
	}	
	return 0;
}

UVa : 784 :: Maze Exploration

Problem : Please find the problem here.

Explanation : Simple DFS. 

Code :

#include <bits/stdc++.h>
using namespace std;

const int mxN = 30, di[4]={1, 0, -1, 0}, dj[4]={0, 1, 0, -1};
int n, m, si, sj;
string s[mxN], in, br;

bool isok(int i, int j){
    return i>=0 && i<n && j>=0 && j<(int)s[i].length() && s[i][j] == ' ';
}

void dfs(int i, int j){
    s[i][j] = '#'; 
    for(int k = 0; k < 4; k++){
	int ni = i+di[k], nj = j+dj[k];
	if(isok(ni, nj)){
	    dfs(ni, nj);
	}
    }
}

int main()
{	
    ofstream fout("out");
    int t;
    cin >> t;
    getline(cin, in); // read the newline character at the start
    for(int k = 0; k < t; k++){
	in.clear();
	getline(cin, in);  // read maze from here
	int i = 0;
	while(1){
	    if(in[0] == '_'){ // break if hit the partition line
		br = in;
		break;
	    }
	    s[i] = in;
	    for(int j = 0; j < (int)s[i].length(); j++){
		if(s[i][j] == '*'){
		    si = i, sj = j; s[i][j] = '#';
		}
	    }
	    i++;
	    getline(cin, in);
	}
	n = i
	dfs(si, sj);
	for(int j = 0; j < n; j++){
	    cout << s[j] << '\n';
	}
	cout << br << '\n';
    }
    return 0;
}

UVa : 10946 :: You want what filled?

Problem : Please find the problem here.

Explanation : We are basically asked to find the connected components in the graph and print them in non increasing order along with the charcter in each connected component printed in increasing order. 

Code : Used DFS.

#include <bits/stdc++.h>
using namespace std;
#define mp make_pair
#define debug

const int mxN = 60, di[4]={1, 0, -1, 0}, dj[4]={0, 1, 0, -1};
int n, m, cnt;
string s[mxN];
vector<pair<char, int>> ans;
bool vis[mxN][mxN];

bool isok(int i, int j){
    return i>=0 && i<n && j>=0 && j<m && s[i][j] != '.' && !vis[i][j];
}

bool cmp(pair<char, int> a, pair<char, int> b){
    if(a.second == b.second){
        return a.first < b.first;
    }
    return a.second >= b.second;
}

void dfs(int i, int j){
    vis[i][j] = 1;
    for(int k = 0; k < 4; k++){
        int ni = i+di[k], nj = j+dj[k];
        if(isok(ni, nj) && s[i][j] == s[ni][nj]){
	    dfs(ni, nj);
	}
    }
    cnt++;
}

int main()
{
    ofstream fout("out");
    int t = 1;
    while(cin >> n >> m && (n||m)){
        for(int i = 0; i < n; i++){
	    cin >> s[i];
        }
	memset(vis, false, sizeof(vis));
	ans.clear();
	for(int i = 0; i < n; i++){
	    for(int j = 0; j < m; j++){
		if(isok(i, j)){
		    cnt = 0;
		    dfs(i, j);
		    ans.push_back(mp(s[i][j], cnt));
	        }
	    }
	}
	sort(ans.begin(), ans.end(), cmp);
	cout << "Problem " << t++ << ":\n";
	for(auto a : ans){
	    cout << a.first << ' ' << a.second << '\n';
	}
    }
    return 0;
}

UVa : 785 :: Grid Colouring

Problem : Please find the problem here.

Explanation : Just need to traverse the graph, could be solved by DFS or BFS. Input parsing is troublesome though. Good for practice.  

Code : Used DFS.

#include <bits/stdc++.h>
using namespace std;

const int mxN = 32, di[4]={1, 0, -1, 0}, dj[4]={0, 1, 0, -1};
int n, m;
string s[mxN], in, br;
char contour;

bool isok(int i, int j){
    return i>=0 && i<n && j>=0 && j<(int)s[i].length() && s[i][j] == ' ';
}

void dfs(int i, int j, char c){
    s[i][j] = c;
    for(int k = 0; k < 4; k++){
        int ni = i+di[k], nj = j+dj[k];
	if(s[ni][nj] != c && isok(ni, nj)){ 
	    dfs(ni, nj, c);
	}
    }
}

int main()
{
    ofstream fout("out");
    while(getline(cin, in)){
	int i = 0;
	do{
	    if(in[0] == '_'){
	        br = in;
	        break;
	    }
	    s[i] = in;

	    // lazy way to find contour,
	    if(i == 0){
		for(int j = 1; j < (int)in.length(); j++){
		    if(in[j]!=' ' && in[j]!='_' && in[j]==in[j-1]){
			contour = in[j];
			break;
		    }
		}
	    }
	    // ends here
	    i++;
	}while(getline(cin, in));
	

        n = i;

        // now look for marker and apply dfs from that point.
	for(int i = 0; i < n; i++){
	    for(int j = 0; j < (int)s[i].length(); j++){
		if(s[i][j] != contour && s[i][j] != ' ' && s[i][j] != '\t'){
		    dfs(i, j, s[i][j]);
		}
	    }
	}

	// finally print the grid.
	for(int i = 0; i < n; i++){
	    cout << s[i] << '\n';
	}

	// dont forget the seperation line -> "_______"
	cout << br << '\n';
    }	
    return 0;
}