Ultraviolet A 544 heavy cargo (variant of the minimal spanning tree)

This question is the maximum Spanning Tree and the minimum weight edge on the bottleneck.

The Code is as follows:

#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <map>#include <iostream>using namespace std;const int N = 210;const int INF = -1;const int INFB = 100000;int n, r, g[N][N], p[N], f[N][N], d[N];bool mark[N];string S, T;void prim() {    bool vis[N];    memset(vis, 0, sizeof(vis));    for ( int i = 1; i <= n; ++i ) d[i] = INF, p[i] = INF;    d[1] = 0;    int mi, v, ans = 0;    for ( int u = 0; u < n; ++u ) {        mi = INF;        for ( int i = 1; i <= n; ++i ) if ( !vis[i] && d[i] > mi ) v = i, mi = d[i];        ans += mi;        //cout << mi << endl;        vis[v] = true;        for ( int i = 1; i <= n; ++i ) if ( !vis[i] && d[i] < g[v][i] )  p[i] = v, d[i] = g[v][i];     }    //cout << ans << endl;}void dfs( int v ) {    for ( int u = 1; u <= n; ++u )         if ( !mark[u] && p[u] == v ) {           mark[u] = true;           //cout << u << ' ';           for ( int x = 1; x <= n; ++x ) if ( mark[x] && x != u ){           //   cout << f[x][v] << ' ' << g[u][v] << endl;               f[x][u] = f[u][x] = min( f[x][v], g[u][v] );           }           dfs(u);        }} int main(){    int icase = 1;    while ( scanf("%d%d", &n, &r), !(!n&&!r) ) {        getchar();        map <string,int> mp;        int id = 1;        for ( int i = 1; i <= n; ++i ) for ( int j = 1; j <= n; ++j ) g[i][j] = INF;        mp.clear();        while ( r-- ) {            string s1, s2;            int w;            cin >> s1 >> s2 >> w;            if ( !mp[s1] ) mp[s1] = id++;            if ( !mp[s2] ) mp[s2] = id++;            int u = mp[s1], v = mp[s2];            g[u][v] = g[v][u] = w;        }        cin >> S >> T;        prim();        for ( int i = 1; i <= n; ++i ) for ( int j = i; j <= n; ++j ) f[i][j] = f[j][i] = INFB;        memset( mark, 0, sizeof(mark));        mark[1] = true;        dfs( 1 );        //for ( int i = 1; i <= n; printf("\n"),++i ) for ( int j = 1; j <= n; ++j ) cout << f[i][j] << ' ';        int u = mp[S], v = mp[T];        printf("Scenario #%d\n%d tons\n\n", icase++, f[u][v]);    }}