UVa 534 Frogger (最小瓶颈路）

很裸的一道最小瓶颈路的模板题，先求最小生成树，然后在求点对点的最小瓶颈路

代码如下：

#include 
#include 
#include 
#include 
using namespace std;

const int N = 210;
const double INF = 1000000;
int n, p[N];
double x[N], y[N], map[N][N], f[N][N], s1;
bool mark[N];
void prim() {
    bool vis[N];
    memset(vis, 0, sizeof(vis));
    double d[N], mi;
    for ( int i = 1; i <= n; ++i ) d[i] = INF, p[i] = -1;
    d[1] = 0.0, s1 = 0;
    int v;
    for ( int u = 0; u < n; ++u ) {
        mi = INF;
        for ( int i = 1; i <= n; ++i ) if ( !vis[i] && d[i] < mi ) mi = d[i], v = i;
        vis[v] = true;
        s1 += mi;
        for ( int i = 1; i <= n; ++i ) if ( !vis[i] && d[i] > map[v][i] ) d[i] = map[v][i], p[i] = v;
    }
    //printf("%lf\n", s1);
}
void dfs( int v ) {
    for ( int u = 1; u <= n; ++u ) {
        if ( !mark[u] && p[u] == v ) {
           mark[u] = true;
           for ( int x = 1; x <= n; ++x ) 
               if ( mark[x] && x != u )  f[x][u] = f[u][x] = max( f[x][v], map[u][v] );  
           dfs(u);
        }
    }
}

int main()
{
    int icase = 1; 
    while ( scanf("%d", &n), n) {
        for ( int i = 1; i <= n; ++i ) 
            scanf("%lf%lf", &x[i], &y[i]);
        for ( int i = 1; i <= n; ++i ) for ( int j = i+1; j <= n; ++j ) 
            map[i][j] = map[j][i] = sqrt( (x[i]-x[j])*(x[i]-x[j]) + (y[i]-y[j])*(y[i]-y[j]));
        prim();
        memset(mark, 0, sizeof(mark));
        for ( int i = 1; i <= n; ++i ) for ( int j = 1; j <= n; ++j ) f[i][j] = 0.0;
        mark[1] = true;
        dfs( 1 );
        //for ( int i = 1; i <= n; printf("\n"), ++i ) for ( int j = 1; j <= n; ++j ) printf("%lf ", f[i][j]);
        printf("Scenario #%d\n", icase++);
        printf("Frog Distance = %.3lf\n\n", f[1][2]);
    }
}