学步园

这是一道很典型的单源点最短路径问题，用Dij来解，记录路径，通过dfs来输出路径

代码如下：

#include <cstdio>
#include <cstring>

const int N = 20;
const int INF = 100000000;
int n, s, t, ans, icase;
int g[N][N], d[N], p[N];

void dij() {
    int mi, v; 
    for ( int i = 1; i <= n; ++i ) d[i] = INF;
    d[s] = 0, ans = 0;
    bool vis[N];
    memset( vis, 0, sizeof(vis) );
    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;
        for ( int i = 1; i <= n; i++ ) if ( !vis[i] && g[v][i] >= 0 && d[i] > d[v] + g[v][i] ) d[i] = d[v] + g[v][i], p[i] = v;
    }
}
void output( int x ) {
    if ( x == -1 ) return;
    output( p[x] );
    if ( p[x] != -1 ) printf(" ");
    printf("%d", x);
    return;
}
int main()
{
    icase = 1;
    while ( scanf("%d", &n) != EOF && n ) {
        for ( int i = 1; i <= n; p[i] = -1, ++i ) for ( int j = i; j <= n; ++j ) g[i][j] = g[j][i] = -1;
        for ( int i = 1; i <= n; ++i ) {
            int num;
            scanf("%d", &num);
            for ( int j = 1; j <= num; ++j ) {
                int v, del;
                scanf("%d %d", &v, &del);
                g[i][v] = del;
            }
        }
            scanf("%d%d", &s, &t);
            dij();
            printf("Case %d: Path = ", icase++);
            output( t );
            printf("; %d second delay\n", d[t]);
    }
}