【POJ 1062】昂贵的聘礼（最短路）

#include <cstdio>
#define N 105
#define INF 1e9
int m, n;
int p[N], l[N], x, t, v, g[N][N];
int w, minc, d[N], u[N], ans;
void Dijkstra(int q)
{
    for(int i = 1; i <= n; i++){
        d[i] = g[1][i];
        if(!d[i])d[i]=INF;
        u[i]=0;
    }
    u[1] = 1;
    for(int i = 2; i <= n; i++)
    {
        minc = 1e9;
        w = 0;
        for(int j = 2; j <= n; j++)
            if(!u[j] && minc >= d[j] && l[j] >= q && l[j] <= q + m)
            {
                minc = d[j];
                w = j;
            }
        if(!w) break;
        u[w] = 1;
        for(int j = 2; j <= n; j++)
            if(!u[j] && g[w][j] && d[j] > d[w] + g[w][j] && l[j] >= q && l[j] <= q + m)
                d[j] = d[w] + g[w][j];
    }
    for(int i = 2; i <= n; i++)
        if(d[i] + p[i] < ans &&l[i] >= q && l[i] <= q + m)ans = d[i] + p[i];
 
}
int main()
{
    scanf("%d%d", &m, &n);
    for(int i = 1; i <= n; i++)
    {
        scanf("%d%d%d", &p[i], &l[i], &x);
        for(int j = 1; j <= x; j++)
        {
            scanf("%d%d", &t, &v);
            g[i][t] = v;
        }
    }
    ans = p[1];
    for(int i = l[1] - m; i <= l[1] ; i++)
        Dijkstra(i);
    printf("%d\n",ans);
    return 0;
}