hdu 2680(最短路bellmanFord)

843MS，差点就超时!

/*
  Name: 最短路(bellmanFord) 
  Copyright: 
  Author: Try_86
  Date: 11/04/12 20:39
  Description: 建反向图，求终点到各点的最短路 
*/

#include <cstdio>
#include <iostream>

using namespace std;

const int N = 1005;
const int M = 40005;
const int MAX = 1000000000;

int dis[N];
struct edge {
    int u;
    int v;
    int w;
}e[M];

void init(int vs, int s) {
    for (int i=1; i<=vs; ++i) dis[i] = MAX;
    dis[s] = 0;
    return ;
}

void relax(int u, int v, int w) {
    if (dis[v] > dis[u] + w) dis[v] = dis[u] + w;
    return ;
}

void bellmanFord(int es, int vs, int s) {
    init(vs, s);
    for (int i=1; i<vs; ++i) {
        for (int j=0; j<es; ++j) relax(e[j].u, e[j].v, e[j].w);
    }
    return ;
}

int main() {
    int n, m, end;
    while (scanf("%d%d%d", &n, &m, &end) != EOF) {
        for (int i=0; i<m; ++i)    scanf ("%d%d%d", &e[i].v, &e[i].u, &e[i].w);
        bellmanFord(m, n, end);
        int cs, s, maxs;
        maxs = MAX;
        scanf ("%d", &cs);
        for (int i=0; i<cs; ++i) {
            scanf ("%d", &s);
            if (maxs > dis[s]) maxs = dis[s];
        }
        if (maxs != MAX) printf ("%d\n", maxs);
        else printf ("-1\n");
    }
    return 0;
}