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;

}