hdu 1874(最短路Bellman_Ford)

/*

  Name: 最短路(bellmanFord) 

  Copyright: 

  Author: Try_86

  Date: 11/04/12 19:03

  Description: 求一对顶点间的最短路

  注意：建立反向边时两端点的顺序 

*/



#include <cstdio>

#include <climits>

#include <iostream>



using namespace std;



const int N = 205;

const int M = 2005;

const int MAX = 100000000;



int dis[N];  

struct edge {//边结点 

    int u;

    int v;

    int w;

}e[M];



void init(int vs, int s) {//初始化 

    for (int i=0; i<vs; ++i) dis[i] = MAX;

    dis[s] = 0;

    return ;

}



void relax(int u, int v, int w) {//松弛操作 

    if (dis[v] > dis[u] + w) dis[v] = dis[u] + w;

    return ;

}



void bellmanFord(int es, int vs, int s) {//题目给出的数据不会出现负权回路 

    init(vs, s);

    for (int i=0; i<vs-1; ++i) {

        for (int j=0; j<es; ++j) relax(e[j].u, e[j].v, e[j].w);

    }

    return ;

}



int main() {

    int n, m, s, t;

    while (scanf("%d%d", &n, &m) != EOF) {

        for (int i=0; i<m; ++i) {

            scanf ("%d%d%d", &e[i].u, &e[i].v, &e[i].w);

            e[i+m].v = e[i].u;  //注意，建立反向边的端点顺序 

            e[i+m].u = e[i].v;

            e[i+m].w = e[i].w;

        }

        scanf ("%d%d", &s, &t);

        bellmanFord(m<<1, n, s);

        if (dis[t] == MAX) printf ("-1\n");

        else printf ("%d\n", dis[t]);

    }

    return 0;

}