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;
}