/* 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; }