poj 3259 Wormholes（bellman-ford判断负环）

题目链接：http://poj.org/problem?id=3259

题目就是问你能否回到原点而且时间还倒回去了。题目中有些路中有单向的虫洞能让时间回到过去

所以只要将虫洞这条边的权值赋为负然后再判断有没有负环就行了。

#include <iostream>
#include <cstring>
using namespace std;
const int inf = 10001;
int f , n , m , w ,dis[1001] , counts;
struct TnT {
    int u , v , weight;
}T[5200];
void relax(int u , int v , int weight) {
    if(dis[v] > dis[u] + weight)
        dis[v] = dis[u] + weight;
}
bool bellman_ford() {
    for(int i = 1 ; i < n ; i++) {
        for(int j = 1 ; j <= counts ; j++) {
            relax(T[j].u , T[j].v , T[j].weight);
        }
    }
    bool flag = true;
    for(int i = 1 ; i <= counts ; i++) {
        if(dis[T[i].v] > dis[T[i].u] + T[i].weight) {
            flag = false;
            break;
        }
    }
    return flag;
}
int main() {
    cin >> f;
    int s , e , t;
    while(f--) {
        cin >> n >> m >> w;
        counts = 0;
        memset(dis , inf , sizeof(dis));
        for(int i = 1 ; i <= m ; i++) {
            cin >> s >> e >> t;
            T[++counts].u = s , T[counts].v = e , T[counts].weight = t;
            T[++counts].u = e , T[counts].v = s , T[counts].weight = t;
        }
        for(int i = 1 ; i <= w ; i++) {
            cin >> s >> e >> t;
            T[++counts].u = s , T[counts].v = e , T[counts].weight = -t;
        }
        if(bellman_ford()) {
            cout << "NO" << endl;
        }
        else {
            cout << "YES" << endl;
        }
    }
    return 0;
}