POJ 3259 Wormholes【Bellman_ford判断负环】

题意：给出n个点，m条正权的边，w条负权的边，问是否存在负环

因为Bellman_ford最多松弛n-1次， 因为从起点1终点n最多经过n-2个点，即最多松弛n-1次，如果第n次松弛还能成功的话，则说明存在有负环

#include<iostream>  
#include<cstdio>  
#include<cstring> 
#include <cmath> 
#include<stack>
#include<vector>
#include<map> 
#include<set>
#include<queue> 
#include<algorithm>  
using namespace std;

typedef long long LL;
const int INF = (1<<30)-1;
const int maxn=10005;
int d[maxn];
int flag;
int nodenum,edgenum,ecnt;

struct Edge{
    int u,v,cost;
} e[maxn];

int Bellman_ford(){
    for(int i=1;i<=nodenum;i++) d[i]=INF;
    d[1]=0;
    
    for(int i=1;i<nodenum;i++){
        for(int j=1;j<=ecnt;j++){
            if(d[e[j].v]>d[e[j].u]+e[j].cost)
            d[e[j].v]=d[e[j].u]+e[j].cost;
        }
    }
    
    flag=1;
    for(int i=1;i<=ecnt;i++){
        if(d[e[i].v]>d[e[i].u]+e[i].cost){
            flag=0;
            break;
        }
    }
    return flag;    
}

void addedges(int a,int b,int c){
    e[++ecnt].u=a;
    e[ecnt].v=b;
    e[ecnt].cost=c;    
}

int main(){
    int m,w,ncase;
    int a,b,c;
    scanf("%d",&ncase);
    while(ncase--){
        ecnt=0;
        scanf("%d %d %d",&nodenum,&m,&w);
        while(m--){
            scanf("%d %d %d",&a,&b,&c);
            addedges(a,b,c);
            addedges(b,a,c);            
        }
        while(w--){
            scanf("%d %d %d",&a,&b,&c);
            c=-c;
            addedges(a,b,c);
        }
        if(Bellman_ford()) printf("NO
");
        else printf("YES
");    
    }
    return 0;
}