POJ 3259 Wormholes

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题意：判断是否存在负环

#include<iostream>#include<cstdio>#include<cstring>using namespace std;#define Max 0xfffffffstruct node{    int from;    int to;    int t;}s[10010];int dis[10010];bool bellmanford(int n,int p){    int i,j;    bool flag;    for(i = 0; i <= n; i++)    {        flag = false;        for(j = 0; j < p; j++)        {            if(dis[s[j].to] > dis[s[j].from] + s[j].t)            {                dis[s[j].to] = dis[s[j].from] + s[j].t;                flag = true;            }        }        if(!flag)            break;    }    if(!flag)        return false;    else        return true;}int main(){    int f;    int i,a,b,c;    int n,m,w;    scanf("%d",&f);    while(f--)    {        int n,m,w;        scanf("%d%d%d",&n,&m,&w);        int p = 0;        memset(dis,Max,sizeof(dis));        //边是双向的        for(i = 0; i < m; i++)        {            scanf("%d%d%d",&a,&b,&c);            s[p].from = a;            s[p].to = b;            s[p+1].from = b;            s[p+1].to = a;            s[p++].t = c;            s[p++].t = c;        }        //虫洞传送的路径，注意权值为负        for(i = 0; i < w; i++)        {            scanf("%d%d%d",&a,&b,&c);            s[p].from = a;            s[p].to = b;            s[p++].t = -c;        }        if(bellmanford(n,p))            printf("YES\n");        else            printf("NO\n");    }    return 0;}

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