POJ 3259 bell_man求负环 || SPFA（bell

POJ 3259 bell_man求负环 || SPFA（bell_man模板+SPFA）

题目：https://vjudge.net/problem/POJ-3259

While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Each of FJ's farms comprises N (1 ≤ N ≤ 500) fields conveniently numbered 1..N, M (1 ≤ M ≤ 2500) paths, and W (1 ≤ W ≤ 200) wormholes.

As FJ is an avid time-traveling fan, he wants to do the following: start at some field, travel through some paths and wormholes, and return to the starting field a time before his initial departure. Perhaps he will be able to meet himself :) .

To help FJ find out whether this is possible or not, he will supply you with complete maps to F (1 ≤ F ≤ 5) of his farms. No paths will take longer than 10,000 seconds to travel and no wormhole can bring FJ back in time by more than 10,000 seconds.

Input

Line 1: A single integer, F. F farm descriptions follow.
Line 1 of each farm: Three space-separated integers respectively: N, M, and W
Lines 2.. M+1 of each farm: Three space-separated numbers ( S, E, T) that describe, respectively: a bidirectional path between S and E that requires T seconds to traverse. Two fields might be connected by more than one path.
Lines M+2.. M+ W+1 of each farm: Three space-separated numbers ( S, E, T) that describe, respectively: A one way path from S to E that also moves the traveler back T seconds.

Output

Lines 1.. F: For each farm, output "YES" if FJ can achieve his goal, otherwise output "NO" (do not include the quotes).

Sample Input

Sample Output

NO
YES

Hint

For farm 1, FJ cannot travel back in time.
For farm 2, FJ could travel back in time by the cycle 1->2->3->1, arriving back at his starting location 1 second before he leaves. He could start from anywhere on the cycle to accomplish this.

题意：

给你三个数n,m,w，分别表示有n个农场，m条路，w个黑洞，接下来m行分别有三个数s，e，val表示s到e有一条权值为val 的边，接下来的w行每行有三个数s，e，val表示s到e有一条权值为0-val的边，因为黑洞可以回到过去，所以其权值为负。现在问你给你这些点，求每组数据是否存在一条回路能使时间倒退。

思路：求负环就行 bell——man模板题有SPFA的优化版代码来自https://blog.csdn.net/iceiceicpc/article/details/51986857

因为自己的代码死都过不了的那种，，查错到失望

代码：

  //    bell_man模板


#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<set>
#include<algorithm>
#include<map>
#define maxn 200005
using namespace std;
const int maxv=500+5;
const int maxe=6000+5;
const int INF=0x3f3f3f3f;
struct proc
{
    int s,e,t;    //起点  终点  边权
}edge[maxe];
int n,m,w,dis[maxv];
int cnt;
void addEdge(int s,int e,int t)   //建立图
{
    edge[cnt].s=s;
    edge[cnt].e=e;
    edge[cnt++].t=t;
}
int bell_man()
{
    bool flag;
    //松弛
    for(int i=1;i<=n;i++)
    {
        flag=false;
        for(int j=1;j<=cnt;j++)
        {
            if(dis[edge[j].e]>dis[edge[j].s]+edge[j].t)
            {
                dis[edge[j].e]=dis[edge[j].s]+edge[j].t;
                flag=true;
            }
        }
        if(!flag) break;
    }
    //寻找负环，判断条件是负环可以无限松弛。
    for(int i=1;i<=cnt;i++)
    {
        if(dis[edge[i].e]>dis[edge[i].s]+edge[i].t)
        {
            return true;
        }
    }
    return false;
}

int main()
{
    int s,e,v;
    int cas;
    scanf("%d",&cas);
    while(cas--)
    {
        memset(dis,INF,sizeof(dis));
        cnt=1;
        scanf("%d %d %d",&n,&m,&w);
        for(int i=1;i<=m;i++)
        {
            scanf("%d %d %d",&s,&e,&v);
            addEdge(s,e,v);
            addEdge(e,s,v);
        }
        for(int i=1;i<=w;i++)
        {
            scanf("%d %d %d",&s,&e,&v);
            addEdge(s,e,0-v);
        }
        int ans=bell_man();
        if(ans) printf("YES
");
        else printf("NO
");
    }
    return 0;
}

      //   SPFA优化模板
//
//#include <iostream>
//#include <cstdio>
//#include <cstring>
//#include <algorithm>
//#include <queue>
//
//using namespace std;
//const int maxn=500+5;
//const int maxe=6000+5;
//const int INF=0x7fffffff;
//
//struct proc
//{
//    int v,w;
//    int next;
//};
//
//proc edge[maxe];
//int dis[maxn],vis[maxn],f,n,m,w,head[maxe];
//int cnt[maxn];
//
//int k;
//void addEdge(int u,int v,int w)
//{
//    edge[k].v=v;
//    edge[k].w=w;
//    edge[k].next=head[u];
//    head[u]=k++;
//}
//
//bool spfa()
//{
//    for(int i=0;i<=n;i++)
//    {
//        dis[i]=INF;
//    }
//    memset(cnt,0,sizeof(cnt));
//    memset(vis,0,sizeof(vis));
//    queue<int> q;
//    vis[1]=1;
//    dis[1]=0;
//    cnt[1]=1;
//    q.push(1);
//    while(!q.empty())
//    {
//        int cur=q.front();
//        q.pop();
//        vis[cur]=false;
//        for(int i=head[cur];i+1;i=edge[i].next)
//        {
//            int id=edge[i].v;
//            if(dis[cur]+edge[i].w<dis[id])
//            {
//                dis[id]=dis[cur]+edge[i].w;
//                if(!vis[id])
//                {
//                    cnt[id]++;
//                    if(cnt[cur]>=n)
//                        return false;
//                    vis[id]=true;
//                    q.push(id);
//                }
//            }
//        }
//    }
//    return true;
//}
//
//int main()
//{
//    while(~scanf("%d",&f))
//    {
//        while(f--)
//        {
//            scanf("%d %d %d",&n,&m,&w);
//            k=0;
//            memset(head,-1,sizeof(head));
//            int s,e,val;
//            for(int i=0;i<m;i++)
//            {
//                scanf("%d %d %d",&s,&e,&val);
//                addEdge(s,e,val);
//                addEdge(e,s,val);
//            }
//            for(int i=0;i<w;i++)
//            {
//                scanf("%d %d %d",&s,&e,&val);
//                addEdge(s,e,0-val);
//            }
//            if(spfa()) printf("NO
");
//            else printf("YES
");
//        }
//    }
//}

SPFA的详解： https://blog.csdn.net/acm_1361677193/article/details/48211319