poj 3259 Wormholes(最短路)

Description

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

2
3 3 1
1 2 2
1 3 4
2 3 1
3 1 3
3 2 1
1 2 3
2 3 4
3 1 8

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

 

题目大意：虫洞问题，现在有n个点，m条边，代表现在可以走的通路，比如从a到b和从b到a需要花费c时间，现在在地上出现了w个虫洞，虫洞的意义就是你从a到花费的时间是-c(时间倒流,并且虫洞是单向的)，现在问你从某个点开始走，能回到从前

本质就是求该图是否存在负环。也就是如何求出一个图是否含有负环。

注意：M是正权，双向，W是负权，单向

 

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn=5300;
const int INF=10001;
struct node
{
    int s;
    int e;
    int t;
} path[maxn];
int n,m,w;
int val[maxn]; //源点到各点的权值
int index; //边的总数
void bellman()
{
    memset(val,INF,sizeof(val));
    bool flag;
    for(int i=0; i<n-1; i++)
    {
        flag=false;
        for(int j=0; j<index; j++)
        {
            if(val[path[j].e]>val[path[j].s]+path[j].t)
            {
                val[path[j].e]=val[path[j].s]+path[j].t;
                flag=true;
            }
        }
        if(!flag)
            break;
    }
    for(int i=0; i<index; i++)
    {
        if(val[path[i].e]>val[path[i].s]+path[i].t)
        {
            cout<<"YES"<<endl;
            return ;
        }
    }
    cout<<"NO"<<endl;
    return ;
}
int main()
{
    int t;
    int a,b,c;//临时变量
    cin>>t;
    while(t--)
    {
        index=0;
        cin>>n>>m>>w;
        for(int i=1; i<=m; i++) //正权边，双向
        {
            cin>>a>>b>>c;
            path[index].s=a;
            path[index].e=b;
            path[index++].t=c;
            path[index].s=b;
            path[index].e=a;
            path[index++].t=c;
        }
        for(int i=1; i<=w; i++) //负权边，单向
        {
            cin>>a>>b>>c;
            path[index].s=a;
            path[index].e=b;
            path[index++].t=-c;
        }
        bellman();
    }
    return 0;
}

 这样也可以

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn=5300;
const int INF=10001;
struct node
{
    int s;
    int e;
    int t;
} path[maxn];
int n,m,w;
int val[maxn]; //源点到各点的权值
int index; //边的总数
void bellman()
{
    for(int i=1;i<=n;i++)
        val[i]=INF;
    val[1]=0;
    bool flag;
    for(int i=1; i<n; i++)
    {
        flag=false;
        for(int j=1; j<index; j++)
        {
            if(val[path[j].e]>val[path[j].s]+path[j].t)
            {
                val[path[j].e]=val[path[j].s]+path[j].t;
                flag=true;
            }
        }
        if(!flag)
            break;
    }
    if(flag)
        cout<<"YES"<<endl;
    else
        cout<<"NO"<<endl;
    return ;
}
int main()
{
    int t;
    int a,b,c;//临时变量
    cin>>t;
    while(t--)
    {
        index=1;
        cin>>n>>m>>w;
        for(int i=1; i<=m; i++) //正权边，双向
        {
            cin>>a>>b>>c;
            path[index].s=a;
            path[index].e=b;
            path[index++].t=c;
            path[index].s=b;
            path[index].e=a;
            path[index++].t=c;
        }
        for(int i=1; i<=w; i++) //负权边，单向
        {
            cin>>a>>b>>c;
            path[index].s=a;
            path[index].e=b;
            path[index++].t=-c;
        }
        bellman();
    }
    return 0;
}