F - Wormholes （spfa处理带有负权值的问题）

滴答滴答---题目链接

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 backT 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.

题目大意：

农民约翰在农场散步的时候发现农场有大量的虫洞，这些虫洞是非常特别的因为它们都是单向通道，为了方便现在把约翰的农田划分成N快区域，M条道路，W的虫洞。

约翰是时空旅行的粉丝，他希望这样做，在一个区域开始，经过一些道路和虫洞然后回到他原来所在的位置，这样也许他就能见到他自己了。

穿越虫洞会回到以前。。。。。（穿越者约翰）

题意：给出n个点和m条双向边且权值为正，w条单向边，权值为负。要求给定一个图，判断图中是否有负环

迪杰斯特拉算法不能处理带有负权值的问题，佛洛依德倒是可以，不过复杂度很明显太高，所以spfa是不错的选择，只需要判断出发点时间是否变小.

#include<algorithm>
#include<queue>
#include<stdio.h>
#include<string.h>
#include<vector>
#include<math.h>
using namespace std;
const int N=505;
const int inf=0x3f3f3f;
struct st
{
    int y,time;
    st(int y,int time):y(y),time(time) {}
};
vector<st>g[N];
int v[N];
int spfa(int s)
{
    queue<int>Q;
    Q.push(s);
    while(Q.size())
    {
        s=Q.front();
        Q.pop();
        int len=g[s].size();
        for(int i=0; i<len; i++)
        {
            st q=g[s][i];
            if(v[s]+q.time<v[q.y])
            {
                v[q.y]=v[s]+q.time;
                Q.push(q.y);
            }
        }
        if(v[1]<0)
            return 1;
    }
    return 0;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m,w;
        scanf("%d%d%d",&n,&m,&w);
        for(int i=1; i<=n; i++)
        {
            v[i]=inf;
            g[i].clear();
        }
        v[1]=0;
        for(int i=0; i<m; i++)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            g[a].push_back(st(b,c));
            g[b].push_back(st(a,c));
        }
        for(int i=0; i<w; i++)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            g[a].push_back(st(b,-c));
        }
        int ans=spfa(1);

        if(ans==1)printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}

F - Wormholes （spfa处理带有负权值的问题）

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