poj 3259 Wormholes SPFA&&推断负环

Posted 2020-09-20 jzssuanfa

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 36852		Accepted: 13502

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

分析：

英语不好。就不在翻译了。

意思就是求最短路。假设最短路中有负环，输出yes，没有输出no。

代码：

#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
#include<cstdlib>
#define INF 0x3f3f3f3f
#define maxn 10010
using namespace std;

int N,M,W;
int pnum;
int dis[maxn];
int vis[maxn];
int head[maxn];
int used[maxn];
bool cnt;
struct node{
    int from;
    int to;
    int val;
    int next;
};
node pp[2*maxn];

void addpp(int u,int v,int w)
{
    node E ={u,v,w,head[u]};
    pp[pnum]=E;
    head[u]=pnum++;
}

void init()
{
    int a,b,c;
    while(M--)
    {
        scanf("%d%d%d",&a,&b,&c);
        addpp(a,b,c);
        addpp(b,a,c);
    }
}

void gmap()
{
    int a,b,c;
    while(W--)
    {
        scanf("%d%d%d",&a,&b,&c);
        addpp(a,b,-c);
    }
}

void SPFA(int s)
{
    queue<int>q;
    memset(dis,INF,sizeof(dis));
    memset(vis,0,sizeof(vis));
    memset(used,0,sizeof(used));
    dis[s]=0;
    vis[s]=1;
    used[s]++;
    q.push(s);
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        vis[u]=0;
        for(int i=head[u];i!=-1;i=pp[i].next)
        {
            int v=pp[i].to;
            if(dis[v]>dis[u]+pp[i].val)
            {
                dis[v]=dis[u]+pp[i].val;
                if(!vis[v])
                {

                   vis[v]=1;
                   q.push(v);
                   used[v]++;
                   if(used[v]>N)
                   {
                     cnt=true;
                     return;
                    }
                }
            }
        }
    }

}

int main(){
    int T;
    scanf("%d",&T);
    while(T--)
    {
        pnum=0;
        cnt=false;
        memset(head,-1,sizeof(head));
        scanf("%d%d%d",&N,&M,&W);
        init();
        gmap();
        SPFA(1);
        if(cnt)
        printf("YES\n");
        else
        printf("NO\n");
    }
    return 0;
}