poj 3259 Wormholes 【spfa判负环---求最短路】

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 45543 Accepted: 16808

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

23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8

Sample Output

NOYES

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：

有n个点，m个双向路径，w个虫洞。

路径花费正数时间，虫洞花费负数时间。

求能否经过若干条路径回来过去---

思路：

spfa判负环...

spfa思想----

不断地向队列中添加队列中没有的距离变小的点（再次被优化的）。

然后拿队列中的点去优化其他路径-----

如果有某一点多次进去队列说明路径形成了环，次数超过N次，一定是负环。

代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int ma[1000][1000],dis[1000],ci[1000];int que[400000],kp,top,n,m,w;bool falg[1000];int main(){    int t;scanf("%d",&t);    while (t--)    {        scanf("%d%d%d",&n,&m,&w);        for (int i=1;i<=n;i++)        {            for (int j=1;j<=n;j++)                ma[i][j]=99999999;            dis[i]=99999999;            falg[i]=false;            ci[i]=0;        }        int a,b,c;        for (int i=0;i<m;i++)        {            scanf("%d%d%d",&a,&b,&c);            ma[a][b]=ma[b][a]=min(ma[a][b],c);        }        for (int i=0;i<w;i++)        {            scanf("%d%d%d",&a,&b,&c);            ma[a][b]=min(ma[a][b],-c);        }        bool fafe=false;        top=0;kp=0;        dis[1]=0;        que[kp++]=1;        while (top<kp)        {            int p=que[top++];            falg[p]=false;            if (ci[p]>n)            {                fafe=true;                break;            }            for (int i=1;i<=n;i++)            {                if (dis[p]+ma[p][i]<dis[i])                {                    dis[i]=dis[p]+ma[p][i];                    if (!falg[i])                    {                        falg[i]=true;                        ci[i]++;                        que[kp++]=i;                    }                }            }        }        if (fafe)            printf("YES\n");        else            printf("NO\n");    }    return 0;}