poj——3259——Wormholes

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 43434 Accepted: 15955

题意：有n个点，m条无向边，w条虫洞（可以回退到进洞前），问是否可以利虫洞见到以前的自己。

思路：找负全环；

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 toF (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 toE 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.

Source

USACO 2006 December Gold

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <queue>#include <stack>#include <algorithm>using namespace std;struct node{    int u,v,w;}que[10010];int k;int dis[1010];  int t,n,m1,m2;const int INF=0x3f3f3f3f;void bell_man(){    for(int i=1;i<=n;i++)        dis[i]=INF;    dis[1]=0;    int u,v,w;    int mark;    for(int i=0;i<n;i++)  //多跑一次，存在松弛？ 有负权环？：无；    {        mark=1;        for(int j=0;j<k;j++)        {            u=que[j].u,v=que[j].v;            w=que[j].w;            if(dis[u]+w<dis[v])            {                mark=0;                dis[v]=dis[u]+w;            }        }        if(mark)            break;    }    if(mark)    printf("NO\n");    else        printf("YES\n");}int main(){    scanf("%d",&t);    int u,v,w;    while(t--)    {        scanf("%d%d%d",&n,&m1,&m2);         k=0;         for(int i=0;i<m1;i++)         {             scanf("%d%d%d",&u,&v,&w);             que[k].u=u;que[k].v=v,que[k++].w=w;             que[k].u=v;que[k].v=u,que[k++].w=w;         }         for(int i=0;i<m2;i++)         {             scanf("%d%d%d",&que[k].u,&que[k].v,&que[k].w);             que[k].w=-que[k].w;;             k++;         }         bell_man();    }    return 0;}