poj 3259 Wormholes （spfa求最短路）

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 27964 Accepted: 10058

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 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

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.

 

解题思路：首先，应知道图不一定是连通的，只有图中存在负环就输出“YES”。因此，用spfa算法时，如果一次从源点的松弛不能到达所有边，必须把剩下的再用一次spfa算法。

#include<stdio.h>#include<string.h>#include<queue>using namespace std;#define N 505const int inf=10005;int map[N][N];         //记录点之间关系int dis[N];       //记录当前点到源点的距离int mark[N];     //标记该点是否在队列内int num[N],n;       //记录一次搜索中某点入队的次数int Min(int a,int b){return a<b?a:b;}int spfa(){int i,s=1;queue<int>q;memset(num,0,sizeof(num));while(1){if(q.empty()){for(i=1;i<=n;i++)        if(!num[i])      //若源点没有和所有的点连通，break;if(i<=n){q.push(i);    //则把后面的点入队，在进行一次spfa算法memset(mark,0,sizeof(mark));memset(num,0,sizeof(num));for(;i>0;i--)       num[i]=1;      //先把前面的不可到点标记}elsereturn 0; }s=q.front();q.pop();mark[s]=0;for(i=1;i<=n;i++){if(dis[i]>dis[s]+map[s][i]){dis[i]=dis[s]+map[s][i];if(!mark[i]){mark[i]=1;num[i]++;q.push(i);if(num[i]>n)      //存在负环return 1;}}}}return 0;}int main(){int T,a,b,c,m,w,i,j;scanf("%d",&T);while(T--){scanf("%d%d%d",&n,&m,&w);for(i=1;i<=n;i++){for(j=1;j<=n;j++)map[i][j]=inf;           dis[i]=inf;            }while(m--){scanf("%d%d%d",&a,&b,&c);map[a][b]=Min(map[a][b],c);    //两点存在多条路map[b][a]=Min(map[b][a],c);     //bidirectional双向边}while(w--){scanf("%d%d%d",&a,&b,&c);map[a][b]=Min(map[a][b],0-c);}if(spfa())printf("YES\n");elseprintf("NO\n");}return 0;}