POJ-3259-Wormholes

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 35294 Accepted: 12881

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.

题目大意是说John从出发点通过一些路径和昆虫洞，最后能返回出发点，并且时间早于出发时间，要求你判断所给的路径和昆虫洞是否能够满足这一要求

其实想明白了以后就是判断是否存在负权回路，我用的Spfa做的；

这里注意一下：path是双向的，wormhole是单向的。

Source

#include <stdio.h>#include <stdlib.h>#include <queue>#include <string.h>using namespace std;const int nmax=2505;const int inf=0xfffffff;queue<int>q;struct edge{int to;int val;int next;}edge[nmax*nmax];int dis[nmax];int cout[nmax];int hand[nmax];int top;int n,m,w;void addedge(int f,int t,int v) //静态邻接表{edge[top].to=t;edge[top].val=v;edge[top].next=hand[f];hand[f]=top++;}int spfa(int start){int i;int inqueue[nmax];memset(inqueue,0,sizeof(inqueue));for(i=0;i<=n;i++)dis[i]=inf;dis[start]=0;memset(cout,0,sizeof(cout));int now=1;q.push(start);cout[start]++;inqueue[start]=1;while(!q.empty()){now=q.front();q.pop();inqueue[now]=0;for(i=hand[now];i!=-1;i=edge[i].next){int t=edge[i].to;int val=edge[i].val;if(dis[now]+val<dis[t]){dis[t]=dis[now]+val;if(!inqueue[t]){cout[t]++;q.push(t);inqueue[t]=1;if(cout[t]>n)return 1;}}}}return 0;}int main(){int t,i,j;scanf("%d",&t);while(t--){scanf("%d%d%d",&n,&m,&w);int a,b,c;memset(hand,-1,sizeof(hand));top=0;while(m--){scanf("%d%d%d",&a,&b,&c);//双向的；addedge(a,b,c);addedge(b,a,c);}while(w--){scanf("%d%d%d",&a,&b,&c);//单向的，把它负的就可以了；addedge(a,b,-c);}if(spfa(1))printf("YES\n");//存在负环说明能提前返回。elseprintf("NO\n");}return 0;}

USACO 2006 December Gold