Wormholes
Time Limit: 2000MSMemory Limit: 65536K
Total Submissions: 36641Accepted: 13405
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 WLines 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
题目链接：?id=3259
题目大意：时空旅行，前m条路是双向的，旅行时间为正值，w条路是虫洞，单向的，旅行时间是负值，也就是能回到过去。求从一点出发，判断能否在”过去“回到出发点，即会到出发点的时间是负的。
解题思路：裸的负权最短路问题，SPAF解决。枚举出发点即可。
代码如下：
#include <cstdio>#include <cstring>#include <vector>#include <queue>#include <algorithm>#define inf 1e9using namespace std;int const maxn=505;int n,p;int dist[maxn],vis[maxn],num[maxn]; struct node {int v,w;node(int vv,int ww){v=vv;w=ww;}};vector <node> vt[maxn];void SPAF(int v0){memset(vis,0,sizeof(vis));memset(num,0,sizeof(num));for(int i=0;i<maxn;i++)dist[i]=-inf;queue <int>q;q.push(v0);dist[v0]=0;while(!q.empty()){int u=q.front();q.pop();if(num[u]>n)continue;num[u]++;if(num[u]>=n)dist[u]=inf;vis[u]=0;int len=vt[u].size();for(int i=0;i<len;i++){int v=vt[u][i].v;int w=vt[u][i].w;if(dist[v]<dist[u]+w){dist[v]=dist[u]+w;if(v==v0&&dist[v]>0){p=1;return ;}if(!vis[v]){vis[v]=1;q.push(v);}}}}}int main(void){int m1,m2,u,v,w,t;scanf("%d",&t);while(t–){scanf("%d%d%d",&n,&m1,&m2);int ans=0;p=0;for(int i=0;i<maxn;i++)vt[i].clear();for(int i=0;i<m1;i++){scanf("%d%d%d",&u,&v,&w); //双向路径vt[u].push_back(node(v,-w));vt[v].push_back(node(u,-w));}for(int i=0;i<m2;i++){scanf("%d%d%d",&u,&v,&w);vt[u].push_back(node(v,w)); //单向路径}for(int i=1;i<=n;i++){SPAF(i);if(p)break;}if(p)printf("YES\n");elseprintf("NO\n");}}
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