POJ - 3259
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 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
2 3 3 1 1 2 2 1 3 4 2 3 1 3 1 3 3 2 1 1 2 3 2 3 4 3 1 8Sample Output
NO YESHint
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：有n个农场，农场间有m条双向路，且有w个单向虫洞连接农场，第一行输入一个数字T，代表有T组测试数据，紧跟一行输入三个整数n,m,w，其下m行输入a,b两个农场之间有一条需要花费c时间的路，其后w行代表有一个从a到b的单向虫洞可以返回到c时间时间之前，问是否能够通过虫洞返回到开始出发之前。
解决方法：最短路问题，可以把正常的路存为正值，而通过虫洞的改为负值，判断是否会存在负边权即可，可以使用flod算法。
AC代码：
#include <cstdio> #include <iostream> #include <algorithm> #include <cmath> #include <cstdlib> #include <cstring> #include <map> #include <stack> #include <queue> #include <vector> #include <bitset> #include <set> #include <utility> using namespace std; typedef long long ll; #define inf 0x3f3f3f3f #define rep(i,l,r) for(int i=l;i<=r;i++) #define lep(i,l,r) for(int i=l;i>=r;i--) #define ms(arr) memset(arr,0,sizeof(arr)) //priority_queue<int,vector<int> ,greater<int> >q; const int maxn = (int)1e3 + 5; const ll mod = 1e9+7; int n,m,w; int mapp[510][510]; bool vis[maxn]; int dist[maxn]; int cnt[maxn]; bool flod() { rep(k,1,n) { rep(i,1,n) { rep(j,1,n) { if(mapp[i][j]>mapp[i][k]+mapp[k][j]) mapp[i][j]=mapp[i][k]+mapp[k][j]; } if(mapp[i][i]<0) return true; } } return false; } int main() { //freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); ios::sync_with_stdio(0),cin.tie(0); int T; cin>>T; while(T--) { cin>>n>>m>>w; rep(i,1,n) { rep(j,1,n) { mapp[i][j]=inf; } dist[i]=inf; vis[i]=false; mapp[i][i]=0; } rep(i,1,m) { int a,b,c; cin>>a>>b>>c; if(c<mapp[a][b]) mapp[a][b]=mapp[b][a]=c; } rep(i,1,w) { int a,b,c; cin>>a>>b>>c; mapp[a][b]=-c; } bool ju=flod(); if(!ju) cout<<"NO"<<endl; else cout<<"YES"<<endl; } return 0; }
