POJ3259 Wormholes(bellman)

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 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 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.

n个点，m条路，w个虫洞(虫洞的权值是-w)，问可不可以回到从前。

bellman判断负环即可。

AC代码：

#include "iostream"#include "cstdio"#include "cstring"#include "algorithm"using namespace std;const int maxn = 5010;const int inf = 0x3f3f3f3f;int n, m, w_m, dis[maxn];struct node{/* data */int s, e, t;}e[maxn];bool Bellman(int cnt){bool flag;for(int i = 0; i < n - 1; ++i) {flag = false;for(int j = 0; j < cnt; ++j)if(dis[e[j].e] > dis[e[j].s] + e[j].t) {dis[e[j].e] = dis[e[j].s] + e[j].t;flag = true;}if(!flag) break;}for(int i = 0; i < cnt; ++i)if(dis[e[i].e] > dis[e[i].s] + e[i].t) return true;return false;}int main(int argc, char const *argv[]){int t;scanf("%d", &t);while(t--) {memset(dis, inf, sizeof(dis));scanf("%d%d%d", &n, &m, &w_m);int cnt = 0;for(int i = 1; i <= m; ++i) {int u, v, w;scanf("%d%d%d", &u, &v, &w);e[cnt].s = e[cnt + 1].e = u;e[cnt].e = e[cnt + 1].s = v;e[cnt++].t = w;e[cnt++].t = w;}for(int i = 1; i <= w_m; ++i) {int u, v, w;scanf("%d%d%d", &u, &v, &w);e[cnt].s = u;e[cnt].e = v;e[cnt++].t = -w;}if(Bellman(cnt)) printf("YES\n");else printf("NO\n");}return 0;}