poj_3259_spfa

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 27177 Accepted: 9784

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.

<pre name="code" class="cpp">#include <iostream>#include <cstdio>#include <cstring>#include <queue>#include <vector>using namespace std;int n, m, w;int map[510][510];bool visit[510];int path[510], in[510];bool spfa(){queue<int>q;visit[1] = 1;path[1] = 0;in[1]++;q.push(1);while (!q.empty()){int v = q.front();q.pop();if (in[v] >= n)return 1;visit[v] = 0;for (int i = 1; i <= n; i++){if (path[v] + map[v][i] < path[i]){path[i] = path[v] + map[v][i];if (!visit[i]){visit[i] = 1;if (in[i] >= n)return 1;q.push(i);in[i]++;}} }visit[v] = 1;}return 0;}int main(){int t;scanf("%d", &t);while (t--){memset(path, 0x3f, sizeof (path));memset(visit, 0, sizeof (visit));memset(in, 0, sizeof (in));memset(map, 0x3f, sizeof(map));scanf("%d%d%d", &n, &m, &w);for (int i = 0; i < m; i++){int a, b, c;scanf("%d%d%d", &a, &b, &c);if (map[a][b] > c){map[a][b] = c;map[b][a] = c;}}for (int i = 0; i < w; i++){int a, b, c;scanf("%d%d%d", &a, &b, &c);c = -c;if (map[a][b] > c)map[a][b] = c;}if (spfa())printf("YES\n");else printf("NO\n");}return 0;}