poj3259

Wormholes

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.

Source

USACO 2006 December Gold

spfa算法中，当出现负环的时候某个节点进出队列的次数一定大于或等于节点数。

#include<iostream>#include<queue>#include<algorithm>using namespace std;int map[5500][5500];int dist[5500];int cnt[5500];int inqueue[5500];int spfa(int n) {    queue<int> Q;    while (!Q.empty())        Q.pop();    memset(inqueue, 0, sizeof (inqueue));    memset(cnt, 0, sizeof (cnt));    for (int i = 0; i < 5500; i++)        dist[i] = 100000000;    dist[1] = 0;    inqueue[1] = 1;    cnt[1]++;    Q.push(1);    while (!Q.empty()) {        int tmp = Q.front();        Q.pop();        inqueue[tmp] = 0;        for (int i = 1; i <= n; i++) {            if (dist[tmp] + map[tmp][i] < dist[i]) {                dist[i] = dist[tmp] + map[tmp][i];                if (!inqueue[i]) {                    cnt[i]++;                    if (cnt[i] >= n)                        return 1;                    inqueue[i] = 1;                    Q.push(i);                }            }        }    }    return 0;}int main() {    int test_cases;    cin >> test_cases;    while (test_cases--) {        int n, m, w;        cin >> n >> m >> w;        for (int i = 0; i <= n; i++)            for (int j = 0; j <= n; j++)                map[i][j] = 100000000;        for (int i = 0; i < m; i++) {            int s, e, t;            cin >> s >> e >> t;            if (map[s][e] > t)                map[s][e] = map[e][s] = t;        }        for (int i = 0; i < w; i++) {            int s, e, t;            cin >> s >> e >> t;            map[s][e] = -t;        }        if (spfa(n))            cout << "YES\n";        else            cout << "NO\n";    }    return 0;}