Wormholes

时限：1000ms 内存限制：10000K  总时限：3000ms

描述：

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.

输入：

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.

输出：

Lines 1..F For each farm, output YES if FJ can achieve his goal, otherwise output NO (do not include the quotes).

输入样例：

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 8

输出样例：

NO
YES

提示：

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.
#include<iostream>#define INF 100000using namespace std;struct Border{int start;int end;int time;};Border border[5500];int N, M, W, dis[500];bool BellmanFord(int count){int i, j;for (i = 0; i < N; ++i)dis[i] = INF;dis[0] = 0;for (i = 0; i < N; ++i){bool flag = false;for (j = 0; j < count; ++j){if (dis[border[j].end] > dis[border[j].start] + border[j].time){dis[border[j].end] = dis[border[j].start] + border[j].time;flag = true;}}if (!flag)break;}for (j = 0; j < count; ++j){if (dis[border[j].end] > dis[border[j].start] + border[j].time)return true;}return false;}int main(void){int F, a, b, c, count;cin >> F;while (F--){count = 0;cin >> N >> M >> W;while (M--){cin >> a >> b >> c;border[count].start = a;border[count].end = b;border[count].time = c;++count;border[count].start = b;border[count].end = a;border[count].time = c;++count;}while (W--){cin >> a >> b >> c;border[count].start = a;border[count].end = b;border[count].time = -c;++count;}if (BellmanFord(count))cout << "YES" << endl;elsecout << "NO" << endl;}return 0;}