poj 3259 Wormholes (判負環)
題目連結:http://poj.org/problem?id=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 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
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.我用的floyd判負環,直接給程式碼。
說一下spfa判負環,spfa不能直接判負環,但是這個題不是直接給的負環。所以我們只需要用一個數組記錄每個點出現的次數,如果大於n,則說明存在負環,因為正環只會判斷一次。
#pragma GCC optimize(2) #include<stdio.h> #include<algorithm> #include<iostream> #include<string.h> #include<set> #include<vector> #include<string> #include<queue> using namespace std; const int maxn = 505; const int inf = 0x3f3f3f3f; typedef long long ll; int dis[maxn][maxn]; int t; int n, m, k; int floyd() { for (int p = 1; p <= n; p++) { for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { if (dis[i][j] > dis[i][p] + dis[p][j]) { dis[i][j] = dis[i][p] + dis[p][j]; } } if (dis[i][i] < 0) { return 1; } } } return 0; } int main() { //freopen("C://input.txt", "r", stdin); scanf("%d", &t); while (t--) { memset(dis, inf, sizeof(dis)); scanf("%d%d%d", &n, &m, &k); for (int i = 0; i <= n; i++) { dis[i][i] = 0; } for (int i = 1; i <= m; i++) { int u, v, w; scanf("%d%d%d", &u, &v, &w); if (w < dis[u][v]) { dis[u][v] = dis[v][u] = w; } } for (int i = 1; i <= k; i++) { int u, v, w; scanf("%d%d%d", &u, &v, &w); dis[u][v] = -w; } if (floyd()) { printf("YES\n"); } else { printf("NO\n"); } } return 0; }