Wormholes//spfa 负环判断 链式向前星
题目:
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 60744 | Accepted: 22692 |
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 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
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 8
Sample Output
NO YES
Hint
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.
思路:
代码:
#include <iostream> #include <queue> #include <cstring> using namespace std; const int size1 = 500 + 5; const int size2 = 5200 + 5; struct Edge { int to; int len; int next; }; //链式向前星 Edge edge[size2]; int head[size1]; int cnt = 0; queue<int> que; bool onque[size1];//是否在队列上 int intoque[size1];//>N则说明有可达负环(可达!!!) int dis[size1]; bool marked[size1];//标记可达边 void init(int N)//初始化 { cnt = 0; memset(head, -1, sizeof(head)); for(int i = 1; i <= N; i++) { dis[i] = 0x3f3f3f3f; marked[i] = false; onque[i] = false; intoque[i] = 0; } } void addEdge(int u, int v, int w)//加边 { edge[cnt].to = v; edge[cnt].len = w; edge[cnt].next = head[u]; head[u] = cnt++; } bool spfa(int x, int N) { dis[x] = 0; //marked[x] = true; que.push(x); onque[x] = true; marked[x] = true; while(!que.empty()) { int c = que.front(); onque[c] = false; que.pop(); for(int h = head[c]; h != -1; h=edge[h].next) { if(dis[c] + edge[h].len < dis[edge[h].to])//边松弛 { marked[edge[h].to] = true;//该边可达 dis[edge[h].to] = dis[c] + edge[h].len; if(!onque[edge[h].to])//入队列 { que.push(edge[h].to); onque[edge[h].to] = true; intoque[edge[h].to]++; if(intoque[edge[h].to] > N)//检测负环 return true; } } } } return false;//无负环 } int main() { int F; int N, M, W; int S, E, T; cin >> F; while(F--) { scanf("%d%d%d", &N, &M, &W); init(N); for(int i = 0; i < M; i++) { scanf("%d%d%d", &S, &E, &T); //双向边 addEdge(S,E,T); addEdge(E,S,T); } for(int i = 0; i < W; i++) { scanf("%d%d%d", &S, &E, &T); addEdge(S,E,-T);//负权 } int flag = 0; for(int i = 1; i <= N; i++)//对每一个不可达的地方进行检测 { if(!marked[i]) { if(spfa(i,N)) { flag = 1; break; } } } if(flag) printf("YES\n"); else printf("NO\n"); } return 0; }