POJ3259 Wormholes —— spfa求负环

Posted 2020-10-08 h_z_cong

题目链接：http://poj.org/problem?id=3259

 

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 55082		Accepted: 20543

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

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 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

 

 

 

 

题解：

 

 

 

代码如下：

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <algorithm>
  5 #include <vector>
  6 #include <cmath>
  7 #include <queue>
  8 #include <stack>
  9 #include <map>
 10 #include <string>
 11 #include <set>
 12 #define rep(i,a,n) for(int (i) = a; (i)<=(n); (i)++)
 13 #define ms(a,b) memset((a),(b),sizeof((a)))
 14 using namespace std;
 15 typedef long long LL;
 16 const double EPS = 1e-8;
 17 const int INF = 2e9;
 18 const LL LNF = 9e18;
 19 const int MOD = 1e9+7;
 20 const int MAXN = 1e3+10;
 21 
 22 int N, M, W;
 23 
 24 struct edge
 25 {
 26     int to, w, next;
 27 }edge[MAXN*MAXN];
 28 int cnt, head[MAXN];
 29 
 30 void add(int u, int v, int w)
 31 {
 32     edge[cnt].to = v;
 33     edge[cnt].w = w;
 34     edge[cnt].next = head[u];
 35     head[u] = cnt++;
 36 }
 37 
 38 void init()
 39 {
 40     cnt = 0;
 41     memset(head, -1, sizeof(head));
 42 }
 43 
 44 int dis[MAXN], times[MAXN], inq[MAXN];
 45 bool spfa(int st)
 46 {
 47     memset(inq, 0, sizeof(inq));
 48     memset(times, 0, sizeof(times));
 49     for(int i = 1; i<=N; i++)
 50         dis[i] = INF;
 51 
 52     queue<int>Q;
 53     Q.push(st);
 54     inq[st] = 1;
 55     dis[st] = 0;
 56     while(!Q.empty())
 57     {
 58         int u = Q.front();
 59         Q.pop(); inq[u] = 0;
 60         for(int i = head[u]; i!=-1; i = edge[i].next)
 61         {
 62             int v = edge[i].to;
 63             if(dis[v]>dis[u]+edge[i].w)
 64             {
 65                 dis[v] = dis[u]+edge[i].w;
 66                 if(!inq[v])
 67                 {
 68                     Q.push(v);
 69                     inq[v] = 1;
 70                     if(++times[v]>N) return true;
 71                 }
 72             }
 73         }
 74     }
 75     return false;
 76 }
 77 
 78 int main()
 79 {
 80     int T;
 81     scanf("%d",&T);
 82     while(T--)
 83     {
 84         init();
 85         scanf("%d%d%d", &N, &M, &W);
 86         for(int i = 1; i<=M; i++)
 87         {
 88             int s, e, t;
 89             scanf("%d%d%d", &s, &e, &t);
 90             add(s, e, t);
 91             add(e, s, t);
 92         }
 93 
 94         for(int i = 1; i<=W; i++)
 95         {
 96             int s, e, t;
 97             scanf("%d%d%d", &s, &e, &t);
 98             add(s, e, -t);
 99         }
100 
101         int flag = 0;
102         for(int i = 1; i<=N; i++)
103         if(spfa(i)) 
104         {
105             flag = 1;
106             break;
107         }
108         
109 
110         if(flag)
111             puts("YES");
112         else
113             puts("NO");
114     }
115 }

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