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
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 }