poj 3259--Wormholes

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 Tseconds 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.
 
bellman-ford判负环比较经典的一道题，当初也练了一下spfa，数据比较小的时候还是前者占便宜--
spfa 
 1 #include <cstdio>
 2 #include <cstring>
 3 #include <queue>
 4 #include <algorithm>
 5 #define INF 0x7f7f7f7
 6 using namespace std;
 7 int dis[510],vis[510],num[510];
 8 int head[1000];
 9 int n,cnt;
10 struct edge
11 {
12     int v,w,next;
13 }e[10000];
14 void addedge(int a,int b,int c)
15 {
16     e[cnt].v=b;
17     e[cnt].w=c;
18     e[cnt].next=head[a];
19     head[a]=cnt++;
20 }
21 int spfa()
22 {
23     int i;
24     for(i=0;i<=n;i++)
25     {
26         dis[i]=INF;
27         vis[i]=num[i]=0;
28     }
29     dis[1]=0;
30     vis[1]=1;
31     num[1]++;
32     deque<int>q;
33     q.push_back(1);//slf优化
34     while(!q.empty())
35     {
36         int x=q.front();
37         q.pop_front();
38         vis[x]=0;
39         for(i=head[x];i!=-1;i=e[i].next)
40         {
41             if(dis[e[i].v]>dis[x]+e[i].w)
42             {
43                 dis[e[i].v]=dis[x]+e[i].w;
44                 if(!vis[e[i].v])
45                 {
46                     vis[e[i].v]=1;            
47                     if(++num[e[i].v]>n)
48                         return 0;
49                     if(!q.empty())
50                     {
51                         if(dis[e[i].v]>dis[q.front()])
52                             q.push_back(e[i].v);
53                         else
54                             q.push_front(e[i].v);
55                     }
56                     q.push_back(e[i].v);
57                 }
58             }
59         }    
60     }
61     return 1;
62 }
63 
64 int main()
65 {
66     int t,m,w;
67     int a,b,c;
68     scanf("%d",&t);
69     while(t--)
70     {
71         cnt=0;
72         scanf("%d%d%d",&n,&m,&w);
73         memset(head,-1,sizeof(head));
74         while(m--)
75         {
76             scanf("%d%d%d",&a,&b,&c);
77             addedge(a,b,c);
78             addedge(b,a,c);
79         }
80         while(w--)
81         {
82             scanf("%d%d%d",&a,&b,&c);
83             addedge(a,b,-c);
84         }
85         if(!spfa())
86             printf("YES\n");
87         else
88             printf("NO\n");
89     }
90     return 0;
91 }
bellman-ford 
 1 #include <stdio.h>
 2 struct edge
 3 {
 4     int u, v, t;
 5 } edges[5220];
 6 int el, n, d[501];
 7 void bellman()
 8 {
 9     int i, j, flag;
10     for (i = 1; i < n; i++)
11     {
12         flag = 0;
13         for (j = 0; j < el; j++)
14             if (d[edges[j].v] > d[edges[j].u] + edges[j].t)
15             {
16                 d[edges[j].v] = d[edges[j].u] + edges[j].t;
17                 flag = 1;
18             }
19             if (!flag)
20             {
21                 puts("NO");
22                 return;
23             }
24     }
25     for (i = 0; i < el; i++)
26         if (d[edges[i].v] > d[edges[i].u] + edges[i].t)
27         {
28             puts("YES");
29             return;
30         }
31         puts("NO");
32 }
33 int main(void)
34 {
35     int f, m, w, s, e, t;
36     int i, j;
37     scanf("%d", &f);
38     while (f--)
39     {
40         el = 0;
41         scanf("%d %d %d", &n, &m, &w);
42         for (i = 0; i < m; i++)
43         {
44             scanf("%d %d %d", &s, &e, &t);
45             edges[el].u = s;
46             edges[el].v = e;
47             edges[el++].t = t;
48             edges[el].u = e;
49             edges[el].v = s;
50             edges[el++].t = t;
51         }
52         for (i = 0; i < w; i++)
53         {
54             scanf("%d %d %d", &s, &e, &t);
55             edges[el].u = s;
56             edges[el].v = e;
57             edges[el++].t = -t;
58         }
59         bellman();
60     }
61     return 0;
62 }

 

posted @ 2013-03-16 20:06  Tamara.c  阅读(211)  评论(0)  编辑  收藏  举报