poj 3259 Wormholes(最短路)

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

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

题目大意：虫洞问题，现在有n个点，m条边，代表现在可以走的通路，比如从a到b和从b到a需要花费c时间，现在在地上出现了w个虫洞，虫洞的意义就是你从a到花费的时间是-c(时间倒流,并且虫洞是单向的)，现在问你从某个点开始走，能回到从前

本质就是求该图是否存在负环。也就是如何求出一个图是否含有负环。

注意：M是正权，双向，W是负权，单向

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 using namespace std;
 6 const int maxn=5300;
 7 const int INF=10001;
 8 struct node
 9 {
10     int s;
11     int e;
12     int t;
13 } path[maxn];
14 int n,m,w;
15 int val[maxn]; //源点到各点的权值
16 int index; //边的总数
17 void bellman()
18 {
19     memset(val,INF,sizeof(val));
20     bool flag;
21     for(int i=0; i<n-1; i++)
22     {
23         flag=false;
24         for(int j=0; j<index; j++)
25         {
26             if(val[path[j].e]>val[path[j].s]+path[j].t)
27             {
28                 val[path[j].e]=val[path[j].s]+path[j].t;
29                 flag=true;
30             }
31         }
32         if(!flag)
33             break;
34     }
35     for(int i=0; i<index; i++)
36     {
37         if(val[path[i].e]>val[path[i].s]+path[i].t)
38         {
39             cout<<"YES"<<endl;
40             return ;
41         }
42     }
43     cout<<"NO"<<endl;
44     return ;
45 }
46 int main()
47 {
48     int t;
49     int a,b,c;//临时变量
50     cin>>t;
51     while(t--)
52     {
53         index=0;
54         cin>>n>>m>>w;
55         for(int i=1; i<=m; i++) //正权边，双向
56         {
57             cin>>a>>b>>c;
58             path[index].s=a;
59             path[index].e=b;
60             path[index++].t=c;
61             path[index].s=b;
62             path[index].e=a;
63             path[index++].t=c;
64         }
65         for(int i=1; i<=w; i++) //负权边，单向
66         {
67             cin>>a>>b>>c;
68             path[index].s=a;
69             path[index].e=b;
70             path[index++].t=-c;
71         }
72         bellman();
73     }
74     return 0;
75 }

View Code

这样也可以

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 using namespace std;
 6 const int maxn=5300;
 7 const int INF=10001;
 8 struct node
 9 {
10     int s;
11     int e;
12     int t;
13 } path[maxn];
14 int n,m,w;
15 int val[maxn]; //源点到各点的权值
16 int index; //边的总数
17 void bellman()
18 {
19     for(int i=1;i<=n;i++)
20         val[i]=INF;
21     val[1]=0;
22     bool flag;
23     for(int i=1; i<n; i++)
24     {
25         flag=false;
26         for(int j=1; j<index; j++)
27         {
28             if(val[path[j].e]>val[path[j].s]+path[j].t)
29             {
30                 val[path[j].e]=val[path[j].s]+path[j].t;
31                 flag=true;
32             }
33         }
34         if(!flag)
35             break;
36     }
37     if(flag)
38         cout<<"YES"<<endl;
39     else
40         cout<<"NO"<<endl;
41     return ;
42 }
43 int main()
44 {
45     int t;
46     int a,b,c;//临时变量
47     cin>>t;
48     while(t--)
49     {
50         index=1;
51         cin>>n>>m>>w;
52         for(int i=1; i<=m; i++) //正权边，双向
53         {
54             cin>>a>>b>>c;
55             path[index].s=a;
56             path[index].e=b;
57             path[index++].t=c;
58             path[index].s=b;
59             path[index].e=a;
60             path[index++].t=c;
61         }
62         for(int i=1; i<=w; i++) //负权边，单向
63         {
64             cin>>a>>b>>c;
65             path[index].s=a;
66             path[index].e=b;
67             path[index++].t=-c;
68         }
69         bellman();
70     }
71     return 0;
72 }

View Code