POJ3259 Wormholes

 

Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 44261   Accepted: 16285

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..NM (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, FF farm descriptions follow. 
Line 1 of each farm: Three space-separated integers respectively: NM, and W 
Lines 2..M+1 of each farm: Three space-separated numbers (SET) 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 (SET) 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

 

普通路是双向边!虫洞是单向边!所以边数组要开到M*2+W!

一定要好好看题呀,不然WA了都不知道为啥,一定要算好数据范围啊,不然RE了都不知道为啥……

 

至于问题本身,判图里有没有负环就行了。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #include<cstring>
 5 #include<cmath>
 6 #include<queue>
 7 using namespace std;
 8 const int mxn=8000;
 9 struct edge{
10     int v,dis;
11     int next;
12 }e[mxn];
13 int hd[mxn],cnt;
14 int n,m,w;
15 void add_edge(int u,int v,int dis){
16     e[++cnt].v=v;e[cnt].next=hd[u];e[cnt].dis=dis;hd[u]=cnt;
17 }
18 int inq[mxn],dis[mxn];
19 int vis[mxn];
20 bool SPFA(int s){
21     memset(vis,0,sizeof vis);
22     memset(dis,0x3f,sizeof dis);
23     memset(inq,0,sizeof inq);
24     queue<int>q;
25     inq[s]=1;dis[s]=0;vis[s]=1;
26     q.push(s);
27     while(!q.empty()){
28         int u=q.front();q.pop();
29         for(int i=hd[u];i;i=e[i].next){
30             int v=e[i].v;
31             if(dis[u]+e[i].dis<dis[v]){
32                 dis[v]=dis[u]+e[i].dis;
33                 vis[v]++;
34                 if(vis[v]>n){
35                     printf("YES\n");
36                     return 0;
37                 }
38                 if(!inq[v]){
39                     inq[v]=1;
40                     q.push(v);
41                 }
42             }
43         }
44         inq[u]=0;
45     }
46     return 1;
47 }
48 int main(){
49     int T;
50     scanf("%d",&T);
51     while(T--){
52         memset(hd,0,sizeof hd);
53         memset(e,0,sizeof e);
54         cnt=0;
55         scanf("%d%d%d",&n,&m,&w);
56         int i,j;
57         int x,y,d;
58         for(i=1;i<=m;i++){
59             scanf("%d%d%d",&x,&y,&d);
60             add_edge(x,y,d);
61             add_edge(y,x,d);
62         }
63         for(i=1;i<=w;i++){
64             scanf("%d%d%d",&x,&y,&d);
65             add_edge(x,y,-d);            
66         }
67         bool flag=0;
68         for(i=1;i<=n;i++){
69             if(!SPFA(i)){flag=1;break;}
70         }
71         if(!flag)printf("NO\n");
72     }
73     return 0;
74 }

 

posted @ 2016-08-12 21:57  SilverNebula  阅读(151)  评论(0编辑  收藏  举报
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