| 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..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 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
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 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 "); 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 "); 72 } 73 return 0; 74 }