| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 36860 | Accepted: 13505 |
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
题解：
这道题意思是这个人的农田里有道路，还有虫洞，道路是双向的，虫洞单向，他喜欢虫洞旅行，想要从起点再回到起点时间在出发的时间之前，经过虫洞时间倒退；
做这道题其实就是判断最短路有没有负环的问题；因为只要有负环，就会无限循环，到起点自然就时间倒退了；
代码：SPFA
1 #include<stdio.h> 2 #include<queue> 3 #include<string.h> 4 using namespace std; 5 const int INF=0x3f3f3f3f; 6 const int MAXN=510; 7 const int MAXM=10010*2; 8 int dis[MAXN],vis[MAXN],used[MAXN],head[MAXM]; 9 int N,W,M,en,flot; 10 queue<int>dl; 11 struct Edge{ 12 int from,to,value,next; 13 }; 14 Edge edg[MAXM]; 15 void initial(){ 16 memset(dis,INF,sizeof(dis)); 17 memset(vis,0,sizeof(vis)); 18 memset(used,0,sizeof(used)); 19 memset(head,-1,sizeof(head)); 20 while(!dl.empty())dl.pop(); 21 en=0;flot=0; 22 } 23 void print(){ 24 if(flot)puts("YES"); 25 else puts("NO"); 26 } 27 void add(int u,int v,int w){ 28 Edge E={u,v,w,head[u]}; 29 edg[en]=E; 30 head[u]=en++; 31 } 32 void SPFA(int sx){ 33 dis[sx]=0;vis[sx]=1;dl.push(sx); 34 used[sx]++; 35 while(!dl.empty()){ 36 int k=dl.front(); 37 dl.pop(); 38 vis[k]=0; 39 if(used[k]>N){ 40 flot=1; 41 break; 42 } 43 for(int i=head[k];i!=-1;i=edg[i].next){ 44 int v=edg[i].to; 45 if(dis[k]+edg[i].value<dis[v]){ 46 dis[v]=dis[k]+edg[i].value; 47 if(!vis[v]){ 48 vis[v]=1; 49 dl.push(v); 50 used[v]++; 51 if(used[v]>N){ 52 flot=1;return ; 53 } 54 } 55 } 56 } 57 } 58 } 59 void get(){ 60 int F,a,b,c; 61 scanf("%d",&F); 62 while(F--){ 63 initial(); 64 scanf("%d%d%d",&N,&M,&W); 65 while(M--){ 66 scanf("%d%d%d",&a,&b,&c); 67 add(a,b,c); 68 add(b,a,c); 69 } 70 while(W--){ 71 scanf("%d%d%d",&a,&b,&c); 72 add(a,b,-c); 73 } 74 SPFA(1); 75 print(); 76 } 77 } 78 int main(){ 79 get(); 80 return 0; 81 }
Bellman:
1 #include<stdio.h> 2 #include<string.h> 3 const int INF=0x3f3f3f3f; 4 const int MAXN=510; 5 const int MAXM=6000; 6 int dis[MAXN]; 7 struct Edge{ 8 int u,v,w; 9 }; 10 Edge edg[MAXM]; 11 int N,M,W,top; 12 bool Bellman(int sx){ 13 int u,v,w; 14 memset(dis,INF,sizeof(dis)); 15 dis[sx]=0; 16 for(int i=1;i<=N;i++){ 17 for(int j=0;j<top;j++){ 18 u=edg[j].u;v=edg[j].v;w=edg[j].w; 19 if(dis[u]+w<dis[v])dis[v]=dis[u]+w; 20 } 21 } 22 for(int i=0;i<top;i++){ 23 u=edg[i].u;v=edg[i].v;w=edg[i].w; 24 if(dis[u]+w<dis[v])return false; 25 } 26 return true; 27 } 28 int main(){ 29 int F; 30 int a,b,c; 31 scanf("%d",&F); 32 while(F--){ 33 top=0; 34 scanf("%d%d%d",&N,&M,&W); 35 while(M--){ 36 scanf("%d%d%d",&a,&b,&c); 37 edg[top].u=a;edg[top].v=b;edg[top++].w=c; 38 edg[top].u=b;edg[top].v=a;edg[top++].w=c; 39 } 40 while(W--){ 41 scanf("%d%d%d",&a,&b,&c); 42 edg[top].u=a;edg[top].v=b;edg[top++].w=-c; 43 } 44 if(Bellman(1))puts("NO"); 45 else puts("YES"); 46 } 47 return 0; 48 }