标签：
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
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
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 }
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 }
标签：
原文地址：http://www.cnblogs.com/cxbky/p/4907140.html
