1227. Rally Championship
Time limit: 1.0 second Memory limit: 64 MB
A high-level international rally championship is about to be held. The rules of the race state that the race is held on ordinary roads and the route has a fixed length. You are given a map of the cities and two-way roads connecting it. To make the race safer it is held on one-way roads. The race may start and finish anyplace on the road. Determine if it is possible to make a route having a given length S.
Input
The first line of the input contains the number of cities M, the number of roads N and the length S of the route. 1 ≤ M ≤ 100; 1 ≤ N ≤ 10000; 1 ≤ S ≤ 106. S is integer.
The following N lines describe the roads as triples of integers: P, Q, R. Here P and Q are cities connected with a road, and R is the length of this road. All numbers satisfy the following restrictions: 1 ≤ P, Q ≤ M; 1 ≤ R ≤ 32000.
Output
Write YES to the output if it is possible to make a required route and NO otherwise. Note that answer must be written in capital Latin letters.
Samples
| input | output |
|---|---|
3 2 20 1 2 10 2 3 5 |
NO |
3 3 1000 1 2 1 2 3 1 1 3 1 |
YES |
Problem Source: 2002-2003 ACM Central Region of Russia Quarterfinal Programming Contest, Rybinsk, October 2002
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判断环是否存在，同时求路径的长度 不用记录路径，可直接深搜，满足条件返回
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1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<string> 5 #include<vector> 6 #include<algorithm> 7 #include<stack> 8 using namespace std; 9 long map[1001][1001];//邻接矩阵存图 10 bool vis[1001]={false};//标记数组 11 bool gs; 12 int n,m,p,q,i,j,k,start; 13 long ans,s,r; 14 void dfs(int x,long st)//深搜，判断是否存在环，st为长度 15 { 16 ans=st; 17 if(ans>=s)//路径>=s满足条件返回 18 { 19 gs=true; 20 return; 21 } 22 23 for(int is=1;is<=m;is++) 24 { 25 if(!vis[is]&&map[x][is]) 26 { 27 vis[is]=true; 28 long temp=map[x][is]; 29 map[x][is]=map[is][x]=0; 30 if(is==start)//存在环返回 31 { 32 gs=true; 33 return; 34 } 35 dfs(is,st+temp); 36 map[x][is]=map[is][x]=temp;//还原 37 vis[is]=false; 38 if(gs==true)//为了保险，可能多此一举 39 return; 40 41 } 42 } 43 } 44 int main() 45 { 46 cin>>m>>n>>s; 47 memset(vis,false,sizeof(vis)); 48 memset(map,0,sizeof(map)); 49 gs=false; 50 for(i=1;i<=n;i++) 51 { 52 cin>>p>>q>>r; 53 if(map[p][q]==0) 54 map[p][q]=map[q][p]=r; 55 else 56 gs=true; 57 } 58 if(gs==true) 59 { 60 cout<<"YES"<<endl; 61 return 0; 62 } 63 int max1=-1; 64 for(i=1;i<=m;i++) 65 if(!vis[i]) 66 { 67 start=i; 68 69 dfs(i,0); 70 memset(vis,false,sizeof(vis)); 71 if(gs==true) 72 { 73 cout<<"YES"<<endl; 74 return 0; 75 } 76 } 77 cout<<"NO"<<endl; 78 return 0; 79 80 81 82 }