Problem Description
Given a set of sticks of various lengths, is it possible to join them end-to-end to form a square?
Input
The first line of input contains N, the number of test cases. Each test case begins with an integer 4 <= M <= 20, the number of sticks. M integers follow; each gives the length of a stick - an integer between 1 and 10,000.
Output
For each case, output a line containing "yes" if is is possible to form a square; otherwise output "no".
Sample Input
3 4 1 1 1 1 5 10 20 30 40 50 8 1 7 2 6 4 4 3 5
Sample Output
yes no yes
Source
University of Waterloo Local Contest 2002.09.21
1 #include<iostream> 2 #include<cstring> 3 using namespace std; 4 int m,edge;//木棍的个数,木棍的边长 5 int stick[25];//木棍的长度 6 bool vis[25];//是否用过木棍 7 bool flag; 8 void dfs(int n,int len,int i)//n是第几条边，len是这条边已有长度,i从第几条边开始查找（防止超时） 9 { 10 if(n==5)//终止条件 11 { 12 flag=true; 13 return; 14 } 15 if(len==edge) 16 { 17 dfs(n+1,0,1); 18 if(flag==true) 19 return ; 20 } 21 for(int j=i;j<=m;j++) 22 { 23 if(!vis[j]) 24 { 25 if(stick[j]+len<=edge) 26 { 27 vis[j]=true; 28 dfs(n,len+stick[j],j+1); 29 if(flag==true) 30 return ; 31 vis[j]=false; 32 } 33 } 34 } 35 36 } 37 int main() 38 { 39 int T; 40 cin>>T; 41 while(T--) 42 { 43 cin>>m; 44 edge=0; 45 for(int i=1;i<=m;i++) 46 { 47 cin>>stick[i]; 48 edge+=stick[i]; 49 } 50 if(edge%4!=0)//第一个剪枝：是否能组成正方形 51 { 52 printf("no "); 53 continue; 54 } 55 edge/=4; 56 int i; 57 for(i=1;i<=m;i++)//第二个剪枝：木棍小于等于边长 58 { 59 if(stick[i]>edge) 60 break; 61 } 62 if(i!=m+1) 63 { 64 printf("no "); 65 continue; 66 } 67 memset(vis,false,sizeof(vis)); 68 flag=false; 69 dfs(1,0,1); 70 if(flag) 71 printf("yes "); 72 else 73 printf("no "); 74 } 75 return 0; 76 }