Square
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7090 Accepted Submission(s): 2298
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
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1 //578MS 328K 998 B G++ 2 /* 3 4 题意： 5 问给出的值能否均分成四分。 6 7 dfs: 8 排序后实际上大大缩短了搜索时间 9 10 */ 11 #include<iostream> 12 #include<algorithm> 13 using namespace std; 14 int a[50],vis[50]; 15 int n,s; 16 int cmp(int x,int y) 17 { 18 return x>y; 19 } 20 int dfs(int pos,int t,int cur) 21 { 22 if(t==4) return 1; 23 for(int i=pos;i<n;i++) 24 if(!vis[i]){ 25 vis[i]=1; 26 if(cur+a[i]==s){ 27 if(dfs(0,t+1,0)) return 1; 28 }else if(cur+a[i]<s){ 29 if(dfs(i+1,t,cur+a[i])) return 1; 30 } 31 vis[i]=0; 32 } 33 return 0; 34 } 35 int main(void) 36 { 37 int t; 38 scanf("%d",&t); 39 while(t--){ 40 memset(a,0,sizeof(a)); 41 scanf("%d",&n); 42 s=0; 43 for(int i=0;i<n;i++){ 44 scanf("%d",&a[i]); 45 s+=a[i]; 46 } 47 sort(a,a+n,cmp); 48 if(s%4 || a[0]>s/4){ 49 puts("no"); 50 }else{ 51 s/=4; 52 memset(vis,0,sizeof(vis)); 53 if(dfs(0,0,0)) puts("yes"); 54 else puts("no"); 55 } 56 } 57 return 0; 58 }