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
题意 :给几组数据 看每组数据 能否所实用完 并组成一个正方形
代码:
代码:
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int q[22]; int vis[22]; int cmp(int a,int b) { return a<b; } int t,n,s; void dfs(int num,int k,int length) //num 是已完毕的边 k是 数组的位置 length是当前边的长度 { if(t==1) return ; if(num==4) { t=1; return ; } if(length==s) { dfs(num+1,0,0); if(t==1) return ; } for(int i=k; i<n; i++) { if(length+q[i]<=s&&vis[i]==0) { vis[i]=1; dfs(num,i+1,length+q[i]); vis[i]=0; } } } int main() { int m; scanf("%d",&m); while(m--) { t=0; s=0; scanf("%d",&n); for(int i=0; i<n; i++) { scanf("%d",&q[i]); s=s+q[i]; } if(s%4!=0) { printf("no "); continue; } sort(q,q+n,cmp); s=s/4; if(q[n-1]>s) { printf("no "); continue; } dfs(0,0,0); if(t==1) printf("yes "); else printf("no "); } return 0; }标准的回溯 可惜 我没有看到这个问题 要运行数据
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