首页 > 代码库 > poj 2362
poj 2362
G - Square
Time Limit:3000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64uDescription
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
34 1 1 1 15 10 20 30 40 508 1 7 2 6 4 4 3 5
Sample Output
yesnoyes
dfs(be,len,cnt)
cnt==3 return true;
be代表当前选择第几个木棒，len,代表当前的长度
剪枝：
1.首先满足 sum%4==0
2.其次 最大边*4应该《=sum
3.当选择的木棒+当前的长度《=sum/4
AC无压力！
#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<algorithm>#include<cstdlib>#include<queue>#include<vector>#include<set>using namespace std;int t,n,a[10010],sum,maxx,side;bool flag;bool vis[10010];bool cmp(int x,int y){ return x<y;}void dfs(int be,int len,int cnt){ if(flag) return ; if(cnt==3) { flag=true; return ; } if(len==side) dfs(0,0,cnt+1); for(int i=be;i<=n;i++) if(!vis[i]&&len+a[i]<=side) { vis[i]=true; dfs(i+1,len+a[i],cnt); if(flag) return ; vis[i]=false; }}int main(){ scanf("%d",&t); while(t--) { sum=0,maxx=0,flag=false; memset(vis,0,sizeof(vis)); scanf("%d",&n); for(int i=1;i<=n;i++) { scanf("%d",&a[i]); sum+=a[i]; maxx=max(maxx,a[i]); } side=sum/4; sort(a+1,a+1+n,cmp); if(sum%4!=0||maxx*4>sum) printf("no\n"); else { dfs(0,0,0); if(flag) printf("yes\n"); else printf("no\n"); } } return 0;}
poj 2362
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。