Square
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11824 Accepted Submission(s): 3794
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
有n支棍子，是否可以组成一个正方形
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int s[21],n,sum,target;
bool used[21];
int cmp(int a,int b)
{
return a>b;
}
bool dfs(int curs,int curl,int pos)
{//curs：找到了几条正方形的边，curl：当前边已经拼凑的长度，只用pos之后的
if(curs==3)
return 1;//有了三条边，第四条肯定也是存在的
for(int i=pos;i<n;i++)
{
if(used[i]==true)
continue;
if(curl+s[i]==target)
{
used[i]=true;
if(dfs(curs+1,0,0)==true)//找到一条边之后就找下一条边
return true;
used[i]=false;
}
else if(curl+s[i]<target)
{
used[i]=true;
if(dfs(curs,curl+s[i],i)==true)
return true;
used[i]=false;//回溯 ，一条边可用可不用
}
}
return false;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
memset(s,0,sizeof(s));
scanf("%d",&n);
sum=0;
for(int i=0;i<n;i++)
{
scanf("%d",&s[i]);
sum+=s[i];
}
target=sum/4;
if(sum%4!=0||n<4)
cout<<"no"<<endl;
else
{
memset(used,false,sizeof(used));
sort(s,s+n,cmp);
if(target<s[0])
cout<<"no"<<endl;
else if(dfs(0,0,0)==true)
cout<<"yes"<<endl;
else
cout<<"no"<<endl;
}
}
return 0;
}