Square
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11099 Accepted Submission(s): 3566
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
题意就是：看这个数组中的数字组合是否能够构成一个正方形
不能分割数字，不能重复组合
代码：
#include<string.h>
#include<stdio.h>
#include<algorithm>
using namespace std;
int n,s,su,a[1010],vis[1010],len;
bool cmp(int a,int b)
{
return a>b;//从大到小排序
}
void dfs(int a1,int a2,int a3)//(0,0,1)
{
if(a1==3)/**只需要筹齐3次，那么剩下的一定能够成len长度**/
{
su=1;
return ;
}
if(su==1)
return ;/**优化时间**/
for(int i=a3; i<=n; i++)
{
/**对于那些用过的和不符合条件的，for那里可以不扫,故从a3开始**/
/**a3前面的对于最初的a2来说一定不符合**/
if(vis[i]==0)
{
vis[i]=1;
if(a2+a[i]==len)
{
dfs(a1+1,0,1);
/**但是换另外一条边的时候a3要改回1,因为那些未用的，对上一条边来说不符合条件的，可能符合这条边的条件**/
}
else if(a2+a[i]<len)
{
dfs(a1,a2+a[i],i+1);
/**没筹齐从i+1继续，前面的不符合**/
while(a[i]==a[i+1])
i++;
//回溯后如果后面的相同那么不需要再DFS
//前面的数和后面的相同就可以跳过这个数，剪枝
}
vis[i]=0;
}
}
}
int main()
{
int i;
scanf("%d",&s);
while(s--)
{
su=0;
len=0;
memset(vis,0,sizeof(vis));
//memset函数在string.h头文件中
scanf("%d",&n);
for(i=1; i<=n; i++)
{
scanf("%d",&a[i]);
len=len+a[i];
}
int mm=len/4;
sort(a+1,a+n+1,cmp);
//在algorithm头文件中
if(len%4==0&&a[1]<=mm&&n>=4)
{
len/=4;
dfs(0,0,1);
if(su==1)
printf("yes\n");
else
printf("no\n");
}
else
{
printf("no\n");
}
}
return 0;
}
