HDU 1518 Square

Square

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19923    Accepted Submission(s): 6272

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

University of Waterloo Local Contest 2002.09.21

 

Recommend

LL

题解：第一行给测试询问数m

接下来m行，第一个数字是棍子数量n，接下来n个数字是棍子长度。

假如这些棍子能够首尾相接，形成一个正方形，则输出yes，否则no。

我一看这题，就感觉和POJ 1011 Sticks非常像。

于是我这么把POJ的代码胡乱一改，果然——只能输出no,我也不知道咋肥四~~~留一个坑。

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;
int a[25];//被截断的棍子长度集 
int vis[25]; //标记是否使用过该棍子
//int num;//被截断后的棍子数 
int cnt,sum,flag;
 int n;
bool cmp(int a,int b)
{
    return a>b;//降序 
}
int dfs(int rest,int n)  
{
    if(rest==0&&n==0)
        {
            cnt++;
            return cnt;
        }
    if(rest==0)
        rest=sum/4;
    for(int i=1;i<=n;i++)
    {
        if(vis[i])//如果这个数已经被检测过了，跳过 
            continue;
        if(rest>=a[i])
        {
            vis[i]=1;
            if(dfs(rest-a[i],n-1))//因为dfs函数有返回值，使得目标长度成立
                {
                    cnt++;
                    return cnt;
                }
            vis[i]=0;
            if(a[i]==rest||sum/4==rest)//
                break;
            while(a[i]==a[i+1])//因为是降序排列的，如果这个数不满足，那么下一位和它相等的数肯定也不满足 
                i++;
        } 
    }
    return 0; 
}


int main()
{
    int m;
    while(scanf("%d",&m))
    {
        while(m--)
        {
            int i;
            cnt=0;
            flag=0;
            sum=0;
            scanf("%d",&n);
            for(i=1;i<=n;i++)
                {
                    scanf("%d",&a[i]);
                    sum=sum+a[i];
                }
                sort(a+1,a+n+1,cmp);
                memset(vis,0,sizeof(vis));
                flag=dfs(0,n);
                if(flag==4)
                  printf("yes
");
                else
                    printf("no
");         
        }
     } 
  //   system("pause");
    return 0;
}