HDU1518——Square（正方形問題）

Square

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 13808    Accepted Submission(s): 4376

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

 

解：

正方形問題：
要求給定的所有木棍能否組成一個正方形
那麼正方形的變長應該是所有木棍長度之和的四分之一 eage，如果不能整除一定不能構成正方形，
而且當給定的數組中最大值大於eage的時候也不能組成正方形 
定義sum[4]數組，用來分別存放每條邊的長度，
每條邊都有 n  種木棍可以選擇，而且一條邊可以有多個木棍，
所以回溯過程中，將每一條木棍加入每個邊
如果，t==n 到達解空間，此時如果 sum[4] 中的每一個都等於開始算出的邊長eage則可組成正方形，反之不可以 

剪枝不夠提交超時（上課代碼）：

#include <stdio.h>
#include <string.h>
int a[100];
int n;
int sum[5]={0};
int eage;
int flag=0; 

void dfs(int t)
{
	if(flag==1)
		return;
	if(t==n)
	{
		if(sum[1]==eage && sum[2]==eage && sum[3]==eage && sum[4]==eage)
			flag=1;
		return;
	}
	
	for(int i=1;i<5;i++)
	{

		sum[i]+=a[t];
		if(sum[1]<=eage && sum[2]<=eage && sum[3]<=eage && sum[4]<=eage){
			dfs(t+1);
		}
		sum[i]-=a[t];
	}
}


int main()
{
	int t;
	scanf("%d",&t);
	while(t--){
		int max=0;
		scanf("%d",&n);
		memset(a,0,sizeof(a));
		memset(sum,0,sizeof(sum));
		eage=0;
		for(int i=0;i<n;i++)
		{
			scanf("%d",&a[i]);
			if(a[i]>max)
            {
                max=a[i];
            }
			eage+=a[i];
		}
		if(eage%4!=0 || max>(eage/4))
			printf("no\n");
		else{
			eage/=4;
			dfs(0);
			if(flag==1)
			{
				printf("yes\n");
			}else
				printf("no\n");
		}
	}
	return 0;
} 

修改之後加強剪枝：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

int tt,num[25],flag[25],aver,f;

void dfs(int length,int count,int k)
{
    if(f)
    {
        return ;
    }
    if(length==aver)
    {
        count++;
        if(count==4)
        {
            f=1;
            return ;
        }
        k=0;
        length=0;
    }
    for(int i=k;i<tt;i++)
    {
        if(!flag[i] && length+num[i]<=aver)
        {
            flag[i]=1;
            dfs(length+num[i],count,i);
            flag[i]=0;
        }
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&tt);
        int sum=0,max=0;
        for(int i=0;i<tt;i++)
        {
            scanf("%d",&num[i]);
            sum+=num[i];
            if(num[i]>max)
            {
                max=num[i];
            }
        }
        if(sum%4!=0||max>(sum/4))
        {
            printf("no\n");
            continue;
        }
        aver=sum/4;
        memset(flag,0,sizeof(flag));
        f=0;
        dfs(0,0,0);
        if(f==1)
        {
            printf("yes\n");
        }else{
            printf("no\n");
        }
    }
    return 0;
}