HDU1518——Square（正方形问题）

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：
正方形问题：
要求给定的所有木棍能否组成一个正方形
那么正方形的变长应该是所有木棍长度之和的四分之一 eage，如果不能整除一定不能构成正方形，
而且当给定的数组中最大值大于eage的时候也不能组成正方形 
定义sum[4]数组，用来分别存放每条边的长度，
每条边都有 n  种木棍可以选择，而且一条边可以有多个木棍，
所以回溯过程中，将每一条木棍加入每个边
如果，t==n 到达解空间，此时如果 sum[4] 中的每一个都等于开始算出的边长eage则可组成正方形，反之不可以 

剪枝不够提交超时（上课代码）：

#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;
}

修改之后加强剪枝：

#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;
}