深搜，hdoj1518square

HDOJ 1518 Square（DFS 深搜）

题目

Square

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

Total Submission(s): 3911 Accepted Submission(s): 1302



[align=left]Problem Description[/align]
Given a set of sticks of various lengths, is it possible to join them end-to-end to form a square?

[align=left]Input[/align]
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.

[align=left]Output[/align]
For each case, output a line containing "yes" if is is possible to form a square; otherwise output "no".

[align=left]Sample Input[/align]

3
4 1 1 1 1
5 10 20 30 40 50
8 1 7 2 6 4 4 3 5

[align=left]Sample Output[/align]

yes
no
yes

题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=1518

思路：

给出的边要全部用到，将所给的边组成正方形，所有边都要搜到，所以要用到深搜的方法，并且需要剪枝，不然可能会爆，代码是参考别人的，首先要判断所有的边是否能组成正方形，再者就是最终要的DFS，结束条件和递归都是DFS的重要部分，当然这道题需要回溯，因为标记后，如果那种方法不行，回溯才能使下次能搜到这些被标记过但又没有用到的边

代码：

[cpp] view
plaincopy

#include<iostream>

using namespace std;

int a[20+10];

bool used[20+10];

int n,m,sum;

int dfs(int d, int w, int count)

{

if(d == sum)

{

w = 0;

d = 0;//

count++;

if(count == 4)

return 1;

}

for(int i=w; i<m; i++)

{

if(!used[i])

{

used[i] = 1;

if((d+a[i] <= sum ) && dfs(d+a[i], i+1, count))

return 1;

used[i] = 0;

}

}

return 0;

}

int main()

{

cin>>n;

while(n--)

{

cin>>m;

sum = 0;

memset(used, 0, sizeof(used));

for(int i=0; i<m; i++)

{

cin>>a[i];

sum += a[i];

}

int flag = 0;

if(sum % 4 ==0)

{

sum /= 4;

flag = dfs(0, 0, 0);

}

if(flag)

{

cout<<"yes"<<endl;

}

else

{

cout<<"no"<<endl;

}

}

}