HDU1518-Square

原文链接：HDU1518-Square

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

原文大意：就是好多棍子，看能不能拼成正方形

• 所有的数据都得用上
• 输入第一行n：在4和20之间，以下n行，每行第一个数据m代表该行有m个数据

试题分析：输入数据总和需要是4的倍数，当拼好3条边的时候就意味着可以。利用深度搜索即可得出答案

AC代码：

#include<iostream>//#include<string>#include<algorithm>using namespace std;bool used[21];int arr[21];int t,m;//t：边长,sum:周长 ,m:输入的根数 bool cmp(int x,int y){    return x>y;}bool dfs(int count,int x,int y){    int i;    //count:完成几条边，x：在遍历第几根，y：距离拼好目标差多少    if(count==3)    {//若是已经拼完3根，意味着成功         return true;        }    for(i=x;i<m;i++)    {        if(used[i])            continue;        used[i]=true;        if(arr[i]==y)//若是第i个数据与y相等则拼成一条边         {            if(dfs(count+1,0,t))                return true;        }        else if(arr[i]<y)//若小于，则遍历下面的边         {            if(dfs(count,i+1,y-arr[i]))                return true;        }        used[i]=false;//未成功拼成一条边             }    return false;}int main(){    int i,n,sum;    cin>>n;    while(n--)    {        cin>>m;        sum=0;        for(i=0;i<m;i++)        {            cin>>arr[i];            sum+=arr[i];        }        if(sum%4)        {            cout<<"no"<<endl;            continue;        }        t=sum/4;        memset(used,false,sizeof(used));//dev一直提示没有定义memset，但是oj可以通过         sort(arr,arr+m,cmp);        if(dfs(0,0,t))        {            cout<<"yes"<<endl;        }        else{            cout<<"no"<<endl;        }    }    return 0;}