HDU 2589 正方形划分

Problem Description

一个边长为L的正方形可以分成 L*L个小正方形. 有N个石子放在 N个小正方形里，能否将它分成 N个 正方形，使得每个正方形里恰有一个石子且这N 个正方形恰好构成整个正方形 .

 

Input

输入数据首先包含一个整数T，表示测试实例的个数，然后是T组数据，每组第一行包含2个整数L,N,接下来有N行每行2个整数 r,c,表示第r行c列的小正方形里有一个石子 .1<L<=20;1<N<=L*L; 1<=r,c<=L.

 

Output

对于每个测试实例，如能将它分成 N个 正方形输出YES, 否则输出 NO

 

Sample Input

3

5 8
2 4
3 3
3 4
3 5
4 2
4 4
4 5
5 5

3 2
1 1
3 3

2 4
1 1
1 2
2 1
2 2

 

Sample Output

YES
NO
YES

 

没什么好说的，直接暴力的枚举每个正方形大小盖上去。

#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int sz=25;
int T,n,m,x,y,s;
int c[sz][sz];
bool f[sz][sz];

bool check(int x,int y,int r)
{
    return c[x+r][y+r]-c[x+r][y-1]-c[x-1][y+r]+c[x-1][y-1]==1;
}

bool dfs()
{
    if (!s) return true;
    for (int i=1;i<=n;i++)
    {
        for (int j=1;j<=n;j++)
        {
            if (f[i][j])
            {
                for (int k=0;i+k<=n&&j+k<=n;k++)
                {
                    if (check(i,j,k))
                    {
                        s-=k*k;
                        for (int r=0;r<=k;r++)
                        {
                            for (int w=0;w<=k;w++)
                            {
                                f[i+r][j+w]=false;
                            }
                        }
                        if (dfs()) return true;
                        s+=k*k;
                        for (int r=0;r<=k;r++)
                        {
                            for (int w=0;w<=k;w++)
                            {
                                f[i+r][j+w]=true;
                            }
                        }
                    }
                }
                return false;
            }
        }
    }
}

int main()
{
    scanf("%d",&T);
    while (T--)
    {
        scanf("%d%d",&n,&m);
        s=n*n;
        memset(c,0,sizeof(c));
        for (int i=1;i<=m;i++)
        {
            scanf("%d%d",&x,&y);
            c[x][y]=1;
        }
        for (int i=1;i<=n;i++)
        {
            for (int j=1;j<=n;j++)
            {
                c[i][j]+=c[i-1][j]+c[i][j-1]-c[i-1][j-1];
                f[i][j]=true;
            }
        }
        printf("%s\n",dfs()?"YES":"NO");
    }
    return 0;
}