ZOJ 3212 K-Nice（满足某个要求的矩阵构造）

H - K-Nice

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Submit Status Practice ZOJ 3212

Description

This is a super simple problem. The description is simple, the solution is simple. If you believe so, just read it on. Or if you don't, just pretend that you can't see this one.

We say an element is inside a matrix if it has four neighboring elements in the matrix (Those at the corner have two and on the edge have three). An element inside a matrix is called "nice" when its value equals the sum of its four neighbors. A matrix is called "k-nice" if and only if k of the elements inside the matrix are "nice".

Now given the size of the matrix and the value of k, you are to output any one of the "k-nice" matrix of the given size. It is guaranteed that there is always a solution to every test case.

Input

The first line of the input contains an integer T (1 <= T <= 8500) followed by T test cases. Each case contains three integers n, m, k (2 <= n, m <= 15, 0 <= k <= (n - 2) * (m - 2)) indicating the matrix size n * m and it the "nice"-degree k.

Output

For each test case, output a matrix with n lines each containing m elements separated by a space (no extra space at the end of the line). The absolute value of the elements in the matrix should not be greater than 10000.

Sample Input

2
4 5 3
5 5 3

Sample Output

2 1 3 1 1
4 8 2 6 1
1 1 9 2 9
2 2 4 4 3
0 1 2 3 0
0 4 5 6 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0

题目意思：
构造出一个含有k个nice点的n*m的矩阵
nice点：周围四个数字之和等于该数字

分析：
不在边缘和角处构造
这样简单一点
先全部初始化-1
0的点其周围4个全为0
这样构造

之和将等于-1的点替换成1

这样就构造完毕

 

#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
#define eps 10e-6
#define INF 999999999
#define max_v 25
int a[max_v][max_v];
using namespace std;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m,k;
        scanf("%d %d %d",&n,&m,&k);
        memset(a,-1,sizeof(a));
        for(int i=1;i<n-1;i++)
        {
            for(int j=1;j<m-1;j++)
            {
                if(k<=0)
                    break;
                if(a[i][j]<=0)
                {
                    k--;
                    a[i][j]=0;
                    a[i+1][j]=0;
                    a[i-1][j]=0;
                    a[i][j+1]=0;
                    a[i][j-1]=0;
                }
            }
            if(k<=0)
                break;
        }
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(a[i][j]<0)
                    a[i][j]=1;
            }
        }
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(j==0)
                    printf("%d",a[i][j]);
                else
                    printf(" %d",a[i][j]);
            }
            printf("\n");
        }
    }
    return 0;
}
/*

题目意思：
构造出一个含有k个nice点的n*m的矩阵
nice点：周围四个数字之和等于该数字

分析：
不在边缘和角处构造
这样简单一点
先全部初始化-1
0的点其周围4个全为0
这样构造

之和将等于-1的点替换成1

这样就是构造完毕

*/