NOJ1538——[1538] K-Submartix

• 问题描述
• You are given a matrix consisting of some integer, its size is n*m.
  You should find a minimum area submatrix that the sum of the submatrix (sum of all elements in the submatrix) is not less k.
  
• 输入
• The input contains multiple test cases. Each test case begins with n, m (1 <= n, m <= 250) and k (-10^9 <= k <= 10^9) on line. Next n lines contain m integer each matrix . The input ends once EOF is met.
• 输出
• Print a single integer -- the area of the minimum obtained submatrix. If we cannot obtain a matrix of sum is not less k, print -1.
• 样例输入

• 4 4 10
  1 2 3 4
  5 6 7 8
  9 10 11 12
  13 14 15 16
  1 3 15
  10 -5 10
  2 2 10
  1 2
  2 3

• 样例输出

• 1
  3
  -1

• 提示

• 无

• 来源

• wk@NJUPT

   这题就是让你求一个满足矩阵元素和大于等于k，然后面积最小的矩阵，输出最小面积，如果不存在，就输出-1，

我的思路是，想办法把二维的问题化成一维。所以我需要一个数组sum[i][j]，表示第j列从第1行到第i行的元素和。

之后我们枚举矩阵的上下界，利用前缀和的思想，另开一维数组array[]，存的就是前i列的和，接下来就是去判断，然后求最小值了。

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>

using namespace std;

const int maxn=270;

int sum[maxn][maxn];
int array[maxn];
int mat[maxn][maxn];
int main()
{
  int n,m,k;
  while(~scanf("%d%d%d",&n,&m,&k))
  {
    memset(sum,0,sizeof(sum));
    array[0]=0;
    int area=0x3f3f3f3f;
    for(int i=1;i<=n;i++)
      for(int j=1;j<=m;j++)
      {
        scanf("%d",&mat[i][j]);
        sum[i][j]=sum[i-1][j]+mat[i][j];
      }
    for(int up=1;up<=n;up++)
      for(int down=up;down<=n;down++)
      {
        for(int i=1;i<=m;i++)
        {
          array[i]=array[i-1]+sum[down][i]-sum[up-1][i];
          if(array[i]<k)//也算是个剪枝了，之前没写直接TLE
            continue;
          for(int j=i-1;j>=0;j--)
            if(array[i]-array[j]>=k)
            {
              area=min(area,(i-j)*(down-up+1));
              break;
            } 
        }
      }   
    if(area==0x3f3f3f3f)
      printf("-1\n");
    else
      printf("%d\n",area );
  }
  return 0;
}