sgu 168 分类： sgu 2015-03-08 18:44 37人阅读 评论(0) 收藏

You are given N*M matrix A. You are to find such matrix B, that B[i,j]=min{ A[x,y] : (y>=j) and (x>=i+j-y) }

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动态规划，枚举对角线，
因为是取最值，
所以取值区域可以重叠，

就可以得出递推式。

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#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<ctime>
#include<iostream>
#include<algorithm>

const int MAXN = 1005;

int n , m;
int A[MAXN][MAXN] = {0};
int B[MAXN][MAXN] = {0};

using namespace std;

int main()
{
#ifndef ONLINE_JUDGE    
    freopen("sgu168.in","r",stdin);
    freopen("sgu168.out","w",stdout);
#endif

    scanf("%d%d",&n,&m);

    for(int i = 1;i <=n; i++)
      for(int j = 1; j <= m; j++)
        {
        scanf("%d",&A[i][j]);
         B[i][j] = A[i][j];
        }

    for(int L = n + m ; L > 1 ; L-- )
      for(int j = min(m , L - 1),lmt = max(L - n ,1); j >= lmt ; j--)
    {
        int i = L - j;

//      B[i][j] = min(B[i-1][j+1] ,B[i+1][j])

        if(i - 1 >0 && j + 1 <= m)
            B[i][j] = min(B[i][j],B[i-1][j+1]);
        if(i + 1 <= n)  
            B[i][j] = min(B[i][j],B[i+1][j]);
        if(j < m) 
            B[i][j] = min(B[i][j],B[i][j+1]);   //1*2矩阵 
    }



    for(int i = 1;i <=n; i++)
    {
      for(int j = 1; j < m; j++)
        printf("%d ",B[i][j]);
      printf("%d\n",B[i][m]);
    }

#ifndef ONLINE_JUDGE
    fclose(stdin);  
    fclose(stdout);
#endif
    return 0;   
}

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