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  • hdu 1559 暴力

    这数据量和这时限,很显然暴力是最好的解法,预处理前缀和,n方复杂度秒过。

     1 #include <iostream>
     2 #include <cstring>
     3 #include <cstdio>
     4 using namespace std;
     5 
     6 const int N = 1001;
     7 int a[N][N];
     8 int m, n, x, y;
     9 
    10 int get( int i, int j )
    11 {
    12     int x1 = a[i + x - 1][j + y - 1];
    13     int x2 = a[i + x - 1][j - 1];
    14     int x3 = a[i - 1][j + y - 1];
    15     int x4 = a[i - 1][j - 1];
    16     return x1 - x2 - x3 + x4;
    17 }
    18 
    19 int main ()
    20 {
    21     int t;
    22     scanf("%d", &t);
    23     while ( t-- )
    24     {
    25         scanf("%d%d%d%d", &m, &n, &x, &y);
    26         for ( int i = 1; i <= m; i++ )
    27             for ( int j = 1; j <= n; j++ )
    28                 scanf("%d", &a[i][j]);
    29         for ( int i = 1; i <= m; i++ )
    30             for ( int j = 1; j <= n; j++ )
    31                 a[i][j] += a[i][j - 1];
    32         for ( int i = 1; i <= m; i++ )
    33             for ( int j = 1; j <= n; j++ )
    34                 a[i][j] += a[i - 1][j];
    35         int ans = -999999999;
    36         for ( int i = 1; i + x - 1 <= m; i++ )
    37             for ( int j = 1; j + y - 1 <= n; j++ )
    38                 ans = max( ans, get( i, j ) );
    39         printf("%d
    ", ans);
    40     }
    41     return 0;
    42 }
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  • 原文地址:https://www.cnblogs.com/huoxiayu/p/4472427.html
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