UVa 10360 - Rat Attack

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题目：在一个矩阵中，寻找一个元素加和最大的子矩阵。

分析：简单题、暴力枚举。如果直接暴力时间复杂度为O (t*n^2*d^2)，所以要利用dp做预处理。矩形用左上和右下两个顶点描述，即{(x1,y1),(x2,y2)}。设f(i,j)为矩形{(0,0),(i,j)}的元素之和，则矩形{(x1,y1),(x2,y2)}的元素和为f(x2,y2)-f(x1,y2)-f(x2,y1)+f(x1,y1)。时间复杂度为O(t*n^2)。

注意：枚举的范围是 [0,max_x] 和 [0,max_y]。

#include <stdio.h>#include <stdlib.h>#include <string.h>int rat[ 1030 ][ 1030 ];int sum[ 1030 ][ 1030 ];int main(){int t,d,n,x,y,r; while ( scanf("%d",&t) != EOF )while ( t -- ) {scanf("%d%d",&d,&n);memset( rat, 0, sizeof( rat ) );int min_x = 1025,max_x = 1;int min_y = 1025,max_y = 1;for ( int i = 0 ; i < n ; ++ i ) {scanf("%d%d%d",&x,&y,&r);++ x; ++ y;//范围[1,1025])rat[ x ][ y ] = r;if ( x < min_x ) min_x = x;if ( x > max_x ) max_x = x;if ( y < min_y ) min_y = y;if ( y > max_y ) max_y = y;}//预处理memset( sum, 0, sizeof( sum ) );for ( int i = min_x ; i <= max_x ; ++ i ) {int sum_y = 0;for ( int j = min_y ; j <= max_y ; ++ j ) {sum_y += rat[ i ][ j ];sum[ i ][ j ] = sum_y;if ( i > min_x )sum[ i ][ j ] += sum[ i-1 ][ j ];}}int max = 0;int xxx = 0;int yyy = 0;for ( int i = 0 ; i <= max_x ; ++ i ) for ( int j = 0 ; j <= max_y ; ++ j ) {int lu_x = i-d-1,rd_x = i+d;int lu_y = j-d-1,rd_y = j+d;if ( lu_x < 0 ) lu_x = 0;if ( lu_y < 0 ) lu_y = 0;if ( rd_x > max_x ) rd_x = max_x;if ( rd_y > max_y ) rd_y = max_y;int now = sum[ rd_x ][ rd_y ] - sum[ rd_x ][ lu_y ]- sum[ lu_x ][ rd_y ]+ sum[ lu_x ][ lu_y ];if ( max < now ) {max = now;xxx = i-1;yyy = j-1;}}printf("%d %d %d\n",xxx,yyy,max);}return 0;}