P2213 [USACO14MAR]懒惰的牛The Lazy Cow_Sliver

题目描述

It's a hot summer day, and Bessie the cow is feeling quite lazy. She wants

to locate herself at a position in her field so that she can reach as much

delicious grass as possible within only a short distance.

The field Bessie inhabits is described by an N by N grid of square cells

(1 <= N <= 400). The cell in row r and column c (1 <= r,c <= N) contains

G(r,c) units of grass (0 <= G(r,c) <= 1000). From her initial square in

the grid, Bessie is only willing to take up to K steps (0 <= K <= 2*N).

Each step she takes moves her to a cell that is directly north, south,

east, or west of her current location.

For example, suppose the grid is as follows, where (B) describes Bessie's

50    5     25*   6     17    
14    3*    2*    7*    21    
99*   10*   1*(B) 2*    80*    
8     7*    5*    23*   11   
10    0     78*   1     9        

initial position (here, in row 3, column 3):

If K=2, then Bessie can only reach the locations marked with *s.

Please help Bessie determine the maximum amount of grass she can reach, if

she chooses the best possible initial location in the grid.

奶牛贝西非常懒惰，她希望在她的地盘内找到一点最佳位置居住，以便在有限的步数内可以吃到尽量多的青草。

她的地盘是一个N行N列(1 <= N <= 400)的矩阵，第r行c列包含G(r,c)单位的青草(0 <= G(r,c) <= 1000)。从她的居住点，她最多愿意走K步(0 <= K <= 2*N)，每一步她可以找到上、下、左、右直接相邻的某个格子。

输入格式

* Line 1: The integers N and K.

* Lines 2..1+N: Line r+1 contains N integers describing row r of the

grid.

输出格式

* Line 1: The maximum amount of grass Bessie can reach, if she chooses

the best possible initial location (the location from which

she can reach the most grass).

输入输出样例

输入 #1

5 2
50 5 25 6 17
14 3 2 7 21
99 10 1 2 80
8 7 5 23 11
10 0 78 1 9

输出 #1

342

说明/提示

OUTPUT DETAILS:

In the example above, Bessie can reach 342 total units of grass if she

locates herself in the middle of the grid.

Source: USACO 2014 March Contest, Silver

代码

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int N=1010;

int n,m,k,x,y,xl,yl,xr,yr;
int ans=-0x7fffffff,a[N][N],b[N*2][N*2];

int main () {
	scanf("%d%d",&n,&k);
	for(int i=1; i<=n; i++)
		for(int j=1; j<=n; j++)
			scanf("%d",&a[i][j]);
	m=n*2-1;
	for(int i=1; i<=n; i++)
		for(int j=1; j<=n; j++)
			b[i+j-1][n-i+j]=a[i][j];
	for(int i=1; i<=m; i++)
		for(int j=1; j<=m; j++)
			b[i][j]+=b[i][j-1];
	for(int i=1; i<=m; i++)
		for(int j=1; j<=m; j++)
			b[i][j]+=b[i-1][j];
	for(int i=1; i<=n; i++)
		for(int j=1; j<=n; j++) {
			x=i+j-1,y=n-i+j;
			xl=x-k,yl=y-k,xr=x+k,yr=y+k;
			if(xl<1)
				xl=1;
			if(yl<1)
				yl=1;
			if(xr>m)
				xr=m;
			if(yr>m)
				yr=m;
			ans=max(ans,b[xr][yr]-b[xr][yl-1]-b[xl-1][yr]+b[xl-1][yl-1]);
		}
	printf("%d\n",ans);
	return 0;
}