poj 1050 To the Max

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 40812		Accepted: 21645

Description

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the
sub-rectangle with the largest sum is referred to as the maximal sub-rectangle.

As an example, the maximal sub-rectangle of the array:

0 -2 -7 0

9 2 -6 2

-4 1 -4 1

-1 8 0 -2

is in the lower left corner:

9 2

-4 1

-1 8

and has a sum of 15.

Input

The input consists of an N * N array of integers. The input begins with a single positive integer N on a line by itself, indicating the size of the square two-dimensional array. This is followed by N^2 integers separated by whitespace (spaces and newlines).
These are the N^2 integers of the array, presented in row-major order. That is, all numbers in the first row, left to right, then all numbers in the second row, left to right, etc. N may be as large as 100. The numbers in the array will be in the range [-127,127].
Output

Output the sum of the maximal sub-rectangle.
Sample Input

4
0 -2 -7 0 9 2 -6 2
-4 1 -4  1 -1

8  0 -2

Sample Output

15

Source

Greater New York 2001

题意：找最大的子矩阵，使其中的每项和最大。

思路：预处理求每列的和，然后枚举宽度为n的所有矩阵做dp。

ps：明天有比赛，刷刷题练练手吧

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int N=105;
const int inf=0x3f3f3f3f;
int n,ans;
int a

,sum

,dp
;
void init()
{
memset(sum,0,sizeof(sum));
scanf("%d",&n);
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++){
scanf("%d",&a[i][j]);
sum[i][j]=sum[i-1][j]+a[i][j];
}
}
void solve()
{
for(int i=1;i<=n;i++){
for(int j=i;j<=n;j++){
dp[0]=-inf;
for(int k=1;k<=n;k++){
dp[k]=max(dp[k-1]+sum[j][k]-sum[i-1][k],sum[j][k]-sum[i-1][k]);
ans=max(dp[k],ans);
}
}
}
printf("%d\n",ans);
}
int main()
{
init();
solve();
return 0;
}