poj 1050 To the Max

To the Max

Time Limit: 1000MS Memory Limit: 10000KTotal 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

40 -2 -7 0 9 2 -6 2-4 1 -4  1 -18  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[N][N],sum[N][N],dp[N];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;}