POJ_1050_To the Max

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 49811		Accepted: 26400

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

 

• 最大子矩阵和
• 是一维的最大子串和的二维扩展
• 那我们把每列做一个前缀和，O1得到从i行到j行单列的和，之后用最大子串和dp求解就行

 

#include <iostream>
#include <string>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <climits>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
typedef long long           LL ;
typedef unsigned long long ULL ;
const int    maxn = 1e2 + 10   ;
const int    inf  = 0x3f3f3f3f ;
const int    npos = -1         ;
const int    mod  = 1e9 + 7    ;
const int    mxx  = 100 + 5    ;
const double eps  = 1e-6       ;
const double PI   = acos(-1.0) ;

int n, ans, a[maxn][maxn], b[maxn], c[maxn][maxn];
int main(){
    // freopen("in.txt","r",stdin);
    // freopen("out.txt","w",stdout);
    while(~scanf("%d",&n)){
        ans=-inf;
        memset(c,0,sizeof(c));
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++){
                scanf("%d",&a[i][j]);
                c[i][j]=c[i-1][j]+a[i][j];
            }
        for(int i=1;i<=n;i++){
            for(int j=i;j<=n;j++){
                b[0]=0;
                for(int k=1;k<=n;k++){
                    if(b[k-1]>=0){
                        b[k]=b[k-1]+c[j][k]-c[i-1][k];
                    }else{
                        b[k]=c[j][k]-c[i-1][k];
                    }
                    ans=max(ans,b[k]);
                }
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}