poj 1050(矩阵求和问题dp)

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 44765		Accepted: 23700

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

 

思路：将其化简成子段求和问题：将第i行到第j行的数累加成一维数列，在用子段求和求解。

　　　

#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3f
#define zero_(x,y) memset(x , y , sizeof(x))
#define zero(x) memset(x , 0 , sizeof(x))
#define MAX(x) memset(x , 0x3f ,sizeof(x))
using namespace std ;
#define N 505
typedef long long LL ;
LL dp[N],a[N][N],s[N][N];
int main(){
    //freopen("in.txt","r",stdin);
    int n;
    scanf("%d",&n);
    zero(dp);zero(a);zero(s);
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            scanf("%I64d",&a[i][j]);
        }
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            s[j][i]=s[j-1][i]+a[j][i];
           
        }
    }
   
    LL sum=0,b=0;
    for(int i=0;i<=n;i++){
        for(int j=i+1;j<=n;j++){
                b=0;
            for(int k=1;k<=n;k++){
                
                if(b>0)
                    b+=(s[j][k]-s[i][k]);
                else
                    b=s[j][k]-s[i][k];
                if(b>sum)
                    sum=b;
                
            }
        }
    }
    printf("%I64d
",sum);
    return 0;
}