poj1050

To the Max

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

转化为一维数组的最大连续子串和。

#include<iostream>
#include<cstring>
using namespace std;
int line[101];
int matrix[101][101];
int n;

void input(void) {
    cin >> n;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            cin >> matrix[i][j];
}

int dp(void) {//一维数组求最大子串和
    int tmp = 0;
    int max = -(1 << 30);
    for (int i = 0; i < n; i++) {
        tmp += line[i];
        if (tmp > max)
            max = tmp;
        if (tmp < 0)
            tmp = 0;
    }
    return max;
}

int find(void) {
    int max = -(1 << 30);
    for (int i = 0; i < n; i++) {//i表示开始行
        memset(line, 0, sizeof (line)); //每次开始数组清零
        for (int j = i; j < n; j++) {//j表示结束行
            for (int k = 0; k < n; k++)//k表示列
                line[k] += matrix[j][k]; //将二维矩阵转化为一维数组
            int t = dp();
            if (t > max)
                max = t;
        }
    }
    return max;
}

int main(void) {
    input(); //读入数据
    cout << find() << endl; //输出答案
    return 0;
}