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

题目大意：求矩阵的最大子矩阵。

思路： 实质是求 一维数组的最大子段和的问题。     将矩阵压缩成一维数组，求其最大字段和。

先附一段错误的最大子段和代码：

int fun(int* temp,int n){
	int maxx = -127;
	int sum = 0;
	for(int i = 0 ; i < n ; i ++){
		sum += temp[i];
		if(sum > 0)
			maxx = max(sum,maxx);
		else
			sum = temp[i];
	}
	return maxx;
}

这段代码错在，不论前面的段和是否为正，都无条件的加了当前元素(dp[n-1]+temp[i]) ,

如果dp[i-1] < 0 ，此时段和应该停止，不应该加到当前元素上。

正确fun函数：

int fun(int* temp, int n) {
	int maxx = -127;
	int sum = 0;
	for (int i = 0; i < n; i++) {
		if (sum > 0)
			sum += temp[i];
		else
			sum = temp[i];
		maxx = max(sum, maxx);
	}
	return maxx;
}

其中   sum = temp[i]， 当 前面的段小于0，舍去，从当前开始。

极端化思想，如果整个数组都为负数，那么子段就是每一个单独的元素。

关键点是段和为负就重新开始。

8 -9 -10

8+（-9） = -1 此时的段应该截止， 从-10开始又是一个新的段。

完整AC代码：

#include <iostream>
#include <cstdio>
#include <string>
#include <vector>
#include <algorithm>
#include <set>
#include <map>
#include <iterator>
#include <cmath>
#include <memory.h>
#include <cstdlib>
#include <ctime>
using namespace std;

#define _for(i,a,b) for(int i = (a) ; i < (b) ; i ++)
const int MAXN = 110;
int grade[MAXN][MAXN];
int ans;
int n;

//把矩阵压缩成一维数组 ， 转换成求最大子段和的问题。
int fun(int* temp) {
	int maxx = -127;
	int sum = 0;
	for (int i = 0; i < n; i++) {
		if (sum > 0)
			sum += temp[i];
		else
			sum = temp[i];
		maxx = max(sum, maxx);
	}
	return maxx;
}
int solve() {
        int temp[MAXN];            //保存压缩矩阵的数组
	int maxx = -127;
	for (int i = 0; i < n; i++) {
                
		memset(temp, 0, sizeof(temp));
		for (int j = i; j < n; j++) {

			for (int k = 0; k < n; k++)
				temp[k] += grade[j][k];
			int sum = fun(temp);

			if (sum > maxx)
				maxx = sum;
		}
	}
	return maxx;
}
int main() {

	memset(grade, 0, sizeof(grade));
	cin >> n;
	_for(i, 0, n) {
		_for(j, 0, n)
			cin >> grade[i][j];
	}
	int ans = solve();
	cout << ans << endl;
}