poj 1050 To the Max & uva 108

原题：
To the Max
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 45039 Accepted: 23848
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
大意：
经典的最大子矩阵和问题，给你一个n*n的矩阵，问你和最大的子矩阵是多少？

#include <bits/stdc++.h>using namespace std;int table[102][102];int sum[102][102];int tmp[102];int n;int SubSeq(int a[])//最大子段和{    int ans = a[1];    int s = a[1];    for(int i = 2 ;i <= n ;i++)    {        if( s > 0)            s += a[i];        else            s = a[i];        ans = max(ans , s);    }    return ans;}int solve(){    memset(sum , 0 ,sizeof(sum));    int ans = INT_MIN;    for(int i = n ;i >= 1; i--)    {        for(int j = n ;j >= 1 ;j--)            sum[i][j] = sum[i+1][j] + table[i][j];    }    for(int i = n ;i >= 1 ;i--)    {        for(int j  = n+1 ;j > i ;j--)        {            for(int k = 1 ;k <= n ;k++)                tmp[k] = sum[i][k] - sum[j][k];                ans = max(ans , SubSeq(tmp));        }    }    return ans;}int main(){    ios::sync_with_stdio(false);    while(cin>>n)    {        for(int i = 1 ;i <= n ;i++)        {            for(int j = 1 ;j <=n ;j++)                cin>>table[i][j];        }        cout<<solve()<<endl;    }    return 0;}

解答：
本来想在小白书上随便找个贪心的题目练练手，结果找了一道做过的题目，权当是复习了。
最大子矩阵问题也不知道算不算是动态规划问题，思路参考最大子段和问题。可以这样想，给你一个n行n列的矩阵，现在问你固定列数为n时候的最大子矩阵是多少？。可以把这个n行n列的矩阵压缩成一个一维数组，也就是把列数全部相加，然后对这个数组计算最大子段和就是行数和列数分别为n的时候的最大矩阵和值，那么设置两个下标i，j代表第i行和第j行之间的矩阵，枚举i和j的值，把它们的对应列加一起算最大子段和，每次更新答案即可。