POJ 1050 To the Max

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 36422		Accepted: 19130

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

传说中经典的DP问题，昨天比赛时为了解一道看起来类似的题直接网搜了此题的结题报告，发现暴力转化为一维数组求最长子序列的思想果然高明（可是比赛那道题还是TLE了T^T)
二维数组转化以为数组举例：
    0 -2 -7 0
    9  2 -6 2
可以转化为：
    9 0 -13 2
也就是把纵向的一列求和，然后再求最大子序列和即为这个最大子矩阵的和

求最大子序列和用了一点DP思想，DP之前也没怎么写过，每次想DP总想着区间，这里的DP是用点来做的
用一个数组f[i]代表数组s中以s[i]结尾的最大子序列的和，这样得到了状态转移方程：
    f[i]=f[i-1]+f[i]>f[i]?f[i-1]+f[i]:f[i];

代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

long g[101][101];
long s[101];
long f[101];

long FindMax(int n)
{
    memset(f,0,sizeof(f));
    for(int i=1;i<=n;i++)
        f[i]=(f[i-1]+s[i]>s[i]?f[i-1]+s[i]:s[i]);
    long maxn=f[1];
    for(int i=2;i<=n;i++)
        if(f[i]>maxn)
            maxn=f[i];
    return maxn;
}

int main()
{
    int n;
    long ans,result;

    while(scanf("%d",&n)==1)
    {
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                scanf("%ld",&g[i][j]);
        for(int len=1;len<=n;len++)
        {
            for(int y=1;y<=n-len+1;y++)
            {
                for(int x=1;x<=n;x++)
                    s[x]=0;
                for(int x=1;x<=n;x++)
                    for(int t=y;t<y+len;t++)
                        s[x]+=g[x][t];
                result=FindMax(n);
                if(len==1&&y==1)
                    ans=result;
                else
                    if(ans<result)
                        ans=result;
            }
        }
        printf("%ld\n",ans);
    }

    return 0;
}