HDU

先上题目：

To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7177    Accepted Submission(s): 3470

Problem Description

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 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 x 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

 

 

　　二维版的求区间最值。

　　一维版地址：http://www.cnblogs.com/sineatos/p/3515309.html

　　如果是一维版的话状态转移方程：

　　dp[i]=w[i]          ans<=0

　　　　 =ans+w[i]   ans>0            其中ans代表当前位置i往前连续的一段的区域最值

　　而maxn=max(ans,maxn)           maxn代表总体最值

　　对于二维的时候，其实和一维大致相同。这里需要在一个方向上(假设是行数增加的方向)枚举区间，然后在另一个方向上(列数增加的方向)上进行想一维的时候那样的操作。这样做的涵义其实是在某一个确定的区间上，求这个区间上的最值。

　　详细需要看代码。

　　分析如图：

上代码：

#include <cstdio>
#include <cstring>
#define MAX 101
#define INF ~(1<<31)
using namespace std;

int s[MAX][MAX];
int h[MAX][MAX];
int dp[MAX][MAX];

int main()
{
    int n;
    int ans,maxn;
    while(scanf("%d",&n)!=EOF)
    {
        memset(s,0,sizeof(s));
        memset(h,0,sizeof(h));
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=n; j++)
            {
                scanf("%d",&s[i][j]);
                h[i][j]=h[i][j-1]+s[i][j];
            }
        }
        maxn=-INF;
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=i; j++)
            {
                ans=-1;
                for(int k=1; k<=n; k++)
                {
                    if(ans>0) ans+=h[k][i]-h[k][j-1];
                    else ans=h[k][i]-h[k][j-1];
                    maxn=ans>maxn ? ans : maxn;
                }
            }
        }
        printf("%d
",maxn);
    }
    return 0;
}