HDU 1081 To The Max （dp）

题目链接

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

题目分析：

一个N*N的矩阵中的，寻找最大的子矩阵。

主要应用到矩阵压缩的思想，如果某个子矩阵的和最大，我们可以把他们的和压缩为一行，则此时的连续子序列和胃最大的。检查所有的压缩组合，从第一行开始，检查所有的包含此行的往下的行的子矩阵，找出包含此行的子矩阵的最大值，接着检查包含下一行的往下的所有的子矩阵。

代码：

    #include<iostream >
    #include<stdio.h>
    #include<string.h>
    using namespace std;
    int main()
    {
        int jz[141][141],dp[141],sum[141],Max;
        int N,i,j;
        while(~scanf("%d",&N))
        {
            Max=-1333;      ///Max的值每次都要刷新
            for( i=0; i<N; i++)
                for( j=0; j<N; j++)
                    scanf("%d",&jz[i][j]);///给矩阵赋值
            for(int k=0; k<N; k++)///循环每行都要作为一个起始行
            {
                memset(dp,0,sizeof(dp));///dp数组刷新
                memset(sum,0,sizeof(sum));///sum数组刷新
                for(int i=k; i<N; i++)///从当前行开始循环
                {
                    for(int j=0; j<N; j++)///每一列都要考虑
                        sum[j]+=jz[i][j];///以前到该列的子矩阵加上该列的
                    dp[0]=sum[0];
                    for(int h=1; h<N; h++)
                    {
                        dp[h]=max(dp[h-1]+sum[h],sum[h]);///以前的和现在的取大值
                        if(dp[h]>Max)
                            Max=dp[h];
                    }
                }
            }
            printf("%d
",Max);
        }
    }