PKU1050：To the Max

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 21624		Accepted: 11192

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

40 -2 -7 0 9 2 -6 2-4 1 -4  1 -18  0 -2

Sample Output

15

原题地址：http://acm.pku.edu.cn/JudgeOnline/problem?id=1050

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题解

DP， 最大子字段和，最大子矩阵和。

方程不难，只是没遇到过。

  

1、最大子字段和

题目描述：给定n个整数（可能为负数）组成的序列a[1],a[2],a[3],…,a[n],求该序列如a+a+…+a[j]的子段和的最大值。当所给的整均为负数时定义子段和为0，依此定义，所求的最优值为Max{0,a+a+…+a[j]},1<=i<=j<=n

题解：记b[j]=max(a+..+a[j-1]+a[j]),其中1<=i<=j,并且1<=j<=n。则所求的最大子段和为max b[j]，1<=j<=n。

         由b[j]的定义可易知，当b[j-1]>0时b[j]=b[j-1]+a[j]，否则b[j]=a[j]。故b[j]的动态规划递归式为:b[j]=max(b[j-1]+a[j],a[j])，1<=j<=n。

  

实际上本题要求的是子矩阵的和，可以由字段和推广得到

 

2、最大子矩阵和

矩阵是二维的，将其压缩成一维就可以用上面的方法求出最大子矩阵和了

B[I][J]表示第I列，自第一行加到第J行的和。那么B[I][J]-B[I][K]就是第I列中第K行加到第J行的和。

求第K行到第J行之间的最大子矩阵就只要对所有的A[I]=B[I][J]-B[I][K](1<=I<=N)做遍最大字段和就可以了。

  

DP，共三层循环。

最外层i循环1-M，表明子矩阵是从第i行开始累加的。

第二层j循环i-M，表明子矩阵是从第i行累加到第j行。

第三层k从1到N做一维DP。

 

 

代码：

 

#include<stdio.h>
 int compress[101][101],matrix[101][101],dp[101];
 int i,j,k,n,max;
 int main()
{
    scanf("%d",&n);
    for (i=1;i<=n;i++)
        for (j=1;j<=n;j++)
            scanf("%d",&matrix[i][j]);
    for (i=1;i<=n;i++)
        compress[1][i]=matrix[1][i];
    for (i=2;i<=n;i++)
        for (j=1;j<=n;j++)
            compress[i][j]=compress[i-1][j]+matrix[i][j];
    max=0;
    for (i=1;i<=n;i++)
        for (j=i;j<=n;j++)
        {
            dp[1]=compress[j][1]-compress[i-1][1];
            for (k=2;k<=n;k++)
            {
                if (dp[k-1]>0) dp[k]=dp[k-1]+compress[j][k]-compress[i-1][k];
                else dp[k]=compress[j][k]-compress[i-1][k];
                if (dp[k]>max) max=dp[k];
            }
        }
    printf("%d\n",max);
    return 0;
}