HDU 1081 To The Max

To The Max

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

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

Sample Output

15

题意：给出一个二维的矩阵，求和最大的子矩阵的大小。

可以用区间DP的思想来做，注意，如果全是负值，就可以认为取0X0是最大的矩阵，和为0.这样其实就简便了一些，如果一个子矩阵的和为负的，直接就不取了。

一直分区间，把每个区间内的和都求出来与MAX比较。

#include<stdio.h>#include<string.h>int a[101][101],b[101];int n;int dp(){    int i;    int max=0,sum=0;    for(i=1;i<=n;i++)    {        sum+=b[i];        if(sum<0)        sum=0;        if(sum>max)        max=sum;    }    return max;}int main(){    int i,j,k;    while(~scanf("%d",&n))    {        memset(a,0,sizeof(a));        for(i=1; i<=n; i++)            for(j=1; j<=n; j++)                scanf("%d",&a[i][j]);        int sum=0,max=0;        for(i=1; i<=n; i++)        {            memset(b,0,sizeof(b));            for(j=i;j<=n;j++)            {                for(k=1;k<=n;k++)                {                    b[k]+=a[j][k];                }                sum=dp();                if(sum>max)                max=sum;            }        }        printf("%d\n",max);    }    return 0;}