To The Max
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6022 Accepted Submission(s): 2869
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.
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].
(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
Source
题目大意：给定一个含有正数和负数的矩阵，求其子矩阵的和的最大值。
一维的情况很简单，如何把一维的情况转化为二维情况呢？
例如，对于本题的测试数据：
我们可以每次任选几行，压缩成一行，这样就转化为了一维情况。
例如，我们求1~2行中的最大子矩阵：即矩阵高为2（1~2行），宽为1：4的矩阵，可以先把1~2行相加，得到9 0 -13 2，再求这个单行的最大子段，由此就可以求得1~2行的最大子矩阵。
#include<stdio.h>
#include<string.h>
int map[137][137],dp[137],n;
int dpp()
{
int i,sum=0,flag=-1000000;
for(i=0;i<n;i++)
{
sum+=dp[i];
if(sum<0)sum=0;
if(sum>flag)flag=sum;
}
return flag;
}
int main()
{
int i,j,k,l,max=0,t;
while(scanf("%d",&n)!=EOF)
{
for(i=0; i<n; i++)
for(j=0; j<n; j++)
scanf("%d",&map[i][j]);
int ans=-1;
for(i=0; i<n; i++)//从第几行开始
{
for(j=0; j<n; j++)//行数
{
if(i+j<n)
{
memset(dp,0,sizeof(dp));
for(k=i;k<=i+j;k++)//k和l构成一个列数为n的子矩阵
for(l=0;l<n;l++)
{
dp[l]+=map[k][l];
}
max=dpp();//从列数分别1到n的子矩阵中找出和的最大值
if(max>ans)ans=max;
}
}
}
printf("%d\n",ans);
}
return 0;
}