hdu1081 To The Max

To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11685    Accepted Submission(s): 5649

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
这道题的意思是给一个矩阵，求助最大子矩阵和，很容易联想到一维的最大字段和，所以我们可以把二维的转化为一维的
用上一个辅助数组 s[i][j]表示从第一行到底i行，第j列的所有元素的和，那个第i行到第k行第j列的和就为s[k][j]-s[i-1][j];
用for循环遍历即可转化为一维的情况
下面是ac代码
#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <stack>#include <queue>#include <algorithm>using namespace std;#define inf 0x3f3f3f3fint a[105][105];int n;int s[105][105];//j列1行i行的和int b[105][105];//第i行到j行暂时保存这些行的情况int main(){    while(~scanf("%d",&n))    {        for(int i=1; i<=n; i++)            for(int j=1; j<=n; j++)                scanf("%d",&a[i][j]);        memset(s,0,sizeof(s));        for(int i=1; i<=n; i++)            for(int j=1; j<=n; j++)                s[i][j]=s[i-1][j]+a[i][j];        int mx=-1280000;        memset (b,0,sizeof(b));        for(int i=1; i<=n; i++)        {            for(int j=1; j<=n; j++)                for(int k=j; k<=n; k++)                {                    if(b[j][k]<0)                        b[j][k]=s[k][i]-s[j-1][i];                    else                        b[j][k]+=s[k][i]-s[j-1][i];                    if(mx<b[j][k])                        mx=b[j][k];                }        }        printf("%d\n",mx);    }    return 0;}