POJ 1050 To the Max

To the Max

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 39059 Accepted: 20630

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

题目大意： 在一个矩阵里找一个子矩阵，使子矩阵的和最大。

基本思路是将二维数组转化为一维数组，然后就与一维数组求最大值的方法一样。

转化为一维的思路是：输入数组的每一列从i行到j行化为tmp[k][i][j]，k为当前列。

#include<stdio.h>#include<string.h>int n,a[105][105],tmp[105][105][105]; //a[][]为输入，tmp[k][i][j]中保存第k列，从i行到j行的综合。void Set()    //初始化tmp{    memset(tmp,0,sizeof(tmp));    int i,j,k;    for(i=1; i<=n; i++)        for(j=1; j<=n; j++)            scanf("%d",&a[i][j]);    for(k=1; k<=n; k++)        for(i=1; i<=n; i++)            for(j=i; j<=n; j++)                for(int p=i; p<=j; p++)                    tmp[k][i][j]+=a[p][k];}int main(){    while(~scanf("%d",&n))    {        Set();        int sum,_max=0,i,j,k;//此题不存在每个数都为负的情况。        for(i=1; i<=n; i++)            for(j=i; j<=n; j++)            {                sum=0;    //每计算一组从i行到j行的矩形都要给sum先清零。                for(k=1; k<=n; k++) //同一维数组的寻找连续字串的和。                {                    sum+=tmp[k][i][j];                    if(sum<0) sum=0;                    if(sum>_max) _max=sum;                }            }        printf("%d\n",_max);    }    return 0;}