poj 1050 To the Max 【矩阵压缩 】

To the Max

Time Limit: 1000MS

 

Memory Limit: 10000K

Total Submissions: 42327

 

Accepted: 22514

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

压缩矩阵：将问题转化为   一维数组求连续子列的最大和问题。

#include<stdio.h>#include<string.h>#define MAX 100+10int map[MAX][MAX];int dp[MAX][MAX];int max(int x,int y){    return x>y?x:y;}int main(){    int n,k,i,j;    int sum,s;//sum记录每次所遍历矩阵的最大和,s为中间变量     int maxsum;//记录最大矩阵和     while(scanf("%d",&n)!=EOF)    {        memset(dp,0,sizeof(dp));        for(i=1;i<=n;i++)        {            for(j=1;j<=n;j++)            {                scanf("%d",&map[i][j]);                if(i==1)//压缩成一维                 dp[i][j]=map[i][j];                else                dp[i][j]=dp[i-1][j]+map[i][j];            }        }        maxsum=-200;        for(i=1;i<=n;i++)//起始         {            for(j=i;j<=n;j++)//末尾            {                sum=dp[j][1]-dp[i-1][1];                s=0;                for(k=1;k<=n;k++)                {                    s+=dp[j][k]-dp[i-1][k];//和一维数组求法一样                     if(s>sum)                    sum=s;                    if(s<0)                    s=0;                }                maxsum=max(sum,maxsum);            }         }        printf("%d\n",maxsum);    }    return 0;}