hdu 1081 To The Max 基础dp

To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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
分析：通过转换成最大字段和
第一步:dp[i][j]=a[1][j]+ …… +a[i][j]
第二步:sum+=dp[j][k]-dp[k][i-1];
第三步:求max和sum的较大值
代码：
#include<stdio.h>#include<string.h>int a[102][102],dp[102][102];int main(){    int i,j,k,n,max,sum;    while(scanf("%d",&n)!=EOF)    {        memset(dp,0,sizeof(dp));        for(i=1;i<=n;i++)         for(j=1;j<=n;j++)         {             scanf("%d",&a[i][j]);             dp[i][j]=dp[i][j-1]+a[i][j];         }         int max=-9999;         for(i=1;i<=n;i++) //控制行          for(j=i;j<=n;j++) //控制列          {              int sum=0;              for(k=1;k<=n;k++) //控制连续的层数              {                  sum=sum>0?sum:0;                  sum+=dp[k][j]-dp[k][i-1];                  max=max>sum?max:sum;              }          }          printf("%d\n",max);    }    return 0;}