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To the Max_DP

2024-08-11 07:47:34   219人阅读

Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 46858 Accepted: 24819

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

 

 

 

 

【题意】 给出一个n*n的方阵，选出一个子矩阵使得里面的值相加最大

【思路】先不管上下，先看左右，找出值最大的j列到k列，然后确定了以后，就好似成了一列，对这列上下求最大子段和

dp[i][j][k]=max(dp[i-1][j][k]+tmp,tmp);tmp=第i行j到k列的值之和

#include<iostream>#include<string.h>#include<stdio.h>using namespace std;const int inf=0x7777777;const int N=107;int a[N][N];int dp[N][N][N];int main(){    int n;    while(~scanf("%d",&n))    {        memset(dp,0,sizeof(dp));        for(int i=1; i<=n; i++)        {            for(int j=1; j<=n; j++)            {                scanf("%d",&a[i][j]);            }        }        int ans=-inf;        for(int i=1; i<=n; i++)        {            for(int j=1; j<=n; j++)            {                int tmp=0;                for(int k=j; k<=n; k++)                {                    tmp+=a[i][k];                    dp[i][j][k]=max(dp[i-1][j][k]+tmp,tmp);                    if(dp[i][j][k]>ans)                        ans=dp[i][j][k];                }            }        }        printf("%d\n",ans);    }    return 0;}

 

To the Max_DP


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