POJ 1050 To the Max

To the Max

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 49003 Accepted: 25928

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

思路：动态规划 最大子矩阵和  参见 点击打开链接

代码如下

#include<iostream>#include<stdio.h>#include<vector>#include<algorithm>#include<memory.h>using namespace std;int main() {   // freopen("input.txt","r",stdin);    int N;    cin>>N;    int mat[102][102];    int dp[102][102];    for(int i = 1; i <= N; i++) {        for(int j = 1; j <= N; j++) {            cin>>mat[i][j];        }    }    memset(dp,0,sizeof(dp));    for(int i = 1; i <= N; i++) {        for(int j = 1; j <= N; j++) {            dp[i][j] = dp[i-1][j] + dp[i][j-1] - dp[i-1][j-1]  + mat[i][j];        }    }    int res = 0;    for(int p = 1; p <= N; p++) {        for(int q = p; q <= N; q++) {            for(int r = 1; r <= N; r++) {                for(int s = r; s <= N; s++) {                    int rect = dp[q][s] - dp[q][r-1] - dp[p-1][s] + dp[p-1][r-1];                    if(rect > res)  res = rect;                }            }        }    }    cout<<res<<endl;    return 0;}