To the Max(DP)

To the Max
Time Limit: 1000MS
 
Memory Limit: 10000K
Total Submissions: 40625
 
Accepted: 21525
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

4
0 -2 -7 0 
9 2 -6 2
-4 1 -4  1 
-1 8  0 -2

Sample Output

15

 
제목:
N(1~100)을 주고 NXN이 있는 행렬을 나타낸다. 그 다음에 이 NXN의 행렬을 주고 이 행렬의 최대 하위 행렬과 출력한다.
 
아이디어:
      DP.최대 하위 섹션의 합은 dp[i] = max {dp[i-1] + num[i],num[i]}이며, dp[i]는 i까지 최대 하위 섹션의 합을 나타냅니다.
행렬에 대해 매가 거행하는 처음과 끝의 위치를 농축한 다음에 열을 총화하고 최대 필드와 최대치를 한 번 걸어가면 된다.
 
       AC:

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int INF = -100000000;
int num[105][105], sum[105][105];
int dp[105], ans[105];
int n;

int solve() {
    memset(dp, 0, sizeof(dp));

    int Max = INF;
    for (int i = 1; i <= n; ++i) {
        dp[i] = max(ans[i], dp[i - 1] + ans[i]);
        Max = max(Max, dp[i]);
    }

    return Max;
}

int main() {

    scanf("%d", &n);
    memset(sum, 0, sizeof(sum));
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            scanf("%d", &num[i][j]);
            sum[i][j] = sum[i - 1][j] + num[i][j];
        }
    }

    int Max = INF;
    for (int i = 1; i <= n; ++i) {
        for (int j = i; j <= n; ++j) {

            for (int k = 1; k <= n; ++k) {
                ans[k] = sum[j][k] - sum[i - 1][k];
            }

            Max = max(Max, solve());
        }
    }

    printf("%d
", Max);

    return 0;
}