poj 1050 最大的子矩阵和



To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 44665		Accepted: 23659

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

Source

Greater New York 2001

题意：求最大的子矩阵的和；

思路：首先就是用一种很笨的办法做的，就是说先算出矩阵中各个点到（1，1）的构成矩形的和，然后用枚举法，枚举法表示起点到终点的的所有和，，找到最大的；

 sum[i][j] = sum[i-1][j]+sum[i][j-1] - sum[i-1][j-1]+ gra[i][j];表示矩阵中各个点到（1,1）构成的矩阵和，用sum[k][s] - sum[k-i][s]-sum[k][s-j] + sum[k-i][s-j])表示起点到终点构成的矩阵和；

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>

using namespace std;

int gra[200][200];
int sum[200][200];
int n;

int main() {
    scanf("%d",&n);
    memset(sum,0,sizeof(sum));
    int d = 0;
    int m = 0;
    for(int i = 1; i <= n ; i++) {
        sum[1][i] = d + gra[1][i];
        d = sum[1][i];
    }
    d = 0;
    for(int i = 1; i <= n ; i++) {
        sum[i][1] = d + gra[i][1];
        d = sum[i][1];
    }
    for(int i = 1; i <= n ; i++) {
        for(int j = 1; j <= n; j++) {
            sum[i][j] = sum[i-1][j]+sum[i][j-1] - sum[i-1][j-1]+ gra[i][j];
            m = max(sum[i][j],m);
        }
    }
    for(int i = 1; i <= n ; i++) {
        for(int j = 1; j <= n; j++) {
            for(int k = i; k <= n; k++) {
                for(int s = j; s <= n; s++) {
                     m = max((sum[k][s] - sum[k-i][s]-sum[k][s-j] + sum[k-i][s-j]),m);
                }
            }
        }
    }
    printf("%d\n",m);
    return 0;
}

写的好烦的代码，时间复杂度也高，并且感觉到思路自己也想不通，下面介绍较为简洁的代码，思路就是上升子序列的求法，把每一列的和都加起来写到最下面一行上，然后枚举起始行标和终了行标，在枚举列标，用sum[k][j] - sum[i][j];表示列构成的和，然后形成的序列求连续上升子序列的和的办法

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int sum[200][200];
int n,m;

int main() {
    memset(sum,0,sizeof(sum));
    scanf("%d",&n);
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= n; j++) {
            scanf("%d",&sum[i][j]);
            sum[i][j] += sum[i-1][j];
        }
    }
   int m = -2222;
    for(int i = 1; i <= n; i++) {
        for(int k = i; k <= n; k++) {
            int d = 0;
            int s = -2222;
            for(int j = 1; j <= n; j++) {
                 d += sum[k][j] - sum[i][j];
                 s = max(d,s);
                 if(d < 0){
                    d = 0;
                 }
            }
            m = max(m,s);
        }
    }
      printf("%d\n",m);
        return 0;
    }

现在感觉还是不能很好地做题的感觉，不过比以前好多了，不懂得还可以问问别人，每个人都一定要向阳生长的，相信自己，加油哦！