SDNU——8题——E

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

 #include<cstdio> #include<vector> #include<iostream> #include<memory.h> #include<cmath> using namespace std; int findmax(int a[],int n) { int b = 0,ret=-999; int i; for(i = 1;i <= n;i++) { if(b > 0)//判断的其实是上一次输入的状态//如果上一次的值是小于零的，那不如舍弃//就算上一次7+-1小了，但是max还是保留的最大的情况  { b += a[i];}else{ b = a[i];}ret=max(ret,b);  } return ret;  } int main() { int i,j,k,n,now=0; int v[105][105]; int ret[100];  int cjc=-999; while(cin>>n) { memset(v,0,sizeof(v)); for(i=1;i<=n;i++) {  for(j=1;j<=n;j++) { cin>>v[i][j]; }}for(i=1;i<=n;i++)//行的起始点{memset(ret,0,sizeof(ret));for(j=i;j<=n;j++)//以i为起始点的行开始，遍历所有可能的矩阵的和{for(k=1;k<=n;k++)//列的变化{ret[k] += v[j][k];}now=findmax(ret,k);cjc=max(cjc,now);  }  } cout<<cjc<<endl;cjc=-999;  } return 0; }

可能还是我见的世面太少了，我觉得这个题也非常的棒啊

暴力，遍历所有的矩形

固定行动列，固定列动行都行，只不过后者难以实现一些，在脑子里难以实化

首先循环起始行标——【1，n】

然后计算以当前行标为首的所有子矩阵的最优max【i，n】

让后循环列~~每一行都要循环列【1，n】

我就问了————那这样没法遍历到列在中间的情况啊，你这列都是从第一列开始的，不慌，计算的时候，处理的非常的精彩，真的让我收益颇多啊~~

以上——就可以遍历所有子矩形了~~

然后开始求单个矩形的和，这单个矩形是指行高为J，列宽固定为1的矩形，这样求最优和的时候，才能方便的得知，最优解是要几列

ret[k] += v[j][k];

注意ret中下标为k

让后开始findmax，b算是上一个单个矩形的最优和，如果它小于零，证明前面矩形求得和加起来是负数，只会，让下面的求和更小，不如舍去前面所有矩形不加，然后，起始点更新为当前单个矩形，如果大于零能，那就继续求和，不管后面单个矩形和为正为负，反正这个过程 的最优解已经被记录最后，返回最优解。

注意清空一下数据，防止对下次造成影响