杭电1081——to the max(动态规划和暴力法解)

To The Max

Problem Description
Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 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 x 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

题目我起先用暴力法来解答，不过，通不过，超时，后来参考了一下别人的思想吧！没想到这道题还可以用动归的方法解答，很吃惊啊！
代码如下：
[cpp]
//这个程序可以求出m*n的矩阵某一块的最大值
//这种搜索算法没有错误！但是超时，因此还要进一步优化算法！
//这道题的题意说的比较模糊，题目里没有说要多个测试数据，弄得我只处理了一次，导致老是错误，又找不到问题所在！
/*
#include<iostream>
#include<cstdio>
using namespace std;
const int MAX=10010;
int arr[MAX];

int main()
{
    int s,max=-200;
    int we=0;
    int m,n;
    while(scanf("%d",&arr[0])!=EOF)
    {
    //cin>>arr[0];//输入方阵的维数
    m=n=arr[0];
    for(int i=1;i<=arr[0]*arr[0];i++)
        cin>>arr[i];

    for(int i=1;i<=m;i++)
        for(int j=1;j<=n;j++)//这里是表示块数的大小，i*j
            for(int p=1;m-p>=i-1 ;p++)
                for(int q=1;n-q>=j-1 ;q++)//搜索路线，自上而下，自左向右
                {
                    int cur=p+m*(q-1);
                    int count=0;
                    int temp=cur;
                    int temp1=1;
                    s=0;
                    //cout<<"第"<<we<<"次搜索： "<<endl;
                    //cout<<"起点是行数"<<p<<","<<"列数"<<q<<endl;
                    //cout<<"p="<<p<<",n="<<n<<endl;
                    for(int k=1;k<=i*j;k++)
                    {
                        //cout<<arr[temp]<<" ";
                        s=s+arr[temp];
                        count++;
                        temp++;
                        if(count>=i)
                        {
                            temp=cur+temp1*m;
                            temp1++;
                            count=0;
                        }
                    }
                    //cout<<endl;
                    //we++;
                    if(s>max)
                    max=s;
                }
     cout<<max<<endl;
     //system("pause");
     }
    return 0;
}
*/

#include<iostream>
#include<cstdio>
using namespace std;

int arr[101][101];
int sum[101],dp[101];

int main()
{

    int i,j,k,p,n;
    while(scanf("%d",&n)!=EOF)
    {
        int max=-200;
        memset(sum,0,sizeof(sum));
        memset(dp,0,sizeof(dp));
        for(i=0;i<n;i++)
            for(j=

补充：软件开发 , C++ ,