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

40 -2 -7 0 9 2 -6 2-4 1 -4 1 -18 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：

//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*jfor(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=0;j<n;j++)cin>>arr[i][j];//很难想到，这居然是一道动态规划题//1、最大子字段和/*题目可以转化为两个子问题：1，给定n个整数（可能为负数）组成的序列a[1],a[2],a[3],…,a[n],求该序列如a+a+…+a[j]的子段和的最大值。当所给的整均为负数时定义子段和为0，依此定义，所求的最优值为Max{0,a+a+…+a[j]},1<=i<=j<=n我们把b[j]理解为从前面某项开始包含a[j]（a[j]为最后一个元素）的连续的段的和最大值，易知，开始的a[i]必定为正数！记b[j]=max(a+..+a[j-1]+a[j]),其中1<=i<=j,并且1<=j<=n,需要明确的一点是b[j]的值必须包含a[j]。则所求的最大子段和为max{b[j]}，1<=j<=n。由b[j]的定义可易知，当b[j-1]>0时(即前面的段加上a[j]这一段值会更大，自然把a[j]这一段接上)b[j]=b[j-1]+a[j]，否则（由于前面的段为负值，加上a[j],会使值变小，这样的话，我们将前面的段去掉，a[j]即是最大值）b[j]=a[j]。故b[j]的动态规划递归式为 b[j]=max(b[j-1]+a[j],a[j])，1<=j<=n。2、最大子矩阵和矩阵是二维的，将其压缩成一维就可以用上面的方法求出最大子矩阵和了    即将一层层的数相加，压缩成一维的即可，我们遍历所有的压缩可能，就可以通过上面的方法，求出最大值！*/ for(i=0;i<n;i++){memset(sum,0,sizeof(sum));for(k=i;k<n;k++){for(j=0;j<n;j++)sum[j]=arr[k][j]+sum[j];dp[0]=sum[0];for(p=1;p<n;p++){if (dp[p-1]>0) dp[p]=dp[p-1]+sum[p];                        else dp[p]=sum[p];                        if (dp[p]>max) max=dp[p];/*if(dp[p-1]+sum[p]>sum[p])dp[p]=dp[p-1]+sum[p];elsedp[p]=dp[p-1];if(dp[p]>max) max=dp[p];*/}}}cout<<max<<endl;//system("pause");}return 0;}