杭电1081――to the max(动态规划和暴力法解) - c++编程基础

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
#include
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"< //cout<<"起点是行数"< //cout<<"p="< for(int k=1;k<=i*j;k++)
{
//cout< s=s+arr[temp];
count++;
temp++;
if(count>=i)
{
temp=cur+temp1*m;
temp1++;
count=0;
}
}
//cout< //we++;
if(s>max)
max=s;
}
cout< //system("pause");
}
return 0;
}
*/

#include
#include
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 for(j=0;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]

杭电1081――to the max(动态规划和暴力法解)(一)