杭电1081（动态规划）

题目：

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

 

思路：

　　把二维转换为一维，变成求一个序列的最大子串问题；

如-1 -2；-3 -4先变成-1 -2；-4 -6.

把第一行减第零行就是求矩阵第一行的最大值，第二行减第零行

就是求整个矩阵的最大值；第二行减第一行就是求矩阵第二行的

最大值！

 

代码：

　　

 1 #include<stdio.h>

 2 #include<string.h>

 3 #define inf -0xfffffff

 4 //#include<stdlib.h>

 5 /*#include<iostream>

 6 #include<algorithm>

 7 using namespace std;*/

 8 

 9 

10 int main(){

11     int n, i, j, value[101][101], dp[101][101], max, zmax, k, sum;

12     while(scanf("%d", &n) != EOF){

13         memset(dp, 0, sizeof(dp));

14         for(i = 1; i <= n; i ++){

15             for(j = 1; j <= n; j ++){

16                 scanf("%d", &value[i][j]);

17                 dp[i][j] = dp[i - 1][j] + value[i][j];

18             }

19         }

20         zmax = inf;

21         for(i = 1; i <= n; i ++){

22             for(j = i; j <= n; j ++){

23                 max = inf;

24                 sum = 0;

25                 for(k = 1; k <= n; k ++){

26                     sum += dp[j][k] - dp[i - 1][k];

27                     if(sum > max){

28                         max = sum;

29                     }

30                     if(sum < 0){

31                         sum = 0;

32                     }

33                 }

34                 if(max > zmax){

35                     zmax = max;

36                 }

37             }

38         }

39         printf("%d\n", zmax);

40     }

41     return 0;

42 }

奋进