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杭电1081（动态规划）

2024-07-06 12:15:20   225人阅读

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

 


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