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poj1050-To the Max

时间:2015-01-27 00:12:17      阅读:242      评论:0      收藏:0      [点我收藏+]

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头疼,做道水题。。

Description

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*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 * 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.
主要思想就是降维,源代码如下:
 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 #define M 101
 4 int num[M][M];
 5 int N;
 6 
 7 int submax(int a[M])
 8 {
 9 int i,pre=a[1],max=0;
10 for(i=2;i<=N;i++)
11 {
12 if(a[i]+pre>a[i])
13 pre=a[i]+pre;
14 else
15 pre=a[i];
16 if(pre>max)
17 max=pre;
18 }
19 return max;
20 }
21 
22 int submax2()
23 {
24 int b[M];
25 int i,j,k,max=0;
26 for(i=1;i<=N;i++)
27 {
28 memset(b,0,sizeof(b));
29 for(j=i;j<=N;j++)
30 {
31 for(k=1;k<=N;k++)
32 b[k]+=num[j][k];
33 int ff=submax(b);
34 if(ff>max) max=ff;
35 }
36 }
37 return max;
38 }
39 
40 int main()
41 {
42 int i,j,k;
43 scanf("%d",&N);
44 for(i=1;i<=N;i++)
45 for(j=1;j<=N;j++)
46 scanf("%d",&num[i][j]);
47 printf("%d\n",submax2());
48 
49 return 0;
50 }

 

poj1050-To the Max

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原文地址:http://www.cnblogs.com/xlzhh/p/4251522.html

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