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poj1050-To the Max

2024-11-18 04:20:39   202人阅读

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 else15 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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