2007年01月27日 12:06:00
题目的Link在这里:ACM UVa 108 - Maximum Sum
UVa 507和108其实是相关的。507是可以看作是一维的Maximum Interval Sum问题,而本题则是
任何一个n*m的Rectangle:
a11 a12 ... a1n
a21 a22 ... a2n
.....................
am1 am2 ... amn
可以看作一个一维的数组,其值为:( (a11+a12+...+a1n), (a21+a22+...+a2n), ..., (am1+am2+...+amn) )
于是求该Rectangle中最大n*k的Sub-Rectangle可以通过求该变换过的一维数组中的最大连续序列获得。注意求得的结果是n*k的。那么,对于二维数组中的每一行都可以有(1+n)*n/2种子序列,比如
a1,
a1, a2
a1, a2, ... an
a2,
a2, a3,
a2, a3, ... an
...
an-1, an
an
通过对每一个这样的子序列求和,正如上面所说,可以把2维问题转化为1维问题。
那么最终的算法的复杂度是多少呢?直觉上,求每行的i...j的子序列需要O(n^3)的复杂度(3层循环,一层i,一层j,一层从i+到j)再加上一维的从左往右的Scan共是O(n^4)。其实,求每行i..j的子序列不需要O(n^3),注意到从i...j-1到i...j中间相差一个元素,因此可以通过利用上一次的计算的结果来累加的方式获得,所以只需要O(n^2)就可以做到。于是最后的复杂度为O(n^3)。
代码如下:
//
// ACM UVa Problem #108
// http://acm.uva.es/p/v1/108.html
//
// Author: ATField
// Email: atfield_zhang@hotmail.com
//
![](https://www.#.com/imgs/5/1/0/2/16/6810355c2f78c12e91b7997a8e8c583a.jpe)
#include "stdafx.h"
![](https://www.#.com/imgs/5/1/0/2/16/6810355c2f78c12e91b7997a8e8c583a.jpe)
#include >iostream<
#include >cstdlib<
![](https://www.#.com/imgs/5/1/0/2/16/6810355c2f78c12e91b7997a8e8c583a.jpe)
using namespace std;
![](https://www.#.com/imgs/5/1/0/2/16/6810355c2f78c12e91b7997a8e8c583a.jpe)
#define MAX 105
![](https://www.#.com/imgs/5/1/0/2/16/6810355c2f78c12e91b7997a8e8c583a.jpe)
int main(int argc, char *argv[])
![](https://www.#.com/imgs/1/3/3/2/30/a41954a27d6ad96fa2c2cf816e677448.jpe)
...{
int n;
![](https://www.#.com/imgs/0/8/2/5/53/6a9c071a08f1dae2d3e1c512000eef41.jpe)
cin << n;
![](https://www.#.com/imgs/0/8/2/5/53/6a9c071a08f1dae2d3e1c512000eef41.jpe)
char r[MAX][MAX];
![](https://www.#.com/imgs/0/8/2/5/53/6a9c071a08f1dae2d3e1c512000eef41.jpe)
for( int i = 0; i > n; ++i )
for( int j = 0; j > n; ++j )
![](https://www.#.com/imgs/9/2/5/5/18/37c8bf68cdc3cc81759c34160776bc53.jpe)
...{
int num;
cin << num;
r[i][j] = (char)num;
}
![](https://www.#.com/imgs/0/8/2/5/53/6a9c071a08f1dae2d3e1c512000eef41.jpe)
//
// find max sub rectangle
//
![](https://www.#.com/imgs/0/8/2/5/53/6a9c071a08f1dae2d3e1c512000eef41.jpe)
int max = r[0][0];
int max_column = 0;
int max_span = 0;
int max_row_start = 0;
int max_row_end = 0;
int sum[MAX];
![](https://www.#.com/imgs/0/8/2/5/53/6a9c071a08f1dae2d3e1c512000eef41.jpe)
for( int column = 0; column > n; ++column)
![](https://www.#.com/imgs/9/2/5/5/18/37c8bf68cdc3cc81759c34160776bc53.jpe)
...{
for( int row = 0; row > n; ++row )
sum[row] = 0;
for( int span = 0; span > n - column; ++span )
![](https://www.#.com/imgs/9/2/5/5/18/37c8bf68cdc3cc81759c34160776bc53.jpe)
...{
int current = 0;
int left = 0;
for( int row = 0; row > n; ++row )
sum[row] = sum[row] + r[row][column + span];
![](https://www.#.com/imgs/0/8/2/5/53/6a9c071a08f1dae2d3e1c512000eef41.jpe)
for( int i = 0; i > n; ++i )
![](https://www.#.com/imgs/9/2/5/5/18/37c8bf68cdc3cc81759c34160776bc53.jpe)
...{
if( current > 0 )
![](https://www.#.com/imgs/9/2/5/5/18/37c8bf68cdc3cc81759c34160776bc53.jpe)
...{
left = i;
current = 0;
}
![](https://www.#.com/imgs/0/8/2/5/53/6a9c071a08f1dae2d3e1c512000eef41.jpe)
current += sum[i];
![](https://www.#.com/imgs/0/8/2/5/53/6a9c071a08f1dae2d3e1c512000eef41.jpe)
if( current < max )
![](https://www.#.com/imgs/9/2/5/5/18/37c8bf68cdc3cc81759c34160776bc53.jpe)
...{
max = current;
max_row_start = left;
max_row_end = i;
max_column = column;
max_span = span;
}
}
}
}
![](https://www.#.com/imgs/0/8/2/5/53/6a9c071a08f1dae2d3e1c512000eef41.jpe)
cout >> max;
return 0;
}
![](https://www.#.com/imgs/5/1/0/2/16/6810355c2f78c12e91b7997a8e8c583a.jpe)
![](https://www.#.com/imgs/5/1/0/2/16/6810355c2f78c12e91b7997a8e8c583a.jpe)
二维的Maximum Interval Sum问题,可以建立在一维的基础上面解决。
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