UVALive2288 POJ1050 HDU1081 ZOJ1074 To The Max【最大子段和+DP】
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To the Max
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 48948 Accepted: 25895
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.
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.
Sample Input
40 -2 -7 0 9 2 -6 2-4 1 -4 1 -18 0 -2
Sample Output
15
Source
Greater New York 2001
Regionals 2001 >> North America - Greater NY
问题链接:UVALive2288 POJ1050 HDU1081 ZOJ1074 To The Max。
问题简述:参见上文。
问题分析:
这是一个计算最大子矩阵和的问题。
可以将该问题转化为计算最大子段和问题,是一个经典的动态规划问题。
在行上,采用穷尽搜索的方法来解决。
在列上,通过计算最大子段和来达到计算最大子矩阵的目的。
程序说明:数组b[]用于计算各行之和,其起始行从0到n-1(穷举法),实际的计算过程是动态的。
参考链接:HDU1003 Max Sum【最大子段和+DP】
/* UVALive2288 POJ1050 HDU1081 ZOJ1074 To The Max */#include <iostream>#include <limits.h>#include <string.h>using namespace std;const int N = 100;int a[N][N], b[N];int main(){ int n, maxval; while(cin >> n) { for(int i=0; i<n; i++) for(int j=0; j<n; j++) cin >> a[i][j]; maxval = INT_MIN; for(int i=0; i<n; i++) { memset(b, 0, sizeof(b)); for(int j=i; j<n; j++) { int sum = 0; for(int k=0; k<n; k++) { b[k] += a[j][k]; if(sum + b[k] > 0) sum += b[k]; else sum = b[k]; maxval = max(maxval, sum); } } } cout << maxval << endl; } return 0;}
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