NUC1157 To the Max【最大子段和+DP】
2017-06-26 08:03
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To the Max
时间限制: 1000ms 内存限制: 65536KB
通过次数: 1总提交次数:
1
问题描述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.
输入描述
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].
输出描述
Output the sum of the maximal sub-rectangle.
样例输入
4 0 -2 -7 0 9 2 -6 2 -4 1 -4 1 -1 8 0 -2
样例输出
15
来源
Greater New York 2001
问题分析:(略)
这个问题和《UVALive2288 POJ1050 HDU1081 ZOJ1074 To The Max【最大子段和+DP】》是同一个问题,代码直接用就AC了。
程序说明:参见参考链接。
参考链接:UVALive2288 POJ1050 HDU1081 ZOJ1074 To The Max【最大子段和+DP】
题记:程序做多了,不定哪天遇见似曾相识的。
AC的C++程序如下:
/* UVALive2288 POJ1050 HDU1081 ZOJ1074 To The Max */ #include <iostream> #include <limits.h> #include <string.h> using namespace std; const int N = 100; int a , b ; 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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