poj 1050 最大的子矩阵和
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To the Max
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 44665 Accepted: 23659
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
题意:求最大的子矩阵的和;
思路:首先就是用一种很笨的办法做的,就是说先算出矩阵中各个点到(1,1)的构成矩形的和,然后用枚举法,枚举法表示起点到终点的的所有和,,找到最大的;
sum[i][j] = sum[i-1][j]+sum[i][j-1] - sum[i-1][j-1]+ gra[i][j];表示矩阵中各个点到(1,1)构成的矩阵和,用sum[k][s] - sum[k-i][s]-sum[k][s-j] + sum[k-i][s-j])表示起点到终点构成的矩阵和;
AC代码:
#include <iostream>#include <cstdio>#include <cstring>#include <cmath>using namespace std;int gra[200][200];int sum[200][200];int n;int main() { scanf("%d",&n); memset(sum,0,sizeof(sum)); int d = 0; int m = 0; for(int i = 1; i <= n ; i++) { sum[1][i] = d + gra[1][i]; d = sum[1][i]; } d = 0; for(int i = 1; i <= n ; i++) { sum[i][1] = d + gra[i][1]; d = sum[i][1]; } for(int i = 1; i <= n ; i++) { for(int j = 1; j <= n; j++) { sum[i][j] = sum[i-1][j]+sum[i][j-1] - sum[i-1][j-1]+ gra[i][j]; m = max(sum[i][j],m); } } for(int i = 1; i <= n ; i++) { for(int j = 1; j <= n; j++) { for(int k = i; k <= n; k++) { for(int s = j; s <= n; s++) { m = max((sum[k][s] - sum[k-i][s]-sum[k][s-j] + sum[k-i][s-j]),m); } } } } printf("%d\n",m); return 0;}
写的好烦的代码,时间复杂度也高,并且感觉到思路自己也想不通,下面介绍较为简洁的代码,思路就是上升子序列的求法,把每一列的和都加起来写到最下面一行上,然后枚举起始行标和终了行标,在枚举列标,用sum[k][j] - sum[i][j];表示列构成的和,然后形成的序列求连续上升子序列的和的办法
AC代码:
#include <iostream>#include <cstdio>#include <cstring>using namespace std;int sum[200][200];int n,m;int main() { memset(sum,0,sizeof(sum)); scanf("%d",&n); for(int i = 1; i <= n; i++) { for(int j = 1; j <= n; j++) { scanf("%d",&sum[i][j]); sum[i][j] += sum[i-1][j]; } } int m = -2222; for(int i = 1; i <= n; i++) { for(int k = i; k <= n; k++) { int d = 0; int s = -2222; for(int j = 1; j <= n; j++) { d += sum[k][j] - sum[i][j]; s = max(d,s); if(d < 0){ d = 0; } } m = max(m,s); } } printf("%d\n",m); return 0; }现在感觉还是不能很好地做题的感觉,不过比以前好多了,不懂得还可以问问别人,每个人都一定要向阳生长的,相信自己,加油哦!
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