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[poj]1050 To the Max dp
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
4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1
8 0 -2
Sample Output
15
和最大连续子串和类似,可以把二维的转化为一维,将i行到j行的各列数分别相加,求最大连续子串和即可。
同时可利用前缀和,sum[i][j]表示第j列前i行相加的和,i从1开始,则i-j行相加的和为sum[j][k]-sum[i-1][k]
#include <iostream> #include <stdio.h> #include <cstring> #include <algorithm> using namespace std; #define INF 0x3f3f3f3f int a[110][110]; int t[110], dp[110]; int sum[110][110]; int n; int sovle() { int ans = -INF; for (int i = 1; i <= n; i++) { for (int j = i; j <= n; j++) { int Max; dp[1] = sum[j][1] - sum[i-1][1]; for (int k = 1; k <= n; k++) { int temp = sum[j][k]-sum[i-1][k]; dp[k] = max(dp[k-1]+temp, temp); } Max = *max_element(dp+1, dp+n+1); if (Max > ans) ans = Max; } } return ans; } int main() { //freopen("1.txt", "r", stdin); scanf("%d", &n); memset(sum, 0, sizeof(sum)); for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) { scanf("%d", &a[i][j]); sum[i][j] = sum[i-1][j] + a[i][j]; } printf("%d\n", sovle()); return 0; }
[poj]1050 To the Max dp
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