http://poj.org/problem?id=1050
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.
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
150
代码:
#include <iostream>
#include <stdio.h>
#include <math.h>
#include <algorithm>
using namespace std;
const int maxn = 110;
int N;
int mp[maxn][maxn], sum[maxn][maxn];
int maxx = -1e5;
int Maxnum(int l, int r) {
int c[maxn], dp[maxn];
for(int i = 0; i < N; i ++)
c[i] = sum[i][r] - sum[i][l - 1];
dp[0] = c[0];
int ans = max(ans, dp[0]);
for(int i = 1; i < N; i ++) {
dp[i] = max(dp[i - 1] + c[i], c[i]);
ans = max(ans, dp[i]);
}
return ans;
}
int main() {
scanf("%d", &N);
for(int i = 0; i < N; i ++) {
for(int j = 0; j < N; j ++) {
scanf("%d", &mp[i][j]);
}
}
for(int j = 0; j < N; j ++) {
for(int i = 0; i < N; i ++) {
if(i == 0) sum[j][i] = mp[i][j];
else sum[j][i] = sum[j][i - 1] + mp[i][j];
}
}
for(int i = 0; i < N; i ++) {
for(int j = N - 1; j > i; j --) {
maxx = max(maxx, Maxnum(i, j));
}
}
printf("%d\n", maxx);
return 0;
}
枚举上下边界然后枚举最大子序列的值 $O(N^3)$ 时间复杂度
FH
POJ 1050 To the Max
原文:https://www.cnblogs.com/zlrrrr/p/10764469.html