题目

Description

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 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 x 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

题解

题目让求最大的矩形区域的权值和

不看二维,单看一维,就是最大连续子序列
dp[i] = max(dp[i-1]+a[i] , a[i])

而拓展到二维,可以先将它压缩成一维
用 O(n²) 的时间遍历取起始行和终止行,将其压缩成一行
用 O(n) 的时间复杂付求取最大连续子序列
最终总的时间复杂度是 O(n³)

代码

/*
By:OhYee
Github:OhYee
Blog:http://www.oyohyee.com/
Email:oyohyee@oyohyee.com

かしこいかわいい？
エリーチカ！
要写出来Хорошо的代码哦~
*/
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <functional>
using namespace std;
const int INF = 0x7FFFFFFF;
const int maxn = 105;
const int delta[] = {1,-1,0,0};

int Map[maxn][maxn];
int dp[maxn];
int sum[maxn][maxn];

int n;

int DP(int s,int v) {
    int Max = -INF;
    for(int i = 1;i <= n;i++) {
        int t = sum[v][i] - sum[s][i];
        dp[i] = max(dp[i - 1] + t,t);
        Max = max(Max,dp[i]);
    }
    return Max;
}

bool Do() {
    if(scanf("%d",&n) == EOF)
        return false;

    for(int i = 1;i <= n;i++)
        for(int j = 1;j <= n;j++) {
            scanf("%d",&Map[i][j]);
            sum[i][j] = sum[i - 1][j] + Map[i][j];
        }
    int Max = -INF;
    for(int i = 1;i <= n;i++)
        for(int j = 0;j < i;j++) {
            Max = max(Max,DP(j,i));
        }

    printf("%d\n",Max);

    return true;
}

int main() {
    while(Do());
    return 0;
}