<h1 id="heading">题目</h1>
<blockquote>
<h2 id="description">Description</h2>
<blockquote>
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.
 
</blockquote>
<h2 id="input">Input</h2>
<blockquote>
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].
</blockquote>
<h2 id="output">Output</h2>
<blockquote>
Output the sum of the maximal sub-rectangle.
</blockquote>
<h2 id="sample-input">Sample Input</h2>
<blockquote>
4 
0 -2 -7 0 9 2 -6 2 
-4 1 -4 1 -1 
8 0 -2
</blockquote>
<h2 id="sample-output">Sample Output</h2>
<blockquote>
15
</blockquote>
</blockquote>
<h1 id="heading-1">题解</h1>
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 
<code>dp[i] = max(dp[i-1]+a[i] , a[i])</code>
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)
<h1 id="heading-2">代码</h1>
<pre style="background-color:#fff">/*
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 &lt;cstdio&gt;
#include &lt;algorithm&gt;
#include &lt;cstring&gt;
#include &lt;cmath&gt;
#include &lt;string&gt;
#include &lt;iostream&gt;
#include &lt;vector&gt;
#include &lt;list&gt;
#include &lt;queue&gt;
#include &lt;stack&gt;
#include &lt;map&gt;
#include &lt;set&gt;
#include &lt;functional&gt;
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 &lt;= 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(&#34;%d&#34;,&amp;n) == EOF)
 return false;

 for(int i = 1;i &lt;= n;i++)
 for(int j = 1;j &lt;= n;j++) {
 scanf(&#34;%d&#34;,&amp;Map[i][j]);
 sum[i][j] = sum[i - 1][j] + Map[i][j];
 }
 int Max = -INF;
 for(int i = 1;i &lt;= n;i++)
 for(int j = 0;j &lt; i;j++) {
 Max = max(Max,DP(j,i));
 }

 printf(&#34;%d\n&#34;,Max);

 return true;
}

int main() {
 while(Do());
 return 0;
}
</pre>

HDU 1081.To The Max