To the Max
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 52281 Accepted: 27633
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
15
题意：给你一个n*n 的矩形，要你求和最大的一个子矩形
题解：由一维的最大子段和变成了二维的最大子矩阵和，思想还是一样的，那就是保存每一段的最大和，然后更新最大值就行
将二维的看做一维，即控制第二维的深度去求最大子段和
代码如下：
#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <stack>
#include <queue>
#include <cstdio>
#include <cctype>
#include <bitset>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
#define fuck(x) cout<<"["<<x<<"]";
#define FIN freopen("input.txt","r",stdin);
#define FOUT freopen("output.txt","w+",stdout);
//#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int maxn = 1e3+;
const int INF = 0x3f3f3f3f;
int dp[maxn];
int mp[maxn][maxn];
int maxx; int main(){
#ifndef ONLINE_JUDGE
FIN
#endif
int n;
scanf("%d",&n);
maxx=-;
for(int i=;i<=n;i++){
for(int j=;j<=n;j++){
scanf("%d",&mp[i][j]);
if(mp[i][j]>maxx){
maxx=mp[i][j];
}
}
}
if(maxx<=){
printf("%d\n",maxx);
}else{
maxx=-;
int l,r;
for(int i=;i<=n;i++){
for(int j=;j<=n-i+;j++){ //控制所求子段的深度
l=i,r=j+i-;
dp[]=;
for(int k=;k<=n;k++){ //控制所求子段的长度
int tmp=;
for(int s=l;s<=r;s++){
tmp+=mp[k][s];
}
dp[k]=max(dp[k-]+tmp,tmp);
maxx=max(dp[k],maxx);
}
}
}
printf("%d\n",maxx);
}
return ;
}