hdu 1081 To The Max(二维压缩的最大连续序列)(最大矩阵和)

Problem 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

 题意:求n*n的矩阵里和最大的子矩阵。

最大连续序列和找最大和矩阵有着共通点，这里需要把一维的转化为二维的是关键

我们可以假定把每一行看作单个的元素，知道每个元素的值，那我们就能够将n列，看作有n个这样的压缩元素，那我们就是在这n个元素中找最大值

（当然压缩列也是可以的，这里我的代码写的是压缩行）

那么我们转化成用二维的sum[i][j]来维护前缀和，表示第i行前j个数的和

#include <cstdio>
#include <map>
#include <iostream>
#include<cstring>
#include<bits/stdc++.h>
#define ll long long int
#define M 6
using namespace std;
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
int moth[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int dir[4][2]={1,0 ,0,1 ,-1,0 ,0,-1};
int dirs[8][2]={1,0 ,0,1 ,-1,0 ,0,-1, -1,-1 ,-1,1 ,1,-1 ,1,1};
const int inf=0x3f3f3f3f;
const ll mod=1e9+7;
int n;
int sum[107][107];
int dp[107];
int main(){
    ios::sync_with_stdio(false);
    while(cin>>n){
        memset(sum,0,sizeof(sum));
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++){
                int t; cin>>t;
                sum[i][j]=sum[i][j-1]+t; //前缀和 
            }
        int ans=-inf;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=i;j++){ //二重循环枚举 把j~i压缩成一个点 
                int s=-inf;
                memset(dp,0,sizeof(dp));
                for(int k=1;k<=n;k++){   //找最大连续序列 
                    dp[k]=max(dp[k-1]+sum[k][i]-sum[k][j-1],sum[k][i]-sum[k][j-1]);
                    s=max(s,dp[k]);
                }
                ans=max(s,ans);
            }
        cout<<ans<<endl;
    }
}