hdu 2845 Beans(最大不连续子序列和)

Problem Description

Bean-eating is an interesting game, everyone owns an M*N matrix, which is filled with different qualities beans. Meantime, there is only one bean in any 1*1 grid. Now you want to eat the beans and collect the qualities, but everyone must obey by the following rules: if you eat the bean at the coordinate(x, y), you can’t eat the beans anyway at the coordinates listed (if exiting): (x, y-1), (x, y+1), and the both rows whose abscissas are x-1 and x+1.

Now, how much qualities can you eat and then get ?

 

Input

There are a few cases. In each case, there are two integer M (row number) and N (column number). The next M lines each contain N integers, representing the qualities of the beans. We can make sure that the quality of bean isn't beyond 1000, and 1<=M*N<=200000.

 

Output

For each case, you just output the MAX qualities you can eat and then get.

 

Sample Input

4 6

11 0 7 5 13 9

78 4 81 6 22 4

1 40 9 34 16 10

11 22 0 33 39 6

 

Sample Output

242

 

题意：在图中取数，例如取了81之后，同一行的相邻两个不能取，还有81的上面那行和下面那行也不能取，问能取到的最大和是多少？

思路：分别求行列的最大不连续子序列和

#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[]={,,,,,,,,,,,,};
int dir[][]={, ,, ,-, ,,-};
int dirs[][]={, ,, ,-, ,,-, -,- ,-, ,,- ,,};
const int inf=0x3f3f3f3f;
const ll mod=1e9+;
int sum[][];
int main(){
    ios::sync_with_stdio(false);
    int m,n;
    while(cin>>m>>n){
        for(int i=;i<=m;i++){
            for(int j=;j<=n;j++){
                cin>>sum[j][];
            }
            for(int j=;j<=n;j++)
                sum[j][]=max(sum[j][]+sum[j-][],sum[j-][]);
            sum[i][]=sum[n][];
        }
        for(int i=;i<=m;i++)
            sum[i][]=max(sum[i][]+sum[i-][],sum[i-][]);
        cout<<sum[m][]<<endl;
    }
}