# 题目

## 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.

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

242

# 背景

`(x,y-1)` , `(x,y + 1)` ,横坐标为 `x - 1` , `x + 1` 的两行

# 输入

`1<=N*M<=200000`

>最大不连续子序列<

# 代码

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

かしこいかわいい？
エリーチカ！

*/

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
using namespace std;

const int maxn = 200005;
int numt[maxn];
int num[maxn];
int dp[maxn];
int N,M;

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

memset(dp,0,sizeof(dp));

for(int i = 1;i <= N;i++) {
for(int j = 1;j <= M;j++) {
int temp;
scanf("%d",&temp);
if(j <= 2) {
if(j == 1)
numt[j] = temp;
else
numt[j] = max(numt[1],temp);
} else {
numt[j] = max(numt[j - 1],numt[j - 2] + temp);
}
}
num[i] = numt[M];
}

dp[1] = num[1];
dp[2] = max(num[1],num[2]);
for(int i = 3;i <= N;i++) {
dp[i] = max(dp[i - 1],dp[i - 2] + num[i]);
}

printf("%d\n",dp[N]);
return true;
}

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