HDU 2845 Beans（dp+求两次最达不连续子序列和）

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

4 611 0 7 5 13 978 4 81 6 22 41 40 9 34 16 1011 22 0 33 39 6

Sample Output

242

题解：

比赛的时候没想出来，看了博客懂了，由题意就是求两次不连续最大子序列和，横着一次竖着一次，讲解看注释

我看的博客：http://www.2cto.com/kf/201408/323478.html

代码：

#include<iostream>#include<stdio.h>#include<string>#include<cstring>#include<map>#include<queue>#include<stack>#include<algorithm>#include<math.h>#include<deque>#include<stack>using namespace std;int a[20005];//数据保存int p1[20005];//i处取int p2[20005];//i处不取int b[20005];//每一行的情况int m,n;int dp(int c[],int len)//求不连续最大子序列和{    int i,j;    p1[0]=0;//初始化一下    p2[0]=0;    for(i=1;i<=len;i++)    {        p1[i]=p2[i-1]+c[i];//取的时候即为不取的时候+当前值        p2[i]=max(p1[i-1],p2[i-1]);//不取即前一个取的情况和不取的情况的最大值    }    return max(p1[len],p2[len]);//len处取与不取的最大值}int main(){    int i,j,x;    while(scanf("%d%d",&n,&m)!=EOF)    {        for(i=1;i<=n;i++)        {            for(j=1;j<=m;j++)            {                scanf("%d",&a[j]);            }            b[i]=dp(a,m);//存下每一行最大不连续子序列和        }        printf("%d\n",dp(b,n));//输出每一列最大不连续子序列和    }    return 0;}