HDU 2845 Beans(最大不连续子序列和 dp)

古城微笑少年丶 2024-02-17 19:36 18阅读 0赞

## Beans ##

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.  
![convip1-1001-1.JPG][]  
  
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：如图中数字81，如果81取了，那么它上面、下面的行就不能取了，左右的也不能取了。

思路：两次dp，一次行，一次列，状态方程相同dp(i)=max(dp(i-1),dp(i-2)+tmp);

AC代码：

#include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    const int maxn=2e5+10;
    int dpx[maxn],dpy[maxn];
    int main(){
    	int n,m;
    	while(scanf("%d%d",&m,&n)==2){
    		
    		memset(dpy,0,sizeof(dpy));
    		for(int i=m;i>=1;i--)
    		{
    		 for(int j=1;j<=n;j++)
    		 {int tmp;
    		 scanf("%d",&tmp);
    		 int t=j+2;  //因为j-2可能小于0所以整体向后移2 
    		 dpy[t]=max(dpy[t-1],dpy[t-2]+tmp);
    		 } 
    		 dpx[i]=dpy[n+2];
    		}
    		
    		if(dpx[2]<dpx[1])dpx[2]=dpx[1];
    		for(int i=3;i<=m;i++)
    		dpx[i]=max(dpx[i-1],dpx[i-2]+dpx[i]);
    		
    		printf("%d\n",dpx[m]);
    	}
    	return 0;
    }

[convip1-1001-1.JPG]: https://image.dandelioncloud.cn/pgy_files/images/2024/01/30/c38429a6998c49de84048272ff1fa14e.png