HDU 2845 Beans(最大不连续子序列和 dp)

2019-11-06 08:15:14

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Beans

PRoblem DescriptionBean-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 ? InputThere 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. OutputFor 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 Output242
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;}

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