Beans HDU - 2845 (dp)

题目链接:点我

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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

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题意:

一个 n * m的矩阵,矩阵的每个点有不同的值,你每次可以选取坐标为(x, y) 的矩阵但是选取了这个点后你不能选取它左右相邻的点,也不能选取上一行和下一行,问最大值是多少.

思路:

动态规划, 对于一行来说，相邻的数不可同时取，容易得到状态转移方程： dp[i]=max(dp[i-2]+a[i],dp[i-1]); 每一行和每一列进行相同的处理.

代码:

#include<cstdio> #include<algorithm> #include<cmath> #include<iostream> using namespace std; const int maxn = 200000+10; int dpx[maxn],dpy[maxn]; int main(){ int n,m; while(scanf("%d %d", &n, &m) != EOF){ for(int i = 2;i<=n+1;++i){ for(int j = 2;j<=m+1;++j){ int s; scanf("%d", &s); dpy[j]=max(dpy[j-1],dpy[j-2]+s); } dpx[i]=max(dpx[i-1],dpx[i-2]+dpy[m+1]); } printf("%d\n",dpx[n+1]); } return 0; }