HDU-2845-Beans-简单dp

http://acm.hdu.edu.cn/showproblem.php?pid=2845

Beans

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3749    Accepted Submission(s): 1792

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.


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

 

 思路：

直接对 每一行，用dp求得当前行能选到的最大值

再对   n行里面用dp求得最大的总值

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <algorithm>#include <iostream>#include <queue>#include <map>#include <set>#include <vector>using namespace std;__int64 max(__int64 a,__int64 b){return a<b?b:a;}vector <__int64 > tm[200005]; __int64 big[200005];__int64 dp[200005];__int64 dp2[200005];int main(){__int64 i,n,m;__int64 tmp;__int64 j;while(scanf("%I64d%I64d",&n,&m)!=EOF){ for (i=1;i<=n;i++){tm[i].clear();for (j=1;j<=m;j++){scanf("%I64d",&tmp); tm[i].push_back(tmp);}} for (i=1;i<=n;i++){ dp2[0]=tm[i][0];dp2[1]=max(tm[i][0],tm[i][1]);for (j=2;j<tm[i].size();j++){dp2[j]=max(dp2[j-2]+tm[i][j],dp2[j-1]);} big[i]=dp2[tm[i].size()-1]; }dp[1]=big[1]; for (i=2;i<=n;i++){dp[i]=max(dp[i-2]+big[i],dp[i-1]);}printf("%I64d\n",dp[n]);}return 0;}