HDU 2845 Beans (最大不連續子序列和)

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Beans

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2637    Accepted Submission(s): 1302

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 Input4 611 0 7 5 13 978 4 81 6 22 41 40 9 34 16 1011 22 0 33 39 6 

 

Sample Output242 

 

Source2009 Multi-University Training Contest 4 - Host by HDU 

 

Recommendgaojie  這道題是說取一個數字，那麼相鄰的兩行的所有數字和同一行的相鄰兩列的數字就不能再取了。 在同一行裡面，相鄰的數字不能同時取，那麼每一行都有會有一個最大值，那麼可以把每一行最大值看成一個數字，再求一次，相當於在一行裡面求最大不連續子序列的和，只是這裡的每一個數字是每一行的最大值。dp[i]=max(dp[i-2]+a[i],dp[i-1])dp[i]代表到i時的最大子序列的和，對於每一個數字，我只有取或者不取兩種狀態，如果取，那麼我的最大值是從dp[i-2]再加上本身的數值，如果不取，那麼我最大值是dp[i-1]。 

 1 #include<cstdio> 2 #include<cstring> 3 #include<stdlib.h> 4 #include<algorithm> 5 using namespace std; 6 int a[200005],dp[200005]; 7 int main() 8 { 9     int n,m,i,j;10     while(scanf("%d %d",&n,&m)!=EOF)11     {12         for(i=1;i<=n;i++)13         {14             for(j=1;j<=m;j++)15                 scanf("%d",&a[j]);16             for(j=2;j<=m;j++)17                 a[j]=max(a[j-1],a[j-2]+a[j]);18             dp[i]=a[m];19         }20         for(i=2;i<=n;i++)21             dp[i]=max(dp[i-2]+dp[i],dp[i-1]);22         printf("%d\n",dp[n]);23     }24     return 0;25 }

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