不能取相邻的行或列对行和列分别DP，【0】：没取的最大值，【1】：取的最大值
Beans
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1963 Accepted Submission(s): 993
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 ?
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
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
Source
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#include <iostream>
#include <cstdio>
using namespace std;
const int maxn=222222;
int a[maxn],dp[maxn][2];
int N,M;
int main()
{
while(scanf("%d%d",&M,&N)!=EOF)
{
for(int i=1;i<=M;i++)
{
for(int j=1;j<=N;j++)
{
int k;
scanf("%d",&k);
if(j<=1)
{
dp[j][1]=k;
}
else
{
dp[j][0]=max(dp[j-1][0],dp[j-1][1]);
dp[j][1]=dp[j-1][0]+k;
}
}
a=max(dp[N][0],dp[N][1]);
}
for(int i=1;i<=M;i++)
{
if(i==0||i==1)
{
dp[0]=0;
dp[1]=a;
}
else
{
dp[0]=max(dp[i-1][0],dp[i-1][1]);
dp[1]=dp[i-1][0]+a;
}
}
int ans=max(dp[0],dp[1]);
cout<<ans<<endl;
}
return 0;
}