HDU2845--Beans

[align=left]Problem Description[/align]
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 ?

[align=left]Input[/align]
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.

[align=left]Output[/align]
For each case, you just output the MAX qualities you can eat and then get.

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

242

 

/*
首先求出每行我们能取得最大值
*/
#include <iostream>
#include <cstdio>
using namespace std;
#define maxn 200008
int key[maxn];
int sum[maxn];//用来存每一行能取得的最大值
int dp[maxn];
inline int max(int a,int b)
{
return a>b?a:b;
}
int main()
{
int r,c;
while(scanf("%d%d",&r,&c)==2)
{
for(int i=1;i<=r;i++)
{
for(int j=1;j<=c;j++)
{
scanf("%d",&key[j]);
if(j==1)dp[j]=max(0,key[j]);
else
{
dp[j]=max(dp[j-1],dp[j-2]+key[j]);
}
}
sum[i]=dp[c];
}
//好吧，这样我就求出每行能拿的最大值了
dp[1]=sum[1];
for(int i=2;i<=r;i++)
{
dp[i]=max(dp[i-1],dp[i-2]+sum[i]);
}
printf("%d\n",dp[r]);
}
return 0;
}