nyoj 234 吃土豆

吃土豆

时间限制：1000 ms  |  内存限制：65535 KB

难度：4

描述 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 ?

输入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<=500.
输出For each case, you just output the MAX qualities you can eat and then get.样例输入

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

样例输出

242

代码：

/*递归超时
#include<stdio.h>
#include<string.h>
int f(int n,int a[])
{
    if(n==0)
            return a[0];
    if(n==1)
            return a[0]>a[1]?a[0]:a[1];
    return f(n-1,a)>(f(n-2,a)+a
)?f(n-1,a):(f(n-2,a)+a
);
}
int main()
{
    int m,n,i,j;
    int ans[510];
    int dp[510];
    while(scanf("%d %d",&m,&n)!=EOF)
    {
            for(i=0;i<m;i++)
            {
                  for(j=0;j<n;j++)
                        scanf("%d",&ans[j]);
                 dp[i]=f(n-1,ans);
            }
            printf("%d\n",f(m-1,dp));
    }
    system("pause");
    return 0;
}
 */ 
#include<stdio.h>
#include<string.h> 
int vis[510];
int m,n;
int f(int aa,int a[])
{
    int i;
    for(i=0;i<aa;i++)
    {
        if(i==0)
           vis[i]= a[0];
        else if(i==1)
           vis[i]= a[0]>a[1]?a[0]:a[1];
        else
           vis[i]=(vis[i-2]+a[i])>vis[i-1]?(vis[i-2]+a[i]):vis[i-1];
    }
    return vis[aa-1];
}       
int main()
{                           
    int ans[510];
    int dp[510];
    int i,j;
    while(scanf("%d %d",&m,&n)!=EOF)
    {
            for(i=0;i<m;i++)
            {
                  for(j=0;j<n;j++)
                  {
                        scanf("%d",&ans[j]);
                  }
                  dp[i]=f(n,ans);
            }
            int t=f(m,dp);
            printf("%d\n",t);
    }
  //  system("pause");
    return 0;
}