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

4 611 0 7 5 13 978 4 81 6 22 41 40 9 34 16 1011 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[n])?f(n-1,a):(f(n-2,a)+a[n]); }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;}