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

//#include<stdio.h>//int max(int m,int *a)//{//    int i,f[550];//    for(i=1; i<=m; i++)//    {//        if(i==1)  f[i]=a[i];//        else if(i==2)  f[i]=a[i]>a[i-1]?a[i]:a[i-1];//        else//        {//            f[i]=(f[i-2]+a[i])>f[i-1]?f[i-2]+a[i]:f[i-1];//        }//    }//    return f[m];//}////非递归推算法///*int max(int m,int *a){//    if(m==1)return a[1];//    if(m==2)return a[2]=a[1]>a[2]?a[1]:a[2];//    return max(m-2,a)+a[m]>max(m-1,a)?max(m-2,a)+a[m]:max(m-1,a);//}*///////递归分制算法//int main()//{//    int a[510],b[510],i,j,m,n;//    while(scanf("%d %d",&n,&m)==2)//    {//        for(i=1; i<=n; i++)//        {//            for(j=1; j<=m; j++)//            {//                scanf("%d",&a[j]);//            }//            b[i]=max(m,a);//        }//        printf("%d\n",max(n,b));//    }//    return 0;//}#include<cmath>#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<algorithm>using namespace std;int map,dp[510],R[510][510];int main(){    int m,n,i,j;    while(~scanf("%d %d",&m,&n))    {        memset(R,0,sizeof(R));        memset(dp,0,sizeof(dp));        for(i=3; i<m+3; ++i) //把n,m都扩大2，方便dp        {            for(j=3; j<n+3; ++j)            {                scanf("%d",&map);                R[i][j]=max(R[i][j-2],R[i][j-3])+map; //累积i行到j列的最大和            }        }        for(i=3; i<m+3; ++i)        {            dp[i]=max(dp[i-2],dp[i-3])+max(R[i][n+1],R[i][n+2]); //累积到i行的最大和        }        printf("%d\n",max(dp[m+1],dp[m+2]));    }    return 0;}