NYOJ 题目234 吃土豆

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

分析：题意看不懂的看图，图给的很清楚（如取81，则他左右的两个数是不能取的，他的上一行下一行也是不能取的），可以先求出每一行的最大值，map+=max（map[i][j-2],map[i][j-3]），然后再把每一行的最大值进行计算找到最合适的那个，dp[i]=max(dp[i-2],dp[i-3])+max(map[i][m+1],map[i][m+2]);。

AC代码：

#include<stdio.h>#include<string.h>#include<iostream>using namespace std;int dp[510];int map [510][510];int main(){    int n,m,i,j;        while(~scanf("%d%d",&n,&m))    {        memset(map,0,sizeof(map));        memset(dp,0,sizeof(dp));        for(i=3; i<n+3; i++)            for(j=3; j<m+3; j++)            {                scanf("%d",&map[i][j]);                map[i][j]+=max(map[i][j-2],map[i][j-3]);            }        for(i=3; i<n+3; i++)        {            dp[i]=max(dp[i-2],dp[i-3])+max(map[i][m+1],map[i][m+2]);        }        printf("%d\n",dp[n+2]);    }    return 0;}