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
来源
/*
吃了某个土豆，那它上面那行和下面那行就不能吃了，左右两个邻居也不能吃了
对于行和列来说是一样的 不能取i-1,i+1行（列）
因此
对于列来说map[i][j]+=max(map[i][j-2],max[i][j-3]);
对于行来讲dp[i]=max(dp[i-2]，dp[i-3])+max(map[i][m+1],map[i][m+2]);每行有m列
*/
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;
int dp[510];
int map[510][510];
int main()
{
int n,m,i,j;
while(scanf("%d%d",&n,&m)!=EOF)
{
memset(dp,0,sizeof(dp));
memset(map,0,sizeof(map));
//因为动态转移方程需要用到i-3，所以从3开始输入，防止访问非法内存
for(i=3;i<n+3;i++)//行 上下
{
for(j=3;j<m+3;j++)//列 左右
{
scanf("%d",&map[i][j]);
map[i][j]+=max(map[i][j-2],map[i][j-3]);
}
}
//先确定每行最大的，再去选列
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;
}
