一道oj的题
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
#include <iostream>
#include <cstring>
#include <cstdlib>
using namespace std;
int a[505][505];
int Max,m,n;
void DFS(int x,int y,int temp)
{
if(x<0||x>m-1||y<0||y>n-1)
return;
if(Max<temp)
Max=temp;
DFS(x+1,y+1,temp+a[x+1][y+1]);//调试时是这一步出错，，求指教、
DFS(x-1,y+1,temp+a[x-1][y+1]);
DFS(x-1,y-1,temp+a[x-1][y-1]);
DFS(x+1,y-1,temp+a[x+1][y-1]);
}
int main()
{
while(cin>>m>>n)
{
freopen("1.txt","r",stdin);
for(int i=0;i<m;i++)
for(int j=0;j<n;j++)
cin>>a[i][j];
Max=0;
DFS(0,0,0);
cout<<Max<<endl;
}
return 0;
}
