cogs——49. 跳马问题

49. 跳马问题

水题

dfs裸基础

#include<cstdio>

using namespace std;
int n,m,mx[5]={0,1,1,2,2},
    ans,my[5]={0,-2,2,-1,1};
inline void dfs(int x,int y){
    if(x==m&&y==n){ ans++; return;}
    for(int i=1;i<=4;i++){
        int tx=mx[i]+x,ty=my[i]+y;
        if(tx>0&&ty>0&&tx<=m&&ty<=n)
            dfs(tx,ty);
    }
}
int main()
{
    freopen("horse.in","r",stdin);
    freopen("horse.out","w",stdout);
    scanf("%d%d",&n,&m);
    dfs(1,1);
    printf("%d",ans);
    return 0;
}