1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 using namespace std;
 5 
 6 int num[505][505];
 7 int Ans=1,n,m;
 8 int fx[4]={0,0,1,-1};
 9 int fy[4]={1,-1,0,0};
10 
11 void dfs(int x,int y,int len){
12     Ans=max(Ans,len);
13     for (int i=0;i<4;i++)
14       if (num[x][y]>num[x+fx[i]][y+fy[i]]&&num[x+fx[i]][y+fy[i]]!=-1)
15         dfs(x+fx[i],y+fy[i],len+1);
16 }
17 
18 int main(){
19     freopen("ski.in","r",stdin);
20     freopen("ski.out","w",stdout);
21     scanf("%d%d",&n,&m);
22     memset(num,-1,sizeof(num));
23     for (int i=1;i<=n;i++)
24       for (int j=1;j<=m;j++)
25         scanf("%d",&num[i][j]);
26     for (int i=1;i<=n;i++)
27       for (int j=1;j<=m;j++){
28           bool is_ok=0;
29           for(int t=0;t<=3;t++)
30             if (num[i][j]>num[i+fx[t]][j+fy[t]]&&num[i+fx[t]][j+fy[t]]!=-1) is_ok=1;
31           if (is_ok==1) dfs(i,j,1);  
32       }
33     printf("%d",Ans);
34     return 0;
35 }

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 using namespace std;
 5 
 6 int f[505][505];//存储从(i,j)出发可滑的最长路径，避免重复搜索
 7 int num[505][505];
 8 int fx[4]={0,0,1,-1};
 9 int fy[4]={1,-1,0,0};
10 
11 int dfs(int x,int y){
12     int tmp=0;
13     if (f[x][y]!=-1) return f[x][y];
14     for (int i=0;i<4;i++){
15         if (num[x][y]>num[x+fx[i]][y+fy[i]]&&num[x+fx[i]][y+fy[i]]!=-1)
16             tmp=max(tmp,dfs(x+fx[i],y+fy[i])+1);
17     }
18     if (tmp==0) {
19         f[x][y]=1;
20         return 1;
21     }
22     f[x][y]=tmp;
23     return tmp;
24 }
25 int main(){
26     int ans=1;
27     freopen("ski2.in","r",stdin);
28     freopen("ski2.out","w",stdout);
29     int n,m;
30     scanf("%d%d",&n,&m);
31     memset(num,-1,sizeof(num));
32     memset(f,-1,sizeof(f));
33     for (int i=1;i<=n;i++)
34       for (int j=1;j<=m;j++)
35         scanf("%d",&num[i][j]);
36     for (int i=1;i<=n;i++)
37       for (int j=1;j<=m;j++)
38         ans=max(ans,dfs(i,j));
39     printf("%d\n",ans);
40     return 0;
41 }