滑雪（記憶化搜索）

思路：我們用dp[i][j]表示從i，j爲起點可以滑的長度，dp[i][j]可以往4個方向滑（若都合法）所以我們對這4個方向dfs。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include<iostream>
//#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define space putchar(' ')
#define enter putchar('\n')
typedef pair<int,int> PII;
const int mod=1e4+7;
const int N=2e6+10;
const int inf=0x7f7f7f7f;


ll gcd(ll a,ll b)
{
    return b==0?a:gcd(b,a%b);
}

ll lcm(ll a,ll b)
{
    return a*(b/gcd(a,b));
}

template <class T>
void read(T &x)
{
    char c;
    bool op = 0;
    while(c = getchar(), c < '0' || c > '9')
        if(c == '-')
            op = 1;
    x = c - '0';
    while(c = getchar(), c >= '0' && c <= '9')
        x = x * 10 + c - '0';
    if(op)
        x = -x;
}
template <class T>
void write(T x)
{
    if(x < 0)
        x = -x, putchar('-');
    if(x >= 10)
        write(x / 10);
    putchar('0' + x % 10);
}

int fa[N];
int n,m;
int dp[150][150];
int a[150][150];
int judge(int x,int y)
{
    if(x>=1&&x<=n&&y>=1&&y<=m)return 1;
    return 0;
}
int dfs(int x,int y)
{
    if(dp[x][y]!=0)return dp[x][y];
    int f1,f2,f3,f4;
    f1=f2=f3=f4=1;
    if((a[x][y]>a[x-1][y])&&judge(x-1,y))
    {
        f1=dfs(x-1,y)+1;
    }
    if((a[x][y]>a[x+1][y])&&judge(x+1,y))
    {
        f2=dfs(x+1,y)+1;
    }
    if((a[x][y]>a[x][y-1])&&judge(x,y-1))
    {
        f3=dfs(x,y-1)+1;
    }
    if((a[x][y]>a[x][y+1])&&judge(x,y+1))
    {
        f4=dfs(x,y+1)+1;
    }
    return dp[x][y]=max(max(f1,f2),max(f4,f3));
}

int main()
{
   read(n),read(m);
   int ans=0;
   for(int i=1;i<=n;i++)
   {
       for(int j=1;j<=m;j++)
       {
           read(a[i][j]);
       }
   }
   for(int i=1;i<=n;i++)
   {
       for(int j=1;j<=m;j++)
       {
           ans=max(ans,dfs(i,j));
           //cout<<dfs(i,j)<<endl;
          //memset(dp,0,sizeof dp);
       }

   }

   write(ans);






    return 0;
}