poj1088 滑雪 dp+dfs记忆化

简单的搜索，不必多说了，初始状态下每个点能到达的长度是1，它本身。还有，注意关掉文件重定向，被坑好多次了。

代码如下：

 1 #include<cstdio>
 2 #include<algorithm>
 3 using namespace std;
 4 typedef unsigned int ui;
 5 typedef long long ll;
 6 typedef unsigned long long ull;
 7 #define pf printf
 8 #define mem(a,b) memset(a,b,sizeof(a))
 9 #define prime1 1e9+7
10 #define prime2 1e9+9
11 #define pi 3.14159265
12 #define scand(x) scanf("%llf",&x) 
13 #define f(i,a,b) for(int i=a;i<=b;i++)
14 #define scan(a) scanf("%d",&a)
15 #define dbg(args) cout<<#args<<":"<<args<<endl;
16 #define pb(i) push_back(i)
17 #define ppb(x) pop_back(x)
18 #define inf 0x3f3f3f3f
19 #define maxn 105
20 int n,m,t,a[maxn][maxn];
21 int dp[maxn][maxn];//dp[i][j]:从(i,j)位置开始的最长道路的长度 
22 int dir[][2]={0,1,0,-1,1,0,-1,0};
23 bool vis[maxn][maxn];
24 int dfs(int x,int y)
25 {
26     if(vis[x][y])return dp[x][y];
27     vis[x][y]=true;
28     int xx,yy;
29     f(i,0,3)
30     {
31         xx=x+dir[i][0];
32         yy=y+dir[i][1];
33         if(xx<1||xx>n||yy<1||yy>m)continue;
34         if(a[xx][yy]>=a[x][y])continue;
35         dp[x][y]=max(dfs(xx,yy)+1,dp[x][y]);
36     }
37     return dp[x][y];
38 }
39 int main()
40 {
41     //freopen("input.txt","r",stdin);
42     //freopen("output.txt","w",stdout);
43     //std::ios::sync_with_stdio(false);
44     scan(n);
45     scan(m);
46     int ans=0;
47     f(i,1,n)
48         f(j,1,m)
49         {
50             scan(a[i][j]);
51             dp[i][j]=1;
52          } 
53         f(i,1,n)
54             f(j,1,m)
55             {
56                 dfs(i,j);
57                 ans=max(ans,dp[i][j]);
58             }
59                 pf("%d\n",ans);
60  } 

标签：typedef,int,poj1088,long,dfs,ans,dp,define
来源： https://www.cnblogs.com/randy-lo/p/12441060.html