【Foreign】Walk [暴力]

Walk

Time Limit: 20 Sec  Memory Limit: 256 MB

Description

　　

Input

　　

Output

　　

Sample Input

　　3
　　1 2 3
　　1 3 9

Sample Output

　　9
　　3
　　0

HINT

　　

Solution

　　

　　其实吧，就是每次枚举一个d，重新构图，把权值是 d 的倍数的边加入。然后Dfs暴力求一遍直径L，显然 [1, L] 都可以用 d 更新。

　　重点是在于复杂度的证明吧，证明在上面qwq（BearChild当时不敢写qaq）。

Code

 #include<iostream>
 #include<string>
 #include<algorithm>
 #include<cstdio>
 #include<cstring>
 #include<cstdlib>
 #include<cmath>
 #include<vector>
 using namespace std;
 typedef long long s64;

 const int ONE = ;
 const int INF = ;

 int n;
 int x, y, z;
 int Maxval, S[ONE];
 int Ans[ONE];

 vector <int> D[ONE], G[ONE];

 struct power
 {
         int x, y, z;
 }a[ONE];

 int get()
 {
         int res=,Q=;char c;
         while( (c=getchar())< || c> )
         if(c=='-')Q=-;
         res=c-;
         while( (c=getchar())>= && c<= )
         res=res*+c-;
         return res*Q;
 }

 void Dealiv()
 {
         for(int i = ; i <= n; i++)
         {
             int Q = sqrt(a[i].z);
             for(int j = ; j <= Q; j++)
                 if(a[i].z % j == )
                 {
                     D[j].push_back(i);
                     if(a[i].z / j != j) D[a[i].z / j].push_back(i);
                 }
         }
 }

 int vis[ONE], length, A1, Record;
 int next[ONE], first[ONE], go[ONE], tot;

 void Add(int u, int v)
 {
         next[++tot] = first[u]; first[u] = tot; go[tot] = v;
         next[++tot] = first[v]; first[v] = tot; go[tot] = u;
 }

 void Dfs1(int u, int father, int dep)
 {
         if(length < dep) {A1 = u, length = dep;}
         for(int e = first[u]; e; e = next[e])
         {
             int v = go[e];
             if(v == father || vis[v]) continue;
             Dfs1(v, u, dep + );
         }
 }

 void Dfs2(int u, int father, int dep)
 {
         vis[u] = ;
         length = max(length, dep);
         for(int e = first[u]; e; e = next[e])
         {
             int v = go[e];
             if(v == father || vis[v]) continue;
             Dfs2(v, u, dep + );
         }
 }

 int main()
 {
         n = get();
         for(int i = ; i < n; i++)
             a[i].x = get(), a[i].y = get(), a[i].z = get(), Maxval = max(Maxval, a[i].z);

         Dealiv();

         int res = ;

         for(int i = ; i <= Maxval; i++)
         {
             int len = D[i].size(), cnt = ; tot = ;

             for(int j = ; j < len; j++)
             {
                 int id = D[i][j];
                 Add(a[id].x, a[id].y);
                 S[++cnt] = a[id].x,    S[++cnt] = a[id].y;
             }

             Record = ;
             for(int j = ; j <= cnt; j++)
             if(!vis[S[j]])
             {
                 A1 = length = ; Dfs1(S[j], , );
                 length = ;    Dfs2(A1, , );
                 Record = max(Record, length);
             }

             for(int j = ; j <= cnt; j++)
                 first[S[j]] = , vis[S[j]] = ;

             Ans[Record] = i;
         }

         for(int i = n; i >= ; i--)
             Ans[i] = max(Ans[i + ], Ans[i]);

         for(int i = ; i <= n; i++)
             printf("%d\n", Ans[i]);
 }