【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

  1 #include<iostream>    
  2 #include<string>    
  3 #include<algorithm>    
  4 #include<cstdio>    
  5 #include<cstring>    
  6 #include<cstdlib>
  7 #include<cmath>
  8 #include<vector>
  9 using namespace std;  
 10 typedef long long s64;
 11 
 12 const int ONE = 1000005;
 13 const int INF = 2147483640;
 14 
 15 int n;
 16 int x, y, z;
 17 int Maxval, S[ONE];
 18 int Ans[ONE];
 19 
 20 vector <int> D[ONE], G[ONE];
 21 
 22 struct power
 23 {
 24         int x, y, z;
 25 }a[ONE];
 26 
 27 int get()
 28 {
 29         int res=1,Q=1;char c;
 30         while( (c=getchar())<48 || c>57 ) 
 31         if(c=='-')Q=-1; 
 32         res=c-48;     
 33         while( (c=getchar())>=48 && c<=57 )    
 34         res=res*10+c-48;    
 35         return res*Q;
 36 }
 37 
 38 void Dealiv()
 39 {
 40         for(int i = 1; i <= n; i++)
 41         {
 42             int Q = sqrt(a[i].z);
 43             for(int j = 1; j <= Q; j++)
 44                 if(a[i].z % j == 0)
 45                 {
 46                     D[j].push_back(i);
 47                     if(a[i].z / j != j) D[a[i].z / j].push_back(i);
 48                 }
 49         }
 50 }
 51 
 52 int vis[ONE], length, A1, Record;
 53 int next[ONE], first[ONE], go[ONE], tot;
 54 
 55 void Add(int u, int v)
 56 {
 57         next[++tot] = first[u]; first[u] = tot; go[tot] = v;
 58         next[++tot] = first[v]; first[v] = tot; go[tot] = u;
 59 }
 60 
 61 void Dfs1(int u, int father, int dep)
 62 {
 63         if(length < dep) {A1 = u, length = dep;}
 64         for(int e = first[u]; e; e = next[e])
 65         {
 66             int v = go[e];
 67             if(v == father || vis[v]) continue;
 68             Dfs1(v, u, dep + 1);
 69         }
 70 }
 71 
 72 void Dfs2(int u, int father, int dep)
 73 {
 74         vis[u] = 1;
 75         length = max(length, dep);
 76         for(int e = first[u]; e; e = next[e])
 77         {
 78             int v = go[e];
 79             if(v == father || vis[v]) continue;
 80             Dfs2(v, u, dep + 1);
 81         }
 82 }
 83 
 84 int main()
 85 {
 86         n = get();
 87         for(int i = 1; i < n; i++)
 88             a[i].x = get(), a[i].y = get(), a[i].z = get(), Maxval = max(Maxval, a[i].z);
 89         
 90         Dealiv();
 91         
 92         int res = 0;
 93         
 94         for(int i = 1; i <= Maxval; i++)
 95         {
 96             int len = D[i].size(), cnt = 0; tot = 0;
 97             
 98             for(int j = 0; j < len; j++)
 99             {
100                 int id = D[i][j];
101                 Add(a[id].x, a[id].y);
102                 S[++cnt] = a[id].x,    S[++cnt] = a[id].y;
103             }
104             
105             Record = 0;
106             for(int j = 1; j <= cnt; j++)
107             if(!vis[S[j]])
108             {
109                 A1 = length = 0; Dfs1(S[j], 0, 0);
110                 length = 0;    Dfs2(A1, 0, 0);
111                 Record = max(Record, length);
112             }
113             
114             for(int j = 1; j <= cnt; j++)
115                 first[S[j]] = 0, vis[S[j]] = 0;
116             
117             Ans[Record] = i;
118         }
119         
120         for(int i = n; i >= 1; i--)
121             Ans[i] = max(Ans[i + 1], Ans[i]);
122             
123         for(int i = 1; i <= n; i++)
124             printf("%d\n", Ans[i]);
125 }
View Code

 

posted @ 2017-09-09 20:35  BearChild  阅读(193)  评论(0)  编辑  收藏  举报