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  • 【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
    ", Ans[i]);
    125 }
    View Code
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  • 原文地址:https://www.cnblogs.com/BearChild/p/7499242.html
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