【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 = 1000005;
const int INF = 2147483640;

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=1,Q=1;char c;
        while( (c=getchar())<48 || c>57 ) 
        if(c=='-')Q=-1; 
        res=c-48;     
        while( (c=getchar())>=48 && c<=57 )    
        res=res*10+c-48;    
        return res*Q;
}

void Dealiv()
{
        for(int i = 1; i <= n; i++)
        {
            int Q = sqrt(a[i].z);
            for(int j = 1; j <= Q; j++)
                if(a[i].z % j == 0)
                {
                    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 + 1);
        }
}

void Dfs2(int u, int father, int dep)
{
        vis[u] = 1;
        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 + 1);
        }
}

int main()
{
        n = get();
        for(int i = 1; i < n; i++)
            a[i].x = get(), a[i].y = get(), a[i].z = get(), Maxval = max(Maxval, a[i].z);
        
        Dealiv();
        
        int res = 0;
        
        for(int i = 1; i <= Maxval; i++)
        {
            int len = D[i].size(), cnt = 0; tot = 0;
            
            for(int j = 0; 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 = 0;
            for(int j = 1; j <= cnt; j++)
            if(!vis[S[j]])
            {
                A1 = length = 0; Dfs1(S[j], 0, 0);
                length = 0;    Dfs2(A1, 0, 0);
                Record = max(Record, length);
            }
            
            for(int j = 1; j <= cnt; j++)
                first[S[j]] = 0, vis[S[j]] = 0;
            
            Ans[Record] = i;
        }
        
        for(int i = n; i >= 1; i--)
            Ans[i] = max(Ans[i + 1], Ans[i]);
            
        for(int i = 1; i <= n; i++)
            printf("%d
", Ans[i]);
}