Tourists

Tourists

时间限制: 5 Sec  内存限制: 64 MB

题目描述

In Tree City, there are n tourist attractions uniquely labeled 1 to n. The attractions are connected by a set of n − 1 bidirectional roads in such a way that a tourist can get from any attraction to any other using some path of roads.
You are a member of the Tree City planning committee. After much research into tourism, your committee has discovered a very interesting fact about tourists: they LOVE number theory! A tourist who visits an attraction with label x will then visit another attraction with label y if y > x and y is a multiple of x. Moreover, if the two attractions are not directly connected by a road thetourist will necessarily visit all of the attractions on the path connecting x and y, even if they aren’t multiples of x. The number of attractions visited includes x and y themselves. Call this the length of a path.
Consider this city map:

Here are all the paths that tourists might take, with the lengths for each:
1 → 2 = 4, 1 → 3 = 3, 1 → 4 = 2, 1 → 5 = 2, 1 → 6 = 3, 1 → 7 = 4,
1 → 8 = 3, 1 → 9 = 3, 1 → 10 = 2, 2 → 4 = 5, 2 → 6 = 6, 2 → 8 = 2,
2 → 10 = 3, 3 → 6 = 3, 3 → 9 = 3, 4 → 8 = 4, 5 → 10 = 3
To take advantage of this phenomenon of tourist behavior, the committee would like to determine the number of attractions on paths from an attraction x to an attraction y such that y > x and y is a multiple of x. You are to compute the sum of the lengths of all such paths. For the example above, this is: 4 + 3 + 2 + 2 + 3 + 4 + 3 + 3 + 2 + 5 + 6 + 2 + 3 + 3 + 3 + 4 + 3 = 55.

输入

Each input will consist of a single test case. Note that your program may be run multiple times on different inputs. The first line of input will consist of an integer n (2 ≤ n ≤ 200,000) indicating the number of attractions. Each of the following n−1 lines will consist of a pair of space-separated
integers i and j (1 ≤ i < j ≤ n), denoting that attraction i and attraction j are directly connected by a road. It is guaranteed that the set of attractions is connected.

输出

Output a single integer, which is the sum of the lengths of all paths between two attractions x and y such that y > x and y is a multiple of x.

样例输入

10
3 4
3 7
1 4
4 6
1 10
8 10
2 8
1 5
4 9

样例输出

55
分析：LCA裸题；
　　　注意dfs层数太深会爆，所以需要手写栈；
代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <unordered_map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, ls[rt]
#define Rson mid+1, R, rs[rt]
#define sys system("pause")
const int maxn=2e5+10;
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
inline ll read()
{
    ll x=0;int f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m,k,t,p[maxn<<1],all,dep[maxn],tot,h[maxn],vis[maxn<<1],vis1[maxn],fa[maxn],st[20][maxn<<1];
void init()
{
    for(int i=2;i<=all;i++)p[i]=1+p[i>>1];
    for(int i=1;i<=19;i++)
        for(int j=1;(j+(1<<i)-1)<=(ll)all;j++)
            st[i][j]=min(st[i-1][j],st[i-1][j+(1<<(i-1))]);
}
int query(int l,int r)
{
    int x=p[r-l+1];
    return min(st[x][l],st[x][r-(1<<x)+1]);
}
struct node
{
    int to,nxt;
}e[maxn<<1];
void add(int x,int y)
{
    tot++;
    e[tot].to=y;
    e[tot].nxt=h[x];
    h[x]=tot;
}
stack<int>S;
void dfs()
{
    S.push(1);
    while(!S.empty())
    {
        int now = S.top();
        if(vis1[now] == 1)// if node is gray, then color black
        {
            vis1[now] = 2;
            st[0][++all]=dep[fa[now]];
            // do things after dfs children.
            S.pop();
        }
        else if(vis1[now] == 0)// if node is white, then color gray
        {
            vis1[now] = 1;
            st[0][++all]=dep[now];
            vis[now]=all;
            // do things before dfs children.
            for(int i=h[now];i;i=e[i].nxt)
            {
                int to=e[i].to;
                if(!vis1[to])
                {
                    dep[to]=dep[now]+1;
                    fa[to]=now;
                    S.push(to);
                }
            }
        }
    }
}
ll ans;
int main()
{
    int i,j;
    //freopen("in.txt","r",stdin);
    while(~scanf("%d",&n))
    {
        ans=0;
        all=0;
        tot=0;
        memset(dep,0,sizeof(dep));
        memset(vis,0,sizeof(vis));
        memset(h,0,sizeof(h));
        memset(p,0,sizeof(p));
        memset(vis1,0,sizeof(vis1));
        rep(i,1,n-1)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            add(a,b),add(b,a);
        }
        dfs();
        init();
        for(i=1;i<=n;i++)
        {
            for(j=i*2;j<=n;j+=i)
            {
                ans+=dep[i]+dep[j]-2*query(min(vis[i],vis[j]),max(vis[i],vis[j]))+1;
            }
        }
        printf("%lld
",ans);
    }
    //system("Pause");
    return 0;
}