ONTAK 2010 aut

Autostrady

https://szkopul.edu.pl/problemset/problem/f2dSBM7JteWHqtmVejMWe1bW/site/?key=statement

题意：

　　首先给定一棵树，除了n-1条树边以外，还有m条非树边。每次询问两个点的满足以下条件的路径条数。

1. 不能走树上u到v的简单路径的边。
2. 只能走一条非树边。

分析：

　　RMQ求LCA + 线段树合并。

　　问题转化为有多少边的一个端点在u的子树内，另一个在v的子树内。

　　每个询问只在深度大的询问加入深度小的。对每个节点建立一个权值线段树，dfs从叶子节点往上合并，每到一个节点询问一段区间的数。

　　如果询问一个是另一个的祖先，要特判。

代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<iostream>
#include<cctype>
#include<set>
#include<vector>
#include<queue>
#include<map>
#define pa pair<int,int>
#define mp(a,b) make_pair(a,b)
using namespace std;
typedef long long LL;

inline int read() {
    int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
}

const int N = 100005;

int deth[N], id[N], siz[N], ans[N * 5], Time_Index, n; // ans[N * 5] !!!
vector<int> adj[N], ext[N];
vector< pa > q[N];

namespace LCA{
    int ord[N << 1], d[N << 1], p[N], f[N << 1][20], Log[N << 1], tot = 0;
    void dfs(int u,int fa,int dep) {
        ord[++ tot] = u; d[tot] = dep;
        p[u] = tot;
        for (int i=0,sz=adj[u].size(); i<sz; ++i) {
            int v = adj[u][i];
            if (v == fa) continue;
            dfs(v, u, dep + 1);
            ord[++tot] = u; d[tot] = dep;
        }
    }
    void init() {
        Log[0] = -1;
        for (int i=1; i<=tot; ++i) 
            f[i][0] = i, Log[i] = Log[i >> 1] + 1; // i<<1 !!!
        for (int j=1; j<=Log[tot]; ++j) {
            for (int i=1; (i+(1<<j)-1)<=tot; ++i) {
                int x = f[i][j - 1], y = f[i+(1<<(j-1))][j - 1]; // j-1 !!!
                f[i][j] = d[x] < d[y] ? x : y;
            }
        }
    }
    int Lca(int u,int v) {
        u = p[u], v = p[v];
        if (u > v) swap(u, v);
        int k = Log[v - u + 1];
        int x = f[u][k], y = f[v-(1<<k)+1][k];
        return d[x] < d[y] ? ord[x] : ord[y];    
    }
    void Main() {
        tot = 0; dfs(1, 0, 1); init();
    }
}

namespace SegmentTree{
    queue<int> s;
    int sum[N * 9], ls[N * 9], rs[N * 9], tot;
    int NewNode() {
        int k;
        if (!s.empty()) k = s.front(), s.pop();
        else k = ++tot;
        sum[k] = ls[k] = rs[k] = 0;
        return k;
    }
    void Insert(int l,int r,int &rt,int p) {
        if (!rt) rt = NewNode();
        if (l == r) {
            sum[rt] ++; return ;
        }
        int mid = (l + r) >> 1;
        if (p <= mid) Insert(l, mid, ls[rt], p);
        else Insert(mid + 1, r, rs[rt], p);
        sum[rt] = sum[ls[rt]] + sum[rs[rt]];
    }
    int query(int l,int r,int rt,int L,int R) {
        if (!rt) return 0; // !!!
        if (L <= l && r <= R) return sum[rt];
        int mid = (l + r) >> 1, res = 0;
        if (L <= mid) res = query(l, mid, ls[rt], L, R);
        if (R > mid) res += query(mid + 1, r, rs[rt], L, R);
        return res;
    }
    int Merge(int x,int y) {
        if (!x || !y) return x + y;
        ls[x] = Merge(ls[x], ls[y]);
        rs[x] = Merge(rs[x], rs[y]);
        sum[x] = sum[x] + sum[y]; // sum[x] = sum[ls[x]] + sum[rs[x]]; !!!
        s.push(y);
        return x;
    }
    int solve(int u,int fa) {
        int rt = NewNode();
        for (int i=0,sz=adj[u].size(); i<sz; ++i) 
            if (adj[u][i] != fa) rt = Merge(rt, solve(adj[u][i], u));
        for (int i=0,sz=ext[u].size(); i<sz; ++i) 
            Insert(1, n, rt, id[ext[u][i]]);
        for (int i=0,sz=q[u].size(); i<sz; ++i) {
            int v = q[u][i].first;
            if (LCA::Lca(u, v) == v) {
                for (int t,j=0; j<adj[v].size(); ++j) 
                    if ((t=adj[v][j]) == LCA::Lca(u, t) && deth[t] > deth[v]) {// deth[t] > deth[v] !!!
                        ans[q[u][i].second] = query(1, n, rt, 1, n) - query(1, n, rt, id[t], id[t] + siz[t] - 1);
                        break;
                    }
            }
            else ans[q[u][i].second] = query(1, n, rt, id[v], id[v] + siz[v] - 1);
        }
        return rt;
    }
}
void dfs(int u,int fa) {
    deth[u] = deth[fa] + 1; 
    siz[u] = 1;
    id[u] = ++Time_Index;
    for (int i=0,sz=adj[u].size(); i<sz; ++i) {
        int v = adj[u][i];
        if (v == fa) continue;
        dfs(v, u);
        siz[u] += siz[v];
    }
}

int main() {
    n = read();
    for (int i=1; i<n; ++i) {
        int u = read(), v = read();
        adj[u].push_back(v), adj[v].push_back(u);
    }
    int m = read();
    for (int i=1; i<=m; ++i) {
        int u = read(), v = read();
        ext[u].push_back(v), ext[v].push_back(u);
    }
    dfs(1, 0);
    int Q = read();
    for (int i=1; i<=Q; ++i) {
        int u = read(), v = read();
        if (deth[u] > deth[v]) q[u].push_back(mp(v,i));
        else q[v].push_back(mp(u, i));
    }
    LCA::Main();
    SegmentTree::solve(1, 0);
    for (int i=1; i<=Q; ++i) 
        printf("%d
", ans[i] + 1);
    return 0;
}