spoj COT2 - Count on a tree II 树上莫队

题目链接

 

http://codeforces.com/blog/entry/43230树上莫队从这里学的，  受益匪浅..

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
const int maxn = 4e4+5;
int in[maxn], head[maxn], num, out[maxn], cnt, val[maxn], ans[100005], a[maxn];
int f[maxn], d[maxn], p[maxn][20], n, m, vis[maxn], times[maxn], res, id[maxn*2];
struct node
{
    int to, nextt;
}e[maxn*2];
struct query
{
    int l, r, lca, id, block;
    bool operator < (query a) const
    {
        if(block == a.block)
            return r<a.r;
        return block<a.block;
    }
}q[100005];
void add(int u, int v) {
    e[num].to = v, e[num].nextt = head[u], head[u] = num++;
}
void init() {
    mem1(head);
    mem1(p);
}
void dfs(int u, int fa) {
    in[u] = ++cnt;
    id[cnt] = u;
    f[u] = fa;
    for(int i = head[u]; ~i; i = e[i].nextt) {
        int v = e[i].to;
        if(v == fa)
            continue;
        d[v] = d[u]+1;
        dfs(v, u);
    }
    out[u] = ++cnt;
    id[cnt] = u;
}
void initLca() {
    int i, j;
    for(i = 1; i<=n; i++) {
        p[i][0] = f[i];
    }
    for(j = 1; (1<<j)<=n; j++) {
        for(i = 1; i<=n; i++) {
            if(~p[i][j-1])
                p[i][j] = p[p[i][j-1]][j-1];
        }
    }
}
int lca(int u, int v) {
    if(d[u]<d[v])
        swap(u, v);
    int i, j;
    for(i = 0; (1<<i)<=d[u]; i++)
        ;
    i--;
    for(j = i; j>=0; j--)
        if(d[u]-(1<<j)>=d[v])
            u = p[u][j];
    if(u == v)
        return v;
    for(j = i; j>=0; j--) {
        if(p[u][j] != -1 && p[u][j] != p[v][j]) {
            u = p[u][j];
            v = p[v][j];
        }
    }
    return f[u];
}
void check(int x) {
    if(vis[x] && --times[val[x]]==0) {
        res--;
    } else if(vis[x]==0 && times[val[x]]++ == 0) {
        res++;
    }
    vis[x]^=1;
}
void cal() {
    int L = q[0].l, R = q[0].l-1;
    for(int i = 0; i<m; i++) {
        while(L<q[i].l) {
            check(id[L++]);
        }
        while(L>q[i].l) {
            check(id[--L]);
        }
        while(R<q[i].r) {
            check(id[++R]);
        }
        while(R>q[i].r) {
            check(id[R--]);
        }
        if(q[i].lca != id[q[i].l] && q[i].lca != id[q[i].r]) {
            check(q[i].lca);
        }
        ans[q[i].id] = res;
        if(q[i].lca != id[q[i].l] && q[i].lca != id[q[i].r]) {
            check(q[i].lca);
        }
    }
}
int main()
{
    int u, v;
    cin>>n>>m;
    int BLOCK = sqrt(n);
    init();
    for(int i = 1; i<=n; i++) {
        scanf("%d", &val[i]);
        a[i-1] = val[i];
    }
    sort(a, a+n);
    int N = unique(a, a+n)-a;
    for(int i = 1; i<=n; i++) {
        val[i] = lower_bound(a, a+N, val[i])-a+1;
    }
    for(int i = 0; i<n-1; i++) {
        scanf("%d%d", &u, &v);
        add(u, v);
        add(v, u);
    }
    d[1] = 1;
    dfs(1, -1);
    initLca();
    for(int i = 0; i<m; i++) {
        scanf("%d%d", &u, &v);
        if(in[u]>in[v])
            swap(u, v);
        q[i].lca = lca(u, v);
        if(q[i].lca == u) {
            q[i].l = in[u];
            q[i].r = in[v];
        } else {
            q[i].l = out[u];
            q[i].r = in[v];
        }
        q[i].block = q[i].l/BLOCK;
        q[i].id = i;
    }
    sort(q, q+m);
    cal();
    for(int i = 0; i<m; i++) {
        printf("%d\n", ans[i]);
    }
    return 0;
}