題目描述

給定一個n個節點的樹，每個節點表示一個整數，問u到v的路徑上有多少個不同的整數。

輸入格式

第一行有兩個整數n和m（n＝40000，m＝100000）。

第二行有n個整數。第i個整數表示第i個節點表示的整數。

在接下來的n-1行中，每行包含兩個整數u v，描述一條邊（u，v）。

在接下來的m行中，每一行包含兩個整數u v，詢問u到v的路徑上有多少個不同的整數。

輸出格式

對於每個詢問，輸出結果。 貢獻者：つるまる

題目描述

You are given a tree with N nodes. The tree nodes are numbered from 1 to N. Each node has an integer weight.

We will ask you to perform the following operation:

• u v : ask for how many different integers that represent the weight of nodes there are on the path from u to v.

輸入輸出格式

輸入格式：

In the first line there are two integers N and M. (N <= 40000, M <= 100000)

In the second line there are N integers. The i-th integer denotes the weight of the i-th node.

In the next N-1 lines, each line contains two integers u v, which describes an edge (u, v).

In the next M lines, each line contains two integers u v, which means an operation asking for how many different integers that represent the weight of nodes there are on the path from u to v.

輸出格式：

For each operation, print its result.

輸入輸出樣例

8 2
105 2 9 3 8 5 7 7
1 2
1 3
1 4
3 5
3 6
3 7
4 8
2 5
7 8

4
4

SDOI 2018因為沒學過樹上莫隊慘遭爆零，

不過這玩意兒確實定好玩的。

首先建出歐拉序來，然後對於每個詢問，分兩種情況討論。

這裏就不展開講了，

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<vector>
//#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<20,stdin),p1==p2)?EOF:*p1++)
char buf[1 << 21], *p1 = buf, *p2 = buf;
using namespace std;
const int MAXN = 1e5 + 10;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < ‘0‘ || c > ‘9‘) {if(c == ‘-‘) f = -1; c = getchar();}
    while(c >= ‘0‘ && c <= ‘9‘) x = x * 10 + c - ‘0‘, c = getchar();
    return x * f;
}
int N, Q;
int belong[MAXN], block;
struct Query {
    int l, r, ID, lca, ans;
    bool operator < (const Query &rhs) const{
        return belong[l] == belong[rhs.l] ? r < rhs.r : belong[l] < belong[rhs.l];
    //    return belong[l] < belong[rhs.l];
    }
}q[MAXN];
vector<int>v[MAXN];
int a[MAXN], date[MAXN];
void Discretization() {
    sort(date + 1, date + N + 1);
    int num = unique(date + 1, date + N + 1) - date - 1;
    for(int i = 1; i <= N; i++) a[i] = lower_bound(date + 1, date + num + 1, a[i]) - date;    
}
int deep[MAXN], top[MAXN], fa[MAXN], siz[MAXN], son[MAXN], st[MAXN], ed[MAXN], pot[MAXN], tot;
void dfs1(int x, int _fa) {
    fa[x] = _fa; siz[x] = 1;
    st[x] = ++ tot; pot[tot] = x; 
    for(int i = 0; i < v[x].size(); i++) {
        int to = v[x][i];
        if(deep[to]) continue;
        deep[to] = deep[x] + 1;
        dfs1(to, x);
        siz[x] += siz[to];
        if(siz[to] > siz[son[x]]) son[x] = to;
    }
    ed[x] = ++tot; pot[tot] = x;
}
void dfs2(int x, int topfa) {
    top[x] = topfa;
    if(!son[x]) return ;
    dfs2(son[x], topfa);
    for(int i = 0; i < v[x].size(); i++) {
        int to = v[x][i];
        if(top[to]) continue;
            dfs2(to, to);
    }
}
int GetLca(int x, int y) {
    while(top[x] != top[y]) {
        if(deep[top[x]] < deep[top[y]]) swap(x, y);
        x = fa[top[x]];
    }
    return deep[x] < deep[y] ? x : y;
}
void DealAsk() {
    for(int i = 1; i <= Q; i++) {
        int x = read(), y = read();
        if(st[x] > st[y]) swap(x, y);
        int _lca = GetLca(x, y);
        q[i].ID = i;
        if(_lca == x) q[i].l = st[x], q[i]. r = st[y];
        else q[i].l = ed[x], q[i].r = st[y], q[i].lca = _lca;
    }
}
int Ans, out[MAXN], used[MAXN], happen[MAXN];
void add(int x) {
    if(++happen[x] == 1) Ans++;
}
void delet(int x) {
    if(--happen[x] == 0) Ans--;
}
void Add(int x) {
    used[x] ? delet(a[x]) : add(a[x]); used[x] ^= 1;
}
void Mo() {
    sort(q + 1, q + Q + 1);
    int l = 1, r = 0, fuck = 0;
    for(int i = 1; i <= Q; i++) {
        while(l < q[i].l) Add(pot[l]), l++, fuck++;
        while(l > q[i].l) l--, Add(pot[l]), fuck++;
        while(r < q[i].r) r++, Add(pot[r]), fuck++;
        while(r > q[i].r) Add(pot[r]), r--, fuck++;
        if(q[i].lca) Add(q[i].lca);
        q[i].ans = Ans;
        if(q[i].lca) Add(q[i].lca);
    }
    for(int i = 1; i <= Q; i++) out[q[i].ID] = q[i].ans;
    for(int i = 1; i <= Q; i++)
        printf("%d\n", out[i]);
}
int main() {
    N = read(); Q = read();
    //block = 1.5 * sqrt(2 * N) + 1;
    //block = pow(N, 0.66666666666);
    block = sqrt(N);
    for(int i = 1; i <= N; i++) a[i] = date[i] = read();
    for(int i = 1; i <= N * 2; i++) belong[i] = i / block + 1;
    Discretization();
    for(int i = 1; i <= N - 1; i++) {
        int x = read(), y = read();
        v[x].push_back(y); v[y].push_back(x);
    }
    deep[1] = 1; dfs1(1, 0);
    dfs2(1, 1);
/*    for(int i = 1; i <= N; i++)    
        for(int j = 1; j <= i - 1; j++)
            printf("%d %d %d\n", i, j, GetLca(i, j));*/
    DealAsk();
    Mo();
    return 0;
}

SPOJ COT2 - Count on a tree II(樹上莫隊)