SPOJ COT2(树上莫队)

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 perfrom 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.

 

Input

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

In the second line there are N integers.The ith integer denotes the weight of the ith 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.

Output

For each operation,print its result.

题意：给你一颗树，树上每个节点有一个权值，然后给你若干个询问，每次询问让你找出一条链上有多少个不同权值.

解题思路：如果这个问题是在一个序列上做，那么我们可以用莫队做，但是这个是在树上，所以我们先对树进行分块，然后同样也可以用莫队，但是树上莫队的转移就没有序列上莫队的转移那么容易了，这用需要用到Lca,然后用集合的对称差转移，还是很巧妙的。

#include <bits/stdc++.h>using namespace std;const int maxn = 40000 + 10;const int maxm = 100000 + 10;int N, M;int block;//块的大小int nowblock;//当前块的编号int sta[maxn];//存放待分配块的节点int top;//栈顶指针vector<int> g[maxn];//树int pos[maxn];//节点所在的酷块的编号int depth[maxn];//节点的深度int value1[maxn];//节点原本的权值int value2[maxn];//节点离散化之后的权值int H[maxn];//用于离散化节点权值的数组int num[maxn];//表示数字出现过的次数int appear[maxn];//表示节点是否出现过int father[maxn];bool visit[maxn];int Index[maxn<<2];int dp[maxn<<2][25];int First[maxn];int Log[maxn<<2];int res;struct query{    int l, r, u, v, id;    bool operator <(const query &res) const{        if(l == res.l) return r < res.r;        else return l < res.l;    }}Query[maxm];void init(){    res = 1;    top = 0;    father[1] = -1;    memset(visit, false, sizeof(visit));    memset(appear, 0, sizeof(appear));    memset(num, 0, sizeof(num));    block = (int)sqrt(N);    nowblock = 0;    for(int i = 1; i <= N; i++) g[i].clear();}void initRmq(){    Log[0] = -1;    for(int i = 1; i < res; i++)    {        Log[i] = (i&(i - 1)) == 0?Log[i - 1] + 1:Log[i - 1];    }    for(int i = 1; i < res; i++)    {        dp[i][0] = Index[i];    }    for(int j = 1; j < 20; j++)    {        for(int i = 1; i < res&&(i + (1<<j) - 1) < res; i++)        {            dp[i][j] = (depth[dp[i][j - 1]] < depth[dp[i + (1<<(j - 1))][j - 1]])?dp[i][j - 1]:dp[i + (1<<(j - 1))][j - 1];        }    }}int Rmq(int l,int r){    int dis = r - l + 1;    int j = Log[dis];    int result = (depth[dp[l][j]] < depth[dp[r - (1<<j) + 1][j]])?dp[l][j]:dp[r - (1<<j) + 1][j];    return result;}void LCA(int root,int d)//获得{    First[root] = res;    depth[root] = d;    Index[res++] = root;    visit[root] = true;    for(int i = 0; i < g[root].size(); i++)    {        int v = g[root][i];        if(!visit[v])        {            father[v] = root;            LCA(v,d + 1);            Index[res++] = root;        }    }}int getLca(int u, int v){    int f1 = First[u];    int f2 = First[v];    if(f1 > f2) swap(f1, f2);    return Rmq(f1, f2);}int dfs_block(int u){    int sum = 0;    visit[u] = true;    for(int i = 0; i < g[u].size(); i++)    {        int v = g[u][i];        if(!visit[v])        {            sum += dfs_block(v);            if(sum >= block)            {                while(sum--) pos[sta[top--]] = nowblock;                sum = 0;                nowblock++;            }        }    }    sta[++top] = u;    return sum + 1;}void initHash(){    int tot = 0;    for(int i = 1; i <= N; i++)    {        H[tot++] = value1[i];    }    sort(H, H + tot);    tot = unique(H, H + tot) - H;    for(int i = 1; i <= N; i++)    {        value2[i] = lower_bound(H, H + tot, value1[i]) - H + 1;    }}int L, R, ans;void work(int &v){    if(appear[v])    {        if(--num[value2[v]] == 0) ans--;    }    else if(++num[value2[v]] == 1) ans++;    appear[v] ^= 1;    v = father[v];}int result[maxm];int main(){    //freopen("C:\\Users\\creator\\Desktop\\in1.txt","r",stdin) ;    scanf("%d%d", &N, &M);    for(int i = 1; i <= N; i++)    {        scanf("%d", &value1[i]);    }    init();    for(int i = 1; i < N; i++)    {        int u, v;        scanf("%d%d", &u, &v);        g[u].push_back(v);        g[v].push_back(u);    }    LCA(1, 0);    initRmq();    initHash();    memset(visit, false, sizeof(visit));    dfs_block(1);    while(top) pos[sta[top--]] = nowblock;    for(int i = 1; i <= M; i++)    {        int u, v;        scanf("%d%d", &u, &v);        if(pos[u] > pos[v]) swap(u, v);        Query[i].l = pos[u];        Query[i].r = First[v];        Query[i].u = u;        Query[i].v = v;        Query[i].id = i;    }    sort(Query + 1, Query + M + 1);    L = 1;    R = 1;    ans = 0;    memset(visit, false, sizeof(visit));    for(int i = 1; i <= M; i++)    {        int u = Query[i].u;        int v = Query[i].v;        int id = Query[i].id;        int lca = getLca(u, v);        int lca1 = getLca(L, u);        int lca2 = getLca(R, v);        while(L != lca1) work(L);        while(u != lca1) work(u);        while(R != lca2) work(R);        while(v != lca2) work(v);        if(num[value2[lca]] == 0) result[id] = ans + 1;        else result[id] = ans;        L = Query[i].u;        R = Query[i].v;    }    for(int i = 1; i <= M; i++)    {        printf("%d\n", result[i]);    }    return 0;}