spoj cot2 树上莫队

COT2 - Count on a tree II

#tree

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.

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 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.

Output

For each operation, print its result.

Example

Input:8 2105 2 9 3 8 5 7 71 21 31 43 53 63 74 82 57 8

Output:44

题意：给一个树图，每个点的点权（比如颜色编号），m个询问，每个询问是一个区间[a,b]，图中两点之间唯一路径上有多少个不同点权（即多少种颜色）。n<40000，m<100000。

参考http://blog.csdn.net/kuribohG/article/details/41458639

树上莫队：

　　（1）DFS一次，对树进行分块，分成sqrt(n)块，每个点属于一个块。并记录每个点的DFS序。
　　（2）将m个询问区间用所属块号作为第一关键字，DFS序作为第二关键字进行排序。
　　（3）转移都是差不多的，靠具体问题分析转移方式。

转移：

用S(v, u)代表 v到u的路径上的结点的集合。

用root来代表根结点，用lca(v, u)来代表v、u的最近公共祖先。

那么

S(v, u) = S(root, v) xor S(root, u) xor lca(v, u)

其中xor是集合的对称差。

简单来说就是节点出现两次消掉。

lca很讨厌，于是再定义

T(v, u) = S(root, v) xor S(root, u)

观察将curV移动到targetV前后T(curV, curU)变化：

T(curV, curU) = S(root, curV) xor S(root, curU)

T(targetV, curU) = S(root, targetV) xor S(root, curU)

取对称差：

T(curV, curU) xor T(targetV, curU)= (S(root, curV) xor S(root, curU)) xor (S(root, targetV) xor S(root, curU))

由于对称差的交换律、结合律：

T(curV, curU) xor T(targetV, curU)= S(root, curV) xorS(root, targetV)

两边同时xor T(curV, curU)：

T(targetV, curU)= T(curV, curU) xor S(root, curV) xor S(root, targetV)

发现最后两项很爽……哇哈哈

T(targetV, curU)= T(curV, curU) xor T(curV, targetV)

也就是说，更新的时候，xor T(curV, targetV)就行了。

即，对curV到targetV路径(除开lca(curV, targetV))上的结点，将它们的存在性取反即可。

AC代码：

#include<iostream>#include<cstdio>#include<vector>#include<cstring>#include<algorithm>using namespace std;const int maxn = 4e4 + 5;const int BLOCKSIZE = 210;int tid[maxn], dep[maxn], fa[maxn], pos[maxn], st[maxn];//tid:dfs序,dep：深度,fa:父亲节点,pos:块位置int dfsclk, stn, blockclk;struct query{int left, right, id;query(int left,int right,int id):left(left),right(right),id(id){}bool operator <(const query &x) const{return tid[right] < tid[x.right];}};vector<query>v[BLOCKSIZE];struct node{int en, next;}edge[2*maxn];int num1, head[maxn];void init(){num1 = blockclk = dfsclk = stn = 0;memset(head, -1, sizeof(head));}void add(int st, int en){edge[num1].en = en;edge[num1].next = head[st];head[st] = num1++;}int dfs(int u, int p){tid[u] = ++dfsclk;dep[u] = dep[p] + 1;fa[u] = p;int sz = 0;for (int i = head[u];i != -1;i = edge[i].next){int v = edge[i].en;if (v == p) continue;sz += dfs(v, u);if (sz >= BLOCKSIZE){while (sz--)pos[st[--stn]] = blockclk;blockclk++;}}st[stn++] = u;return sz + 1;}int num[maxn], cnt;int val[maxn], t[maxn];int cross;bool in[maxn];void insert(int u){if (!num[val[u]]) cnt++;num[val[u]]++;}void del(int u){num[val[u]]--;if (!num[val[u]]) cnt--;}void inv(int u){if (in[u]){del(u);in[u] = false;}else{insert(u);in[u] = true;}}void move_up(int &u){if (!cross){if (in[u] && !in[fa[u]]) cross = u;else if (!in[u] && in[fa[u]]) cross = fa[u];}inv(u);u = fa[u];}void move_to(int u, int v){if (u == v) return;cross = 0;if (in[v]) cross = v;while (dep[u] > dep[v]) move_up(u);while (dep[u] < dep[v]) move_up(v);while (u != v){move_up(u);move_up(v);}inv(u);inv(cross);}int ans[100005];int main(){init();int n, m;scanf("%d%d", &n, &m);for (int i = 1;i <= n;i++){scanf("%d", &val[i]);t[i] = val[i];}sort(t + 1, t + n + 1);int maxm = unique(t + 1, t + n + 1) - t - 1;for (int i = 1;i <= n;i++)val[i] = lower_bound(t + 1, t + 1 + maxm, val[i]) - t;for (int i = 1;i < n;i++){int st, en;scanf("%d%d", &st, &en);add(st, en), add(en, st);}dfs(1, 0);while (stn--) pos[st[stn]] = blockclk;for (int i = 0;i < m;i++){int st, en;scanf("%d%d", &st, &en);if (tid[st] > tid[en]) swap(st, en);v[pos[st]].push_back(query(st, en, i));}cnt = 0;for (int i = 0;i < BLOCKSIZE;i++){if (!v[i].size()) continue;sort(v[i].begin(), v[i].end());int l = v[i][0].left;int r = l;insert(l);in[l] = true;for (int j = 0;j < (int)v[i].size();j++){move_to(l, v[i][j].left);move_to(r, v[i][j].right);l = v[i][j].left, r = v[i][j].right;ans[v[i][j].id] = cnt;}move_to(l, r);del(r);in[r] = false;}for (int i = 0;i < m;i++)printf("%d\n", ans[i]);//system("pause");return 0;}