HYSBZ 2243 染色 (线段树+树链剖分)

题意：中文题。

析：真是一个好题，但是我TLE了两天，就是因为输入那个询问数，我当作边数了，结果就是一个TLE。。。

大体思路，就是先进行用树链剖分，然后用线段树来维护，维护每个区间的不同数的个数，和每个数的值，在求的时候，在两个端点进行判断，是不是同一种，如果是就减去1，不是则不变。

而且发现一个问题，就是网上的代码所以输出的和我的不一样，但是都AC了，不知道是不是数据水。也不知道谁的对。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 50;
const int mod = 1000;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r > 0 && r <= n && c > 0 && c <= m;
}

struct Edge{
  int to, next;
};
Edge edge[maxn<<1];
int head[maxn], tot, top[maxn], dep[maxn], a[maxn], fp[maxn];
int son[maxn], fa[maxn], num[maxn], p[maxn], pos;
int sum[maxn<<2], setv[maxn<<2], lx[maxn<<2], rx[maxn<<2], value[maxn<<2];

void init(){
  tot = pos = 0;
  ms(head, -1);  ms(son, -1);
  ms(setv, -1);
}

void addEdge(int u, int v){
  edge[tot].to = v;
  edge[tot].next = head[u];
  head[u] = tot++;
}

void dfs1(int u, int f, int d){
  dep[u] = d;  fa[u] = f;
  num[u] = 1;
  for(int i = head[u]; ~i; i = edge[i].next){
    int v = edge[i].to;
    if(v == f)  continue;
    dfs1(v, u, d+1);
    num[u] += num[v];
    if(son[u] == -1 || num[son[u]] < num[v])  son[u] = v;
  }
}

void dfs2(int u, int sp){
  top[u] = sp;  p[u] = ++pos;
  fp[pos] = u;
  if(son[u] == -1)  return ;
  dfs2(son[u], sp);
  for(int i = head[u]; ~i; i = edge[i].next){
    int v = edge[i].to;
    if(v == son[u] || v == fa[u])  continue;
    dfs2(v, v);
  }
}

void push_up(int rt){
  int l = rt<<1, r = rt<<1|1;
  sum[rt] = sum[l] + sum[r];
  if(rx[l] == lx[r])  --sum[rt];
  lx[rt] = lx[l];  rx[rt] = rx[r];
}

void build(int l, int r, int rt){
  if(l == r){
    value[rt] = a[fp[l]];
    sum[rt] = 1;
    lx[rt] = rx[rt] = value[rt];
    return ;
  }
  int m = l + r >> 1;
  build(lson);
  build(rson);
  pu(rt);
}

void push_down(int rt){
  if(setv[rt] < 0)  return ;
  int l = rt<<1, r = rt<<1|1;
  sum[l] = sum[r] = 1;
  value[l] = value[r] = setv[rt];
  lx[l] = rx[l] = lx[r] = rx[r] = setv[rt];
  setv[l] = setv[r] = setv[rt];
  setv[rt] = -1;
}

void update(int L, int R, int val, int l, int r, int rt){
  if(L <= l && r <= R){
    sum[rt] = 1;
    value[rt] = setv[rt] = val;
    lx[rt] = rx[rt] = val;
    return ;
  }
  pd(rt);
  int m = l + r >> 1;
  if(L <= m)  update(L, R, val, lson);
  if(R > m)   update(L, R, val, rson);
  pu(rt);
}

int queryVal(int M, int l, int r, int rt){
  if(l == r)  return value[rt];
  pd(rt);
  int m = l + r >> 1;
  if(M <= m)  return queryVal(M, lson);
  return queryVal(M, rson);
}

int query(int L, int R, int l, int r, int rt){
  if(L <= l && r <= R)  return sum[rt];
  pd(rt);
  int m = l + r >> 1;
  if(L <= m && R > m){
    int ans = query(L, R, lson) + query(L, R, rson);
    if(rx[rt<<1] == lx[rt<<1|1])  --ans;
    return ans;
  }
  else if(L <= m)  return query(L, R, lson);
  else  return query(L, R, rson);
}

void update(int u, int v, int c){
  int f1 = top[u], f2 = top[v];
  while(f1 != f2){
    if(dep[f1] < dep[f2]){
      swap(f1, f2);
      swap(u, v);
    }
    update(p[f1], p[u], c, all);
    u = fa[f1];
    f1 = top[u];
  }
  if(dep[u] > dep[v])  swap(u, v);
  update(p[u], p[v], c, all);
}

int query(int u, int v){
  int f1 = top[u], f2 = top[v];
  int ans = 0;
  while(f1 != f2){
    if(dep[f1] < dep[f2]){
      swap(u, v);
      swap(f1, f2);
    }
    ans += query(p[f1], p[u], all);
    if(queryVal(p[f1], all) == queryVal(p[fa[f1]], all))  --ans;
    u = fa[f1];
    f1 = top[u];
  }
  if(dep[u] > dep[v])  swap(u, v);
  ans += query(p[u], p[v], all);
  return ans;
}

int main(){
  scanf("%d %d", &n, &m);
  init();
  for(int i = 1; i <= n; ++i)  scanf("%d", a+i);
  for(int i = 1; i < n; ++i){
    int u, v;
    scanf("%d %d", &u, &v);
    addEdge(u, v);
    addEdge(v, u);
  }
  dfs1(1, 0, 0);
  dfs2(1, 1);
  build(all);
  char op[5];
  while(m--){
    scanf("%s", op);
    int a, b, c;
    scanf("%d %d", &a, &b);
    if(op[0] == ‘C‘){
      scanf("%d", &c);
      update(a, b, c);
    }
    else  printf("%d\n", query(a, b));
  }
  return 0;
}