传送门：点击打开链接
题意：2种操作，一种路径上的值统一修改，一种是询问路径上数字连续区间个数。
思路：树链剖分+线段树合并。这道题目主要就是难在用树链剖分上套线段树合并后，因为整条链被分成了很多条短的，把这些短的也要按照顺序合并。
又因为其实是从左边和右边链的最底端向上执行的，所以应该把左右的分开，然后就是更接近根节点的点是区间的左区间，远离的点是右区间。
主要的问题就在于路径上的合并，我的代码这一部分写的比较搓。。
#include<map>
#include<set>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
#include<cstdio>
#include<cctype>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define fuck(x) cout<<"["<<x<<"]"
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w+",stdout)
using namespace std;
typedef long long LL;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1const int MX = 1e5 + 5;
const int INF = 0x3f3f3f3f;struct Edge {int v, nxt;
} E[MX << 2];
int Head[MX], h_r;
void edge_init() {h_r = 0;memset(Head, -1, sizeof(Head));
}
void edge_add(int u, int v) {E[h_r].v = v;E[h_r].nxt = Head[u];Head[u] = h_r++;
}int S[MX << 2], col[MX << 2], SL[MX << 2], SR[MX << 2], A[MX], B[MX];
void push_up(int rt, int m) {S[rt] = S[rt << 1] + S[rt << 1 | 1];SL[rt] = SL[rt << 1]; SR[rt] = SR[rt << 1 | 1];if(SR[rt << 1] == SL[rt << 1 | 1]) S[rt]--;
}
void push_down(int rt) {if(col[rt] != -1) {col[rt << 1] = col[rt << 1 | 1] = col[rt];SL[rt << 1] = SR[rt << 1] = col[rt];SL[rt << 1 | 1] = SR[rt << 1 | 1] = col[rt];S[rt << 1] = S[rt << 1 | 1] = 1;col[rt] = -1;}
}
void build(int l, int r, int rt) {col[rt] = -1;if(l == r) {S[rt] = 1;SL[rt] = SR[rt] = A[l];return;}int m = (l + r) >> 1;build(lson);build(rson);push_up(rt, m);
}
void update(int L, int R, int x, int l, int r, int rt) {if(L <= l && r <= R) {S[rt] = 1;SL[rt] = SR[rt] = col[rt] = x;return;}push_down(rt);int m = (l + r) >> 1;if(L <= m) update(L, R, x, lson);if(R > m) update(L, R, x, rson);push_up(rt, m);
}
int query(int L, int R, int l, int r, int rt, int &la, int &ra) {if(L <= l && r <= R) {la = SL[rt]; ra = SR[rt];return S[rt];}push_down(rt);int m = (l + r) >> 1, ret = 0;int lla = -1, lra = -1, rla = -1, rra = -1;if(L <= m) ret += query(L, R, lson, lla, lra);if(R > m) ret += query(L, R, rson, rla, rra);if(L <= m && m < R && SR[rt << 1] == SL[rt << 1 | 1]) ret--;la = lla != -1 ? lla : rla;ra = rra != -1 ? rra : lra;return ret;
}int fa[MX], top[MX], siz[MX], son[MX], dep[MX], id[MX], rear;
/*fa父节点，top重链开头起点，siz子树大小，son重儿子，dep深度，id新编号*//*第一次DFS找到重边，并维护好siz,son,fa,dep*/
void DFS1(int u, int f, int d) {fa[u] = f; dep[u] = d;son[u] = 0; siz[u] = 1;for(int i = Head[u]; ~i; i = E[i].nxt) {int v = E[i].v;if(v == f) continue;DFS1(v, u, d + 1);siz[u] += siz[v];if(siz[son[u]] < siz[v]) {son[u] = v;}}
}
/*将重边编号好，维护id*/
void DFS2(int u, int tp) {top[u] = tp;id[u] = ++rear;if(son[u]) DFS2(son[u], tp);for(int i = Head[u]; ~i; i = E[i].nxt) {int v = E[i].v;if(v == fa[u] || v == son[u]) continue;DFS2(v, v);}
}
/*用来给点编号，以及建立线段树，要用id编号建树*/
void HLD_presolve() {rear = 0;DFS1(1, 0, 1);DFS2(1, 1);for(int i = 1; i <= rear; i++) {A[id[i]] = B[i];}build(1, rear, 1);
}
/*修改，要注意使用id编号修改*/
void HLD_update(int u, int v, int d) {int tp1 = top[u], tp2 = top[v];while(tp1 != tp2) {if(dep[tp1] < dep[tp2]) {swap(u, v);swap(tp1, tp2);}update(id[tp1], id[u], d, 1, rear, 1);u = fa[tp1]; tp1 = top[u];}if(dep[u] > dep[v]) swap(u, v);update(id[u], id[v], d, 1, rear, 1);
}int HLD_query(int u, int v) {int tp1 = top[u], tp2 = top[v], ret = 0;int lastla[2] = { -1, -1}, lastra[2] = { -1, -1}, la[2] = { -1, -1}, ra[2] = { -1, -1};while(tp1 != tp2) {if(dep[tp1] >= dep[tp2]) {ret += query(id[tp1], id[u], 1, rear, 1, la[0], ra[0]);if(ra[0] == lastla[0]) ret--;lastla[0] = la[0]; lastra[0] = ra[0];u = fa[tp1]; tp1 = top[u];} else {ret += query(id[tp2], id[v], 1, rear, 1, la[1], ra[1]);if(ra[1] == lastla[1]) ret--;lastla[1] = la[1]; lastra[1] = ra[1];v = fa[tp2]; tp2 = top[v];}}if(dep[u] <= dep[v]) {ret += query(id[u], id[v], 1, rear, 1, la[1], ra[1]);if(ra[1] == lastla[1]) ret--;if(la[1] == lastla[0]) ret--;} else {ret += query(id[v], id[u], 1, rear, 1, la[0], ra[0]);if(ra[0] == lastla[0]) ret--;if(la[0] == lastla[1]) ret--;}return ret;
}int main() {int n, m; //FIN;while(~scanf("%d%d", &n, &m)) {edge_init();for(int i = 1; i <= n; i++) {scanf("%d", &B[i]);}for(int i = 1; i <= n - 1; i++) {int u, v;scanf("%d%d", &u, &v);edge_add(u, v);edge_add(v, u);}HLD_presolve();while(m--) {char op[10];int a, b, c;scanf("%s", op);if(op[0] == 'Q') {scanf("%d%d", &a, &b);printf("%d\n", HLD_query(a, b));} else {scanf("%d%d%d", &a, &b, &c);HLD_update(a, b, c);}}}return 0;
}