bzoj 2243 染色(树链剖分)

题外话

首先这是个挺裸的题，由于太久没写剖分导致调了好久，前天调了一下午，一直查不到错

昨晚在看春晚的时候突然灵机一动，发现合并的时候出了问题，开电脑把它A掉了= =

感觉自己也蛮拼的

Description

$给定一棵有n个节点的无根树和m个操作，操作有2类：$

$1:将节点a到节点b路径上所有点都染成颜色c$

$2:询问节点a到节点b路径上的颜色段数量(连续相同颜色被认为是同一段)，如"112221"由3段组成:"11"、"222"和"1"$

Solution

很显然这是要剖分的辣，然后用线段树维护信息。然后我们会发现合并的时候有点蛋疼

它要求颜色段数量，所以我们很自然的想到要维护左右端的颜色信息来进行合并

线段树合并时$sum[rt] = sum[ls] + sum[rs] - (rc[ls] == lc[rs])$即可

但是我们正常在查询的时候是要在剖分后的树上”跳来跳去的”

所以每次查询的时候我们都需要维护当前左右端颜色信息，然后在”跳”的过程中合并

注意颜色是$[0,10^9]的， 所以打标记时要开始赋值为-1$

合并的时候想清楚，注意细节就好辣！可以仔细想想或者详情参见代码QvQ

Code

#include <bits/stdc++.h>//树链剖分
using namespace std;
#define ls (rt << 1)
#define rs (rt << 1 | 1)
const int N = 100005, M = N << 2;
int n, m, lcol, rcol, tot, cnt, w[N], q[N], top[N], sz[N], son[N], pre[N], id[N], dep[N], col[N], lc[M], rc[M], sum[M], mark[M], to[M], nxt[M], head[N];
bool vis[N];
inline int readInt() {
    char c;
    while (c = getchar(), c < ‘0‘ || c > ‘9‘);
    int t = c - ‘0‘;
    while (c = getchar(), c >= ‘0‘ && c <= ‘9‘) t = t * 10 + c - ‘0‘;
    return t;
}
void add(int u, int v) {
    to[tot] = v, nxt[tot] = head[u], head[u] = tot++;
    to[tot] = u, nxt[tot] = head[v], head[v] = tot++;
}
void pushdown(int rt) {
    if (~mark[rt]) {
        lc[ls] = lc[rs] = rc[ls] = rc[rs] = mark[ls] = mark[rs] = mark[rt];
        sum[ls] = sum[rs] = 1;
        mark[rt] = -1;
    }
}
void pushup(int rt) {
    lc[rt] = lc[ls], rc[rt] = rc[rs];
    sum[rt] = sum[ls] + sum[rs] - (rc[ls] == lc[rs]);
}
void build(int rt, int l, int r) {
    mark[rt] = -1;
    if (l == r) {
        lc[rt] = rc[rt] = w[l];
        sum[rt] = 1;
        return ;
    }
    int mid = l + r >> 1;
    build(ls, l, mid);
    build(rs, mid + 1, r);
    pushup(rt);
}
void change(int rt, int l, int r, int L, int R, int c) {
    if (L <= l && R >= r) {
        lc[rt] = rc[rt] = mark[rt] = c;
        sum[rt] = 1;
        return;
    }
    pushdown(rt);
    int mid = l + r >> 1;
    if (L <= mid)   change(ls, l, mid, L, R, c);
    if (R > mid)    change(rs, mid + 1, r, L, R, c);
    pushup(rt);
}
int ask(int rt, int l, int r, int L, int R) {
    if (L == l) lcol = lc[rt];
    if (R == r) rcol = rc[rt];
    if (L <= l && R >= r)   return sum[rt];
    pushdown(rt);
    int mid = l + r >> 1, t = 0, t1 = -1, t2 = -1;
    if (L <= mid)   t += ask(ls, l, mid, L, R), t1 = rc[ls];
    if (R > mid)    t += ask(rs, mid + 1, r, L, R), t2 = lc[rs];
    t -= (t1 == t2 && ~t1);
    return t;
}
void work(int a, int b, int c) {
    while (top[a] != top[b]) {
        if (dep[top[a]] < dep[top[b]])  swap(a, b);
        change(1, 1, n, id[top[a]], id[a], c);
        a = pre[top[a]];    
    }
    if (dep[a] < dep[b])    swap(a, b);
    change(1, 1, n, id[b], id[a], c);
}
int work2(int a, int b) {
    int t = 0, t1 = -1, t2 = -1;
    while (top[a] != top[b]) {
        if (dep[top[a]] < dep[top[b]])  swap(a, b), swap(t1, t2);
        t += ask(1, 1, n, id[top[a]], id[a]);
        t -= (t1 == rcol && ~t1);
        t1 = lcol;
        a = pre[top[a]];
    }
    if (dep[a] < dep[b])    swap(a, b), swap(t1, t2);
    t += ask(1, 1, n, id[b], id[a]);
    t -= ((t1 == rcol && ~t1) + (t2 == lcol && ~t2));
    return t;
}
void gao() {
    int r = 0;
    vis[dep[1] = q[0] = 1] = 1;
    for (int i = 0; i <= r; ++i)
        for (int j = head[q[i]]; ~j; j = nxt[j]) 
            if (!vis[to[j]]){
                vis[to[j]] = 1;
                dep[q[++r] = to[j]] = dep[q[i]] + 1;
                pre[q[r]] = q[i];
            }
    for (int i = r; i >= 0; --i) {
        sz[pre[q[i]]] += ++sz[q[i]];
        if (sz[son[pre[q[i]]]] < sz[q[i]])  son[pre[q[i]]] = q[i];
    }
    for (int i = 0; i <= r; ++i)
        if (!top[q[i]]) {
            for (int j = q[i]; j; j = son[j]) {
                top[j] = q[i];
                w[id[j] = ++cnt] = col[j];
            }
        }
    build(1, 1, n);
}
int main() {
    n = readInt(), m = readInt();
    memset(head, -1, sizeof(head));
    for (int i = 1; i <= n; ++i)    col[i] = readInt();
    for (int i = 1, x, y; i < n; ++i) {
        x = readInt(), y = readInt();
        add(x, y);
    }
    gao();
    while (m--) {
        char s[5];
        scanf("%s", s);
        int a, b, c;
        if (s[0] == ‘C‘) {
            a = readInt(), b = readInt(), c = readInt();
            work(a, b, c);
        }
        else {
            a = readInt(), b = readInt();
            printf("%d\n", work2(a, b));
        }
    }
    return 0;
}