Description

題庫鏈接

給定一棵有 \(n\) 個節點的無根樹和 \(m\) 個操作，操作有 \(2\) 類：

1. 將節點 \(a\) 到節點 \(b\) 路徑上所有點都染成顏色 \(c\) ；
2. 詢問節點 \(a\) 到節點 \(b\) 路徑上的顏色段數量（連續相同顏色被認為是同一段）

Solution

線段樹茍題。因為沒有下傳 \(lazy\) 標記調了一上午。

Code

//It is made by Awson on 2018.3.4
#include <bits/stdc++.h>
#define LL long long
#define dob complex<double>
 

#define Abs(a) ((a) < 0 ? (-(a)) : (a))
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
#define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))
#define writeln(x) (write(x), putchar('\n'))
#define lowbit(x) ((x)&(-(x)))
using namespace std;
const
 
 int N = 1e5;
void read(int &x) {
    char ch; bool flag = 0;
    for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());
    for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());
    x *= 1-2*flag;
}
void print(int x) {if (x > 9
 
) print(x/10); putchar(x%10+48); }
void write(int x) {if (x < 0) putchar('-'); print(Abs(x)); }

int n, m, c[N+5], a[N+5], u, v, ca;
struct tt {int to, next; }edge[(N<<1)+5];
int path[N+5], Top;
int size[N+5], id[N+5], son[N+5], top[N+5], dep[N+5], fa[N+5], pos;
char ch[5];
struct node {
    int l, r, cnt;
    node() {}
    node(int _l, int _r, int _cnt) {l = _l, r = _r, cnt = _cnt; }
    node operator + (const node &b) const {node tmp; tmp.l = l, tmp.r = b.r, tmp.cnt = cnt+b.cnt-(r==b.l); return tmp; }
};
struct Segment_tree {
#define lr(o) (o<<1)
#define rr(o) (o<<1|1)
    node sgm[(N<<2)+5]; int lazy[(N<<2)+5];
    void pushdown(int o) {sgm[lr(o)] = sgm[rr(o)] = sgm[o]; lazy[lr(o)] = lazy[rr(o)] = 1; lazy[o] = 0; }
    void build(int o, int l, int r) {
    if (l == r) {sgm[o] = node(a[l], a[l], 1); return; }
    int mid = (l+r)>>1;
    build(lr(o), l, mid); build(rr(o), mid+1, r); sgm[o] = sgm[lr(o)]+sgm[rr(o)];
    }
    node query(int o, int l, int r, int a, int b) {
    if (a <= l && r <= b) return sgm[o];
    if (lazy[o]) pushdown(o); int mid = (l+r)>>1;
    if (b <= mid) return query(lr(o), l, mid, a, b);
    if (a > mid) return query(rr(o), mid+1, r, a, b);
    return query(lr(o), l, mid, a, b)+query(rr(o), mid+1, r, a, b);
    }
    void update(int o, int l, int r, int a, int b, int col) {
    if (a <= l && r <= b) {sgm[o] = node(col, col, 1), lazy[o] = 1; return; }
    if (lazy[o]) pushdown(o); int mid = (l+r)>>1;
    if (a <= mid) update(lr(o), l, mid, a, b, col);
    if (b > mid) update(rr(o), mid+1, r, a, b, col);
    sgm[o] = sgm[lr(o)]+sgm[rr(o)];
    }
}T;

void add(int u, int v) {edge[++Top].to = v, edge[Top].next = path[u], path[u] = Top; }
void dfs1(int o, int depth, int father) {
    dep[o] = depth, size[o] = 1, fa[o] = father;
    for (int i = path[o]; i; i = edge[i].next)
    if (dep[edge[i].to] == 0) {
        dfs1(edge[i].to, depth+1, o); size[o] += size[edge[i].to];
        if (size[edge[i].to] > size[son[o]]) son[o] = edge[i].to;
    }
}
void dfs2(int o, int tp) {
    id[o] = ++pos, a[pos] = c[o], top[o] = tp;
    if (son[o]) dfs2(son[o], tp);
    for (int i = path[o]; i; i = edge[i].next)
    if (edge[i].to != fa[o] && edge[i].to != son[o]) dfs2(edge[i].to, edge[i].to);
}
void update(int u, int v, int c) {
    while (top[u] != top[v]) {
    if (dep[top[u]] < dep[top[v]]) Swap(u, v);
    T.update(1, 1, n, id[top[u]], id[u], c);
    u = fa[top[u]];
    }
    if (dep[u] < dep[v]) Swap(u, v);
    T.update(1, 1, n, id[v], id[u], c);
}
int query(int u, int v) {
    node n1, n2; int f1 = 1, f2 = 1;
    while (top[u] != top[v]) {
    if (dep[top[u]] > dep[top[v]]) {
        if (f1) n1 = T.query(1, 1, n, id[top[u]], id[u]);
        else n1 = T.query(1, 1, n, id[top[u]], id[u])+n1;
        u = fa[top[u]]; f1 = 0;
    }else {
        if (f2) n2 = T.query(1, 1, n, id[top[v]], id[v]);
        else n2 = T.query(1, 1, n, id[top[v]], id[v])+n2;
        v = fa[top[v]]; f2 = 0;
    }
    }
    if (dep[u] > dep[v]) {
    if (f1) n1 = T.query(1, 1, n, id[v], id[u]);
    else n1 = T.query(1, 1, n, id[v], id[u])+n1; f1 = 0;
    }else {
    if (f2) n2 = T.query(1, 1, n, id[u], id[v]);
    else n2 = T.query(1, 1, n, id[u], id[v])+n2; f2 = 0;
    }
    if (f1) return n2.cnt;
    if (f2) return n1.cnt;
    Swap(n1.l, n1.r); n1 = n1+n2; return n1.cnt;
}
void work() {
    read(n), read(m); for (int i = 1; i <= n; i++) read(c[i]);
    for (int i = 1; i < n; i++) read(u), read(v), add(u, v), add(v, u);
    dfs1(1, 1, 0), dfs2(1, 1); T.build(1, 1, n);
    while (m--) {
    scanf("%s", ch);
    if (ch[0] == 'Q') read(u), read(v), writeln(query(u, v));
    else read(u), read(v), read(ca), update(u, v, ca);
    }
}
int main() {
    work(); return 0;
}

[SDOI 2011]染色