P3950部落冲突

题面

(Solution:)

法一：LCT裸题

又好想又好码，只不过常数太大。

法二：树链剖分

每次断边将该边权的值++，连边--，然后边权化点权(给儿子)，询问就查询从x到y的路径上的边权和，树状数组套树链剖分维护.

(Source)

// luogu-judger-enable-o2
#include <stdio.h>
#include <vector>
#include <assert.h>
#include <ctype.h>
#include <set>
#include <cstring>

using namespace std;

namespace io {
    char buf[1<<21], *pos = buf, *end = buf;
    inline char getc() 
    { return pos == end && (end = (pos = buf) + fread(buf, 1, 1<<21, stdin), pos == end) ? EOF : *pos ++; }
    inline int rint() {
        register int x = 0, f = 1; register char c; 
        while (!isdigit(c = getchar())) if (c == '-') f = -1;
        while (x = (x << 1) + (x << 3) + (c ^ 48), isdigit(c = getchar()));
        return x * f;
    }
    template<typename T>
        inline bool chkmin(T &x, T y) { return x > y ? (x = y, 0) : 1; }
    template<typename T>
        inline bool chkmax(T &x, T y) { return x < y ? (x = y, 0) : 1; }
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define ____ debug("go")
#define rep(i, a, b) for (register int i = a; i <= b; ++ i)
#define dep(i, a, b) for (register int i = a; i >= b; -- i)
#define travel(i, u) for (register int i = head[u]; i; i = nxt[i])
#define mem(a, b) memset(a, b, sizeof a)
}
using io::rint;using io::chkmin;using io::chkmax;

const int N = 4e5 + 1;

int n, m;
int tot, ver[N<<1], head[N], nxt[N<<1];
void add(int u, int v) 
{ ver[++tot] = v, nxt[tot] = head[u], head[u] = tot; }

int seg[N], size[N], top[N], son[N], cnt, fa[N], dep[N];
void DFS1(int x, int f) {
    fa[x] = f;
    size[x] = 1;
    dep[x] = dep[f] + 1;
    travel(i, x) {
        if (ver[i] == f) continue;
        DFS1(ver[i], x);
        size[x] += size[ver[i]];
        if (size[son[x]] < size[ver[i]]) son[x] = ver[i];
    }
}
void DFS2(int u, int topf) {
    seg[u] = ++cnt;
    top[u] = topf;
    if (son[u]) DFS2(son[u], topf);
    else return;
    travel(i, u) {
        if (ver[i] != fa[u] && ver[i] != son[u]) DFS2(ver[i], ver[i]);
    }
}

struct BIT {
    int s[N];
    void insert(int x, int c) 
    { for (int i = x; i <= n; i += (i & -i)) s[i] += c; }
    int query(int x) 
    { int res = 0; for (int i = x; i; i -= (i & -i)) res += s[i]; return res; } 
} T;

vector<pair<int, int> > War;

int Query(int x, int y) {
    int ans = 0;
    while (top[x] != top[y]) {
        if (dep[top[x]] < dep[top[y]]) swap(x, y);//打挂的地方
        ans += T.query(seg[x]) - T.query(seg[top[x]] - 1);
        x = fa[top[x]];
    }
    if (dep[x] < dep[y]) swap(x, y);
    ans += T.query(seg[x]) - T.query(seg[y]);
    return ans;
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("P3950.in", "r", stdin);
    freopen("P3950.out", "w", stdout);
#endif
    n = rint(); m = rint();
    for (int i = 1; i < n; ++ i) {
        int  u = rint(), v = rint();
        add(u, v); add(v, u);
    }
    DFS1(1, 0);
    DFS2(1, 1);
    char op[10]; int x, y;
    while (m --) {
        scanf("%s", op);
        if (op[0] == 'Q') {
            x = rint(), y = rint();
            if (Query(x, y)) puts("No");
            else puts("Yes");
        } else if (op[0] == 'U') {
            x = rint();
            x --;
            int u = War[x].first, v = War[x].second;
            if (seg[u] < seg[v]) swap(u, v);
            T.insert(seg[u], -1);
        } else {
            x = rint(), y = rint();
            if (seg[x] < seg[y]) swap(x, y);
            T.insert(seg[x], 1);
            War.push_back(make_pair(x, y));
        }
    }
}