洛谷 P1501 [国家集训队]Tree II

题解

LCT的模板题之一。

像《线段树2》一样，先打乘法标记再打加法标记。

注意 \(51061^2 = 2607225721\)，会爆 int 。可以使用 uint，不过还是推荐 long long。

代码

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;
const int maxn = 100000 + 5;

const int mod = 51061;

namespace LCT
{
    // PartI. Splay
    int fa[maxn], ch[2][maxn], size[maxn], st[maxn], sz; ll val[maxn], sum[maxn], tagsum[maxn], tagmul[maxn]; bool flip[maxn];
    bool inline notroot(int o) {
        return (ch[0][fa[o]] == o) || (ch[1][fa[o]] == o);
    }
    void inline settag(int o, ll add, ll mul) {
        val[o] = (val[o] * mul + add) % mod;
        sum[o] = (sum[o] * mul + size[o] * add) % mod;
        tagsum[o] = (tagsum[o] * mul + add) % mod;
        tagmul[o] = (tagmul[o] * mul) % mod;
    }
    void inline pushdown(int o) {
        if(flip[o]) {
            if(ch[0][o]) flip[ch[0][o]] ^= 1;
            if(ch[1][o]) flip[ch[1][o]] ^= 1;
            swap(ch[0][o], ch[1][o]);
            flip[o] = false;
        }
        if(ch[0][o]) settag(ch[0][o], tagsum[o], tagmul[o]);
        if(ch[1][o]) settag(ch[1][o], tagsum[o], tagmul[o]);
        tagmul[o] = 1, tagsum[o] = 0;
    }
    void inline pushup(int o) {
        sum[o] = (sum[ch[0][o]] + sum[ch[1][o]] + val[o]) % mod;
        size[o] = size[ch[0][o]] + size[ch[1][o]] + 1;
    }
    void inline rotate(int x) {
        int y = fa[x], z = fa[y], d = ch[1][y] == x;
        if(notroot(y)) ch[y == ch[1][z]][z] = x; fa[x] = z;
        ch[d][y] = ch[d^1][x]; if(ch[d][y]) fa[ch[d][y]] = y; ch[d^1][x] = y; fa[y] = x;
        pushup(y); pushup(x);
    }
    void inline splay(int x) {
        int o = x;
        st[sz = 1] = o;
        while(notroot(o)) st[++sz] = o = fa[o];
        while(sz) pushdown(st[sz--]);
        while(notroot(x)) {
            int y = fa[x], z = fa[y];
            if(notroot(y)) rotate(((ch[0][z] == y) ^ (ch[0][y] == x)) ? x : y);
            rotate(x);
        }
        pushup(x);
    }
    // PartII. LCT
    void inline access(int x) {
        for(int y = 0; x; y = x, x = fa[x]) {
            splay(x);
            ch[1][x] = y;
            pushup(x);
        }
    }
    void inline makeroot(int x) {
        access(x); 
        splay(x);
        flip[x] ^= 1;
    }
    void inline split(int x, int y) {
        makeroot(x);
        access(y); splay(y);
    }
    void inline link(int x, int y) {
        makeroot(x);
        fa[x] = y;
    }
    void inline cut(int x, int y) {
        split(x, y);
        fa[x] = ch[0][y] = 0;
        pushup(y);
    }
}

int n, m;

void inline Init()
{
    scanf("%d %d", &n, &m);
    for(int i = 1; i <= n; ++i) {
        LCT::val[i] = 1;
        LCT::tagmul[i] = 1;
        LCT::sum[i] = 1;
        LCT::size[i] = 1;
    }
    int u, v;
    for(int i = 1; i < n; ++i) {
        scanf("%d %d", &u, &v);
        LCT::link(u, v);
    }
}

char opt[5]; int u, v, c, u2, v2;

using namespace LCT;

void inline Solve()
{
    while(m--) {
        scanf("%s %d %d", opt, &u, &v);
        if(opt[0] == '+') {
            scanf("%d", &c);
            split(u, v);
            settag(v, c, 1);
        } else if(opt[0] == '*') {
            scanf("%d", &c);
            split(u, v);
            settag(v, 0, c);
        } else if(opt[0] == '-') {
            scanf("%d %d", &u2, &v2);
            cut(u, v);
            link(u2, v2);
        } else {
            split(u, v);
            printf("%lld\n", sum[v]);
        }
    }
}

int main()
{
    Init();
    Solve();
    return 0;
}