bzoj3637 Query on a tree VI

Posted 2020-09-16 逢山开路 遇水架桥

传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=3637

【题解】

LCT维护连通性问题。

发现如果直接按题目说的复杂度就是错的了（因为依赖于度数）

所以我们分黑白建树，黑的里存白->黑和黑->黑两种边，白的类似。

那么每次只会改变两棵树内的各一个。

因为LCT维护连通性所以需要记录除了主链外的其他size。

在access的时候即可记录。

这个link和cut很迷啊暂时没看懂啊（逃）参考别人代码

留坑

# include <stdio.h>
# include <string.h>
# include <algorithm>
// # include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int M = 2e5 + 10;
const int mod = 1e9+7;

# define RG register
# define ST static

int n, head[M], nxt[M], to[M], tot=0;
bool color[M]; 
inline void add(int u, int v) {
    ++tot; nxt[tot] = head[u]; head[u] = tot; to[tot] = v;
}
inline void adde(int u, int v) {add(u, v), add(v, u);} 

struct LCT {
    int co, ch[M][2], fa[M], sz[M], SZ[M];
    bool rev[M];
    inline void change(int x, int d) {
        sz[x] += d;
        SZ[x] += d;
    }
    inline void set(int n, int _co) {
        co = _co;
        memset(ch, 0, sizeof ch);
        memset(fa, 0, sizeof fa);
        memset(rev, 0, sizeof rev); 
        memset(sz, 0, sizeof sz);
        memset(SZ, 0, sizeof SZ);
        if(!co) for (int i=1; i<=n; ++i) change(i, 1);
    }
    # define ls ch[x][0]
    # define rs ch[x][1]
    inline void up(int x) {
        if(!x) return;
        sz[x] = sz[ls] + sz[rs] + SZ[x];
    }
    # undef ls
    # undef rs 
    inline bool isrt(int x) {
        return ch[fa[x]][0] != x && ch[fa[x]][1] != x;
    }
    inline void rotate(int x) {
        int y = fa[x], z = fa[y], ls = ch[y][1] == x, rs = ls^1;
        if(!isrt(y)) ch[z][ch[z][1] == y] = x;
        fa[ch[x][rs]] = y, fa[y] = x, fa[x] = z; 
        ch[y][ls] = ch[x][rs]; ch[x][rs] = y;
        up(y); up(x);
    }
    inline void splay(int x) {
        while(!isrt(x)) {
            int y = fa[x], z = fa[y];
            if(!isrt(y)) {
                if((ch[y][0] == x)^(ch[z][0] == y)) rotate(x);
                else rotate(y);
            }
            rotate(x);
        }
    }
    inline int access(int x) {
        int t = 0; 
        for (; x; t=x, x=fa[x]) {
            splay(x);
            if(ch[x][1]) SZ[x] += sz[ch[x][1]];
            if(t) SZ[x] -= sz[t];
            ch[x][1] = t;
            up(x);
        }
        return t;
    }
    inline void link(int x, int y) {
        if(!y) return;
        access(y); splay(y);
        splay(x); fa[x] = y; ch[y][1] = x;
        up(y);
    }
    inline void cut(int x, int y) {
        if(!y) return;
        access(x); splay(x);
        fa[ch[x][0]] = 0, ch[x][0] = 0;
        up(x);
    }
    inline int find(int x) {
        access(x); splay(x);
        while(ch[x][0]) x = ch[x][0];
        return x;
    }
    inline void getans(int x) {
        int t = find(x); splay(t);
        if(color[t] == co) printf("%d\n", sz[t]);
        else printf("%d\n", sz[ch[t][1]]);
    }
}T[2];
// T[0]: black, T[1]: white
// color0: black, color1: white

int fa[M]; 
inline void dfs(int x, int father) {
    fa[x] = father; 
    for (int i=head[x]; i; i=nxt[i]) {
        if(to[i] == father) continue;
        T[0].link(to[i], x);
        dfs(to[i], x);
    }
}

int main() {
    scanf("%d", &n);
    for (int i=1, u, v; i<n; ++i) {
        scanf("%d%d", &u, &v);
        adde(u, v);
    }
    T[0].set(n, 0); T[1].set(n, 1); 
    dfs(1, 0); 
//    for (int i=1; i<=n; ++i) T[0].getans(i); puts("\n==============="); 
    int q;
    scanf("%d", &q);
    int opt, x;
    while(q--) {
        scanf("%d%d", &opt, &x); 
        if(opt == 0) T[color[x]].getans(x);
        else {
            T[color[x]].cut(x, fa[x]); 
            T[color[x]].change(x, -1);
            color[x] ^= 1;
            T[color[x]].change(x, 1);
            T[color[x]].link(x, fa[x]);
        }
    }

    return 0;
}

View Code