bzoj2959: 长跑(LCT+并查集)

题解

动态树Link-cut tree(LCT)总结

LCT常数大得真实

没有环,就是\(lct\)裸题吧

有环,我们就可以绕环转一圈,缩点

怎么搞?

当形成环时,把所有点的值全部加到一个点上,用并查集维护加到哪个点上

判断连通性再用一个并查集

Code

#include<bits/stdc++.h>

#define LL long long
#define RG register

using namespace std;
template<class T> inline void read(T &x) {
    x = 0; RG char c = getchar(); bool f = 0;
    while (c != '-' && (c < '0' || c > '9')) c = getchar(); if (c == '-') c = getchar(), f = 1;
    while (c >= '0' && c <= '9') x = x*10+c-48, c = getchar();
    x = f ? -x : x;
    return ;
}
template<class T> inline void write(T x) {
    if (!x) {putchar(48);return ;}
    if (x < 0) x = -x, putchar('-');
    int len = -1, z[20]; while (x > 0) z[++len] = x%10, x /= 10;
    for (RG int i = len; i >= 0; i--) putchar(z[i]+48);return ;
}
const int N = 150010;
struct node {
    int v, s, fa, ch[2];
    bool rev;
}t[N];
int S[N], top;
#define lson (t[x].ch[0])
#define rson (t[x].ch[1])
void pushup(int x) {t[x].s = (t[lson].s + t[rson].s + t[x].v);return ;}
void putrev(int x) {
    swap(lson, rson);
    t[x].rev ^= 1;
    return ;
}
void pushdown(int x) {
    if (t[x].rev) {
        if (lson) putrev(lson);
        if (rson) putrev(rson);
        t[x].rev = 0;
    }
}
int fa[N], Fa[N];
int find(int x) {
    return fa[x] == x ? x : fa[x] = find(fa[x]);
}
int Find(int x) {
    return Fa[x] == x ? x : Fa[x] = Find(Fa[x]);
}
bool isroot(int x) {
    return (t[find(t[x].fa)].ch[0] != x) && (t[find(t[x].fa)].ch[1] != x);
}

void rotate(int x) {
    int y = find(t[x].fa), z = find(t[y].fa), k = t[y].ch[1] == x;
    if (!isroot(y)) t[z].ch[t[z].ch[1] == y] = x;
    t[x].fa = z;
    t[t[x].ch[k^1]].fa = y, t[y].ch[k] = t[x].ch[k^1];
    t[x].ch[k^1] = y; t[y].fa = x;
    pushup(y);
    return ;
}

void splay(int x) {
    S[top = 1] = x;
    for (RG int i = x; !isroot(i); i = find(t[i].fa)) S[++top] = find(t[i].fa);
    for (RG int i = top; i; i--) pushdown(S[i]);
    while (!isroot(x)) {
        int y = find(t[x].fa), z = find(t[y].fa);
        if (!isroot(y))
            (t[y].ch[1] == x) ^ (t[z].ch[1] == y) ? rotate(x) : rotate(y);
        rotate(x);
    }
    pushup(x);
    return ;
}

void access(int x) {
    for (int y = 0; x; y = x, x = find(t[x].fa))
        splay(x), t[x].ch[1] = y, pushup(x);
}

void makeroot(int x) {
    access(x); splay(x); putrev(x);
    return ;
}

void link(int x, int y) {
    makeroot(x);
    t[x].fa = y;
}

void split(int x, int y) {
    makeroot(x), access(y), splay(y);
    return ;
}
int a[N], n, m;

void dfs(int x, int y) {
    fa[x] = y;
    pushdown(x);
    if (lson) dfs(lson, y);
    if (rson) dfs(rson, y);
    return ;
}

int main() {
    read(n), read(m);
    for (int i = 1; i <= n; i++) read(a[i]), fa[i] = Fa[i] = i, t[i].s = t[i].v = a[i];
    while (m--) {
        int x, y, op;
        read(op), read(x), read(y);
        if (op == 1) {
            x = find(x), y = find(y);
            if (x == y) continue;
            if (Find(x) != Find(y)) {
                link(x, y);
                Fa[Find(y)] = Find(x);
            }
            else {
                split(x, y);
                t[y].v = t[y].s;
                dfs(y, y);
                t[y].ch[0] = 0;
            }
        }
        else if (op == 2) {
            int tx = find(x);
            splay(tx);
            t[tx].s += y - a[x];
            t[tx].v += y - a[x];
            a[x] = y;
        }
        else {
            x = find(x), y = find(y);
            if (Find(x) != Find(y)) {
                puts("-1");
                continue;
            }
            split(x, y);
            printf("%d\n", t[y].s);
        }
    }
    return 0;
}

原文地址：https://www.cnblogs.com/zzy2005/p/10324298.html