题目链接

CF487E

题解

圆方树 + 树剖 裸题

建好圆方树维护路径上最小值即可

方点的值为其儿子的最小值，这个用堆维护

为什么只维护儿子？因为这样修改点的时候就只需要修改其父亲的堆

这样充分利用了一对一的特性优化了复杂度

如此询问时如果\(lca\)为方点，再询问一下\(lca\)的父亲即可

复杂度\(O(qlog^2n)\)

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<cmath>
#include<vector>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
#define ls (u << 1)
#define rs (u << 1 | 1)
using namespace std;
const int maxn = 200005,maxm = 100005,INF = 1000000000;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == ‘-‘) flag = -1; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
int hh[maxn],nne = 1,h[maxn],ne = 1;
struct EDGE{int to,nxt;}ed[maxn << 1],e[maxn << 1];
inline void build(int u,int v){
    e[++nne] = (EDGE){v,hh[u]}; hh[u] = nne;
    e[++nne] = (EDGE){u,hh[v]}; hh[v] = nne;
}
inline void add(int u,int v){
    ed[++ne] = (EDGE){v,h[u]}; h[u] = ne;
    ed[++ne] = (EDGE){u,h[v]}; h[v] = ne;
}
struct HEAP{
    priority_queue<int,vector<int>,greater<int> > a,b;
    void ck(){while (!b.empty() && a.top() == b.top()) a.pop(),b.pop();}
    int size(){return a.size() - b.size();}
    void ins(int x){a.push(x);}
    void del(int x){b.push(x);}
    int top(){ck(); return size() ? a.top() : INF;}
}H[maxn];
int n,m,q,N,w[maxn];
int dfn[maxn],low[maxn],st[maxn],Top,cnt;
void dfs(int u,int las){
    dfn[u] = low[u] = ++cnt; st[++Top] = u;
    for (int k = hh[u],to; k; k = e[k].nxt)
        if (k != las){
            if (!dfn[to = e[k].to]){
                dfs(to,k ^ 1);
                low[u] = min(low[u],low[to]);
                if (low[to] >= dfn[u]){
                    add(++N,u);
                    do{add(N,st[Top]);} while (st[Top--] != to);
                }
            }
            else low[u] = min(low[u],dfn[to]);
        }
}
int siz[maxn],top[maxn],dep[maxn],fa[maxn],son[maxn],id[maxn],Hash[maxn],Cnt;
void dfs1(int u){
    siz[u] = 1;
    Redge(u) if ((to = ed[k].to) != fa[u]){
        fa[to] = u; dep[to] = dep[u] + 1;
        dfs1(to);
        if (u > n) H[u - n].ins(w[to]);
        siz[u] += siz[to];
        if (!son[u] || siz[to] > siz[son[u]]) son[u] = to;
    }
    if (u > n) w[u] = H[u - n].top();
}
void dfs2(int u,int flag){
    top[u] = flag ? top[fa[u]] : u;
    id[u] = ++Cnt; Hash[Cnt] = u;
    if (son[u]) dfs2(son[u],1);
    Redge(u) if ((to = ed[k].to) != fa[u] && to != son[u])
        dfs2(to,0);
}
int mn[maxn << 2];
inline void upd(int u){mn[u] = min(mn[ls],mn[rs]);}
void build(int u,int l,int r){
    if (l == r){
        mn[u] = w[Hash[l]];
        return;
    }
    int mid = l + r >> 1;
    build(ls,l,mid);
    build(rs,mid + 1,r);
    upd(u);
}
void modify(int u,int l,int r,int pos,int v){
    if (l == r){mn[u] = v; return;}
    int mid = l + r >> 1;
    if (mid >= pos) modify(ls,l,mid,pos,v);
    else modify(rs,mid + 1,r,pos,v);
    upd(u);
}
int query(int u,int l,int r,int L,int R){
    if (l >= L && r <= R) return mn[u];
    int mid = l + r >> 1;
    if (mid >= R) return query(ls,l,mid,L,R);
    if (mid < L) return query(rs,mid + 1,r,L,R);
    return min(query(ls,l,mid,L,R),query(rs,mid + 1,r,L,R));
}
int solve1(int u,int v){
    int ans = INF;
    while (top[u] != top[v]){
        if (dep[top[u]] < dep[top[v]]) swap(u,v);
        ans = min(ans,query(1,1,N,id[top[u]],id[u]));
        u = fa[top[u]];
    }
    if (dep[u] > dep[v]) swap(u,v);
    ans = min(ans,query(1,1,N,id[u],id[v]));
    if (u > n && fa[u]) ans = min(ans,w[fa[u]]);
    return ans;
}
void solve2(int u,int v){
    modify(1,1,N,id[u],v);
    if (fa[u]){
        H[fa[u] - n].del(w[u]),H[fa[u] - n].ins(v);
        w[fa[u]] = H[fa[u] - n].top();
        modify(1,1,N,id[fa[u]],w[fa[u]]);
    }
    w[u] = v;
}
int main(){
    N = n = read();  m = read(); q = read();
    for (int i = 1; i <= n; i++) w[i] = read();
    for (int i = 1; i <= m; i++) build(read(),read());
    dfs(1,0);
    dfs1(1);
    dfs2(1,0);
    build(1,1,N);
    char opt; int a,b;
    while (q--){
        opt = getchar(); while (opt != ‘A‘ && opt != ‘C‘) opt = getchar();
        a = read(); b = read();
        if (opt == ‘A‘) printf("%d\n",solve1(a,b));
        else solve2(a,b);
    }
    return 0;
}

原文地址：https://www.cnblogs.com/Mychael/p/9203037.html