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luoguP4115 QTREE4 链分治

时间:2018-08-19 10:24:10      阅读:147      评论:0      收藏:0      [点我收藏+]

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具体看$qzc$论文吧......陈年老物了......

主要注意每个链头一棵线段树而不是一棵全局线段树

修改操作写完就是正确的,反而是初始化调了好一会......

跑的还是很快的,有些地方没优化常数也还可以接受

在$luogu$上把$Toptree$给卡下去了,现居$rank1$......

代码的话....借鉴一下思想就行了

实现就没有必要有些地方做得一样了

#include <queue>
#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

extern inline char gc() {
    static char RR[23456], *S = RR + 23333, *T = RR + 23333;
    if(S == T) fread(RR, 1, 23333, stdin), S = RR;
    return *S ++;
}
inline int read() {
    int p = 0, w = 1; char c = gc();
    while(c > 9 || c < 0) { if(c == -) w = -1; c = gc(); }
    while(c >= 0 && c <= 9) p = p * 10 + c - 0, c = gc();
    return p * w;
}

#define sid 100050
#define eid 200050
#define ri register int
const int inf = 1e9;

struct heap {
    priority_queue <int> f, g;
    inline void ins(int v) { if(v != -inf) f.push(v); } 
    inline void era(int v) { if(v != -inf) g.push(v); }
    inline int top() {
        while(1) {
            if(f.empty()) return -inf;
            if(g.empty()) return f.top();
            if(f.top() == g.top()) f.pop(), g.pop();
            else return f.top();
        }
    }
} h[sid], ans;

int n, m, wn, cnp;
int cap[sid], nxt[eid], node[eid], fee[eid];
inline void addeg(int u, int v, int w) {
    nxt[++ cnp] = cap[u]; cap[u] = cnp; node[cnp] = v; fee[cnp] = w;
    nxt[++ cnp] = cap[v]; cap[v] = cnp; node[cnp] = u; fee[cnp] = w;
}

#define cur node[i]
int sz[sid], pre[sid], fa[sid], dep[sid];
inline void dfs(int o) {
    sz[o] = 1;
    for(int i = cap[o]; i; i = nxt[i])
    if(cur != fa[o]) {
        fa[cur] = o; dep[cur] = dep[o] + fee[i]; dfs(cur);
        sz[o] += sz[cur]; if(sz[cur] > sz[pre[o]]) pre[o] = cur;
    }
}

int id, onp, col[sid];
int anc[sid], len[sid], dfn[sid], ord[sid];
inline void dfs(int o, int tp) {
    dfn[o] = ++ id; anc[o] = tp; ord[id] = o; len[tp] ++;
    if(!pre[o]) return; dfs(pre[o], tp);
    for(int i = cap[o]; i; i = nxt[i])
    if(cur != fa[o] && cur != pre[o]) dfs(cur, cur);
}

int tnp, rt[sid];
struct seg { int l, r, v, ls, rs; } t[sid * 4];

#define dis(x) dep[ord[x]] 
inline void upd(int o, int l, int r) {
    int ls = t[o].ls, rs = t[o].rs, mid = (l + r) >> 1;
    t[o].l = max(t[ls].l, t[rs].l + dis(mid + 1) - dis(l));
    t[o].r = max(t[rs].r, t[ls].r + dis(r) - dis(mid));
    t[o].v = max(max(t[ls].v, t[rs].v), t[ls].r + t[rs].l + dis(mid + 1) - dis(mid));
}

inline void build(int &o, int l, int r) {
    if(!o) o = ++ onp;
    if(l == r) {
        int x = ord[l];
        for(ri i = cap[x]; i; i = nxt[i])
        if(cur != fa[x] && cur != pre[x]) h[x].ins(t[rt[cur]].l + dep[cur] - dep[x]);
        int d1 = h[x].top(); h[x].era(d1); int d2 = h[x].top(); h[x].ins(d1);
        t[o].l = t[o].r = max(d1, 0); t[o].v = max(0, max(d1, d1 + d2));
        return;
    }
    int mid = (l + r) >> 1;
    build(t[o].ls, l, mid);
    build(t[o].rs, mid + 1, r);
    upd(o, l, r);
}

inline void mdf(int o, int l, int r, int v, int s) {
    if(l == r) {
        if(v == s) {
            int d1 = h[v].top(); h[v].era(d1); int d2 = h[v].top(); h[v].ins(d1);
            if(col[v]) t[o].l = t[o].r = d1, t[o].v = d1 + d2;
            else t[o].l = t[o].r = max(d1, 0), t[o].v = max(0, max(d1, d1 + d2));
        }
        else {
            h[v].ins(t[rt[s]].l + dep[s] - dep[v]);
            int d1 = h[v].top(); h[v].era(d1); int d2 = h[v].top(); h[v].ins(d1);
            if(col[v]) t[o].l = t[o].r = d1, t[o].v = d1 + d2;
            else t[o].l = t[o].r = max(d1, 0), t[o].v = max(0, max(d1, d1 + d2));
        }
        return;
    }
    int mid = (l + r) >> 1;
    if(dfn[v] <= mid) mdf(t[o].ls, l, mid, v, s);
    else mdf(t[o].rs, mid + 1, r, v, s);
    upd(o, l, r);
}

int main() {
    wn = n = read();
    for(ri i = 1; i < n; i ++) {
        int u = read(), v = read();
        int w = read(); addeg(u, v, w);
    }
    dfs(1); dfs(1, 1); ans.ins(0);            
    for(ri i = n; i; i --) {
        int o = ord[i]; if(o != anc[o]) continue; 
        build(rt[o], dfn[o], dfn[o] + len[o] - 1);
        ans.ins(t[rt[o]].v);
    }
    m = read(); char opt = 0;
    for(ri i = 1; i <= m; i ++) {
        opt = gc();
        while(opt != C && opt != A) opt = gc();
        if(opt == C) {
            int x = read(); col[x] ^= 1;
            if(col[x] == 0) wn ++; else wn --;
            for(ri o = x, p = o; o; o = fa[o]) {
                int f = anc[o];
                int p1 = t[rt[f]].v, d1 = t[rt[f]].l;
                if(fa[f]) h[fa[f]].era(t[rt[f]].l + dep[f] - dep[fa[f]]);
                mdf(rt[f], dfn[f], dfn[f] + len[f] - 1, o, p);
                int p2 = t[rt[f]].v, d2 = t[rt[f]].l;
                if(p1 != p2) ans.era(p1), ans.ins(p2);
                p = o = f;
            }
        }
        else {
            if(wn == 0) printf("They have disappeared.\n");
            else printf("%d\n", ans.top());
        }
    }
    return 0;
}

 

luoguP4115 QTREE4 链分治

原文:https://www.cnblogs.com/reverymoon/p/9499510.html

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