bzoj 3681 Arietta

一棵有根树,每个点有一个音高,有 $m$ 中弹奏方法,每种方法可以弹奏 $d$ 子树中音高在 $[l,r]$ 间的音符,每种方法最多弹 $t$ 次

求最多能弹出多少个音符

$n \leq 10000$

sol:

网络流

暴力连边是

1. $S \rightarrow 每个点$

2. $每个方法 \rightarrow T$

3. $每个点 \rightarrow 每个能用到的方法$

第一种边限制是 $1$ ,第二种边限制是 $t$,第三种边没有限制

第一第二种不好优化,考虑优化第三种

第三种本质上是对树上 $dfn$ 是一段区间(子树),权值也是一段区间的点连边

考虑数据结构优化

可以使用可持久化线段树合并

一开始每个点向它所在的线段树上的点连边

每次合并的时候原来的点向合并出来的新点连边

对于方法我们把它能作用的区域向它连边

这样边数是 $O(2n) + O(可持久化线段树)$ 的,大概是 $O(nlogn)$

#include <bits/stdc++.h>
#define LL long long
#define rep(i, s, t) for (register int i = (s), i##end = (t); i <= i##end; ++i)
#define dwn(i, s, t) for (register int i = (s), i##end = (t); i >= i##end; --i)
using namespace std;
inline int read() {
    int x = 0, f = 1; char ch;
    for (ch = getchar(); !isdigit(ch); ch = getchar()) if (ch == '-') f = -f;
    for (; isdigit(ch); ch = getchar()) x = 10 * x + ch - '0';
    return x * f;
}
const int maxn = 100010, maxm = 1000010, inf = 2147483233;
struct Dinic {
    int cur[maxm], head[maxm], nx[maxm];
    int n, m, s, t;
    struct Edge {
        int from, to, caps;
        Edge(){}
        Edge(int _1, int _2, int _3): from(_1), to(_2), caps(_3) {}
    }es[maxm];
    void AddEdge(int u, int v, int w) {
        es[m] = Edge(u, v, w); nx[m] = head[u]; head[u] = m++;
        es[m] = Edge(v, u, 0); nx[m] = head[v]; head[v] = m++;
    }
    void setn(int _) {n = _;}
    Dinic() {m = 0; memset(head, -1, sizeof(head));}
    queue<int> q; int dis[maxn];
    bool BFS() {
        rep(i, 0, n) dis[i] = 0;
        q.push(t); dis[t] = 1;
        while(!q.empty()) {
            int now = q.front(); q.pop();
            for(int i=head[now];~i;i=nx[i]) {
                Edge &e = es[i^1];
                if(!dis[e.from] && e.caps) {
                    dis[e.from] = dis[now] + 1;
                    q.push(e.from);
                }
            }
        }
        return (dis[s] > 1);
    }
    int DFS(int u, int a) {
        if(u == t || !a) return a;
        int flow = 0, f;
        for(int &i = cur[u]; ~i; i = nx[i]) {
            Edge &e = es[i];
            if(dis[e.to] == dis[u] - 1 && (f = DFS(e.to, min(e.caps, a)))) {
                e.caps -= f; es[i^1].caps += f;
                a -= f; flow += f;
                if(!a) return flow;
            }
        }
        return flow;
    }
    int MaxFlow(int _s, int _t) {
        s = _s, t = _t; int res = 0;
        while(BFS()) {
            memcpy(cur, head, (n + 1) * sizeof(int));
            res += DFS(s, 2147483233);
        }
        return res;
    }
} sol;
int s, t, nodes;
struct Ques {
    int l, r, d, t;
    Ques(){}
    Ques(int _1, int _2, int _3, int _4) : l(_1), r(_2), d(_3), t(_4) {}
}qs[maxn];
int n, m;
int fa[maxn], h[maxn];
int first[maxn], nx[maxn], to[maxn], cnt;
inline void add(int u, int v) {
    to[++cnt] = v;
    nx[cnt] = first[u];
    first[u] = cnt;
}
int root[maxn], ls[maxm << 1], rs[maxm << 1], dfn;
inline void Insert(int &x, int l, int r, int pos, int p) {
    x = ++dfn;
    if(l == r) {
        sol.AddEdge(p, x + nodes, inf);
        return;
    }
    int mid = (l + r) >> 1;
    if(pos <= mid) Insert(ls[x], l, mid, pos, p) ;
    else Insert(rs[x], mid + 1, r, pos, p) ;
    if(ls[x]) sol.AddEdge(ls[x] + nodes, x + nodes, inf);
    if(rs[x]) sol.AddEdge(rs[x] + nodes, x + nodes, inf);
}
inline int merge(int x, int y, int l, int r) {
    if(!x || !y) return x + y;
    int z = ++dfn;
    if(l == r) {
        sol.AddEdge(x + nodes, z + nodes, inf);
        sol.AddEdge(y + nodes, z + nodes, inf);
        return z;
    }
    int mid = (l + r) >> 1;
    ls[z] = merge(ls[x], ls[y], l, mid);
    rs[z] = merge(rs[x], rs[y], mid+1, r);
    if(ls[z]) sol.AddEdge(ls[z] + nodes, z + nodes, inf);
    if(rs[z]) sol.AddEdge(rs[z] + nodes, z + nodes, inf);
    return z;
}
inline void link(int x, int l, int r, int L, int R, int p) {
    if(!x) return;
    if(L <= l && r <= R) {
        sol.AddEdge(x + nodes, p, inf);
        return;
    }
    int mid = (l + r) >> 1;
    if(L <= mid) link(ls[x], l, mid, L, R, p);
    if(R > mid) link(rs[x], mid+1, r, L, R, p);
}
inline void dfs(int x) {
    Insert(root[x], 1, n, h[x], x);
    for(int i=first[x];i;i=nx[i]) {
        dfs(to[i]);
        root[x] = merge(root[x], root[to[i]], 1, n);
    }
}

int main() {
    n = read(), m = read();
    s = n + m + 1, t = n + m + 2, nodes = t + 1;
    rep(i, 2, n) {
        fa[i] = read();
        add(fa[i], i);
    }
    rep(i, 1, n) {
        h[i] = read();
        sol.AddEdge(s, i, 1);
    } dfs(1);
    rep(i, 1, m) {
        int l = read(), r = read(), d = read(), ct = read();
        link(root[d], 1, n, l, r, i+n);
        qs[i] = Ques(l, r, d, ct); sol.AddEdge(i + n, t, ct);
    }
    sol.setn(nodes + dfn + 5);
    cout << sol.MaxFlow(s, t) << endl;
}
View Code

 

posted @ 2019-03-12 10:39  探险家Mr.H  阅读(229)  评论(0编辑  收藏  举报