bzoj3681 Arietta

传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=3681

【题解】

这题和bzoj3218很像，都是有二维限制关系的网络流。

先考虑了bzoj3218是用主席树来建图然后网络流，感觉这题也能这样做（在dfs序上，主席树建图）

后来发现这样好像不兹磁区间减法啊

然后发现可以线段树合并来做，这样就很兹磁了。

加边的时候(u, v, flow)反向边也加成了(v, u, flow)坑了好久。。。

时间复杂度能过（逃）

还有一个坑待解决：

如果bzoj3218加强成不仅有权值区间$[l_i,r_i]$，还有下标区间$[L_i,R_i]$，其中$L_i \leq R_i \leq i$，那么要怎么解决啊？

主席树好像就不行了啊qwq 本题因为有树结构、子树查询所以恰巧比较兹磁线段树合并来维护。

立个flag：这坑noi前解决不了。

upd: 好像一发完博客我就会了。。用树套树就行了（逃

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

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int N = 1e4 + 10, M = 1e6 + 5;
const int mod = 1e9+7, inf = 1e9;

int n, m, par[N], w[N], offset, all;
int head[N], nxt[N], to[N], tot = 0;
inline void add(int u, int v) {
    ++tot; nxt[tot] = head[u]; head[u] = tot; to[tot] = v;
}

int S, T;
namespace MF {
    int head[M], nxt[M], to[M], flow[M], tot = 1; 
    inline void add(int u, int v, int fl) {
        ++tot; nxt[tot] = head[u]; head[u] = tot;
        to[tot] = v, flow[tot] = fl;
    }
    inline void adde(int u, int v, int fl) {
        add(u, v, fl), add(v, u, 0);
    }
    int c[M], cur[M]; queue<int> q;    
    inline bool bfs() {
        while(!q.empty()) q.pop();
        for (int i=1; i<=all; ++i) c[i] = -1;
        c[S] = 0; q.push(S);
        while(!q.empty()) {
            int top = q.front(); q.pop();
            for (int i=head[top]; i; i=nxt[i]) {
                if(c[to[i]] != -1 || !flow[i]) continue;
                c[to[i]] = c[top] + 1;
                q.push(to[i]); 
                if(to[i] == T) return 1;
            }
        }
        return 0;
    }
    inline int dfs(int x, int low) {
        if(x == T) return low;
        int r = low, fl;
        for (int i=cur[x]; i; i=nxt[i]) {
            if(c[to[i]] != c[x]+1 || !flow[i]) continue;
            fl = dfs(to[i], min(r, flow[i])); 
            r -= fl, flow[i] -= fl, flow[i^1] += fl;
            if(flow[i] > 0) cur[x] = i; 
            if(!r) return low;
        }
        if(low == r) c[x] = -1;
        return low-r;
    }
    inline int main() {
        int ans = 0;
        while(bfs()) {
            for (int i=1; i<=all; ++i) cur[i] = head[i];
            ans += dfs(S, inf);
        }
        return ans;
    } 
}

int rt[N]; 
struct SMT {
    int ch[M][2], siz;
    inline void set() {siz = 0;} 
    inline int newnode() {return ++siz;}
    inline int build(int l, int r, int ps, int p) {
        int x = newnode();
//        cout << x + offset << "   [" << l << "," << r <<"]\n"; 
        if(l == r) {
            MF :: adde(p, x + offset, inf);
            return x;
        }
        int mid = l+r>>1;
        if(ps <= mid) ch[x][0] = build(l, mid, ps, p);
        else ch[x][1] = build(mid+1, r, ps, p);
        if(ch[x][0]) MF :: adde(ch[x][0] + offset, x + offset, inf);
        if(ch[x][1]) MF :: adde(ch[x][1] + offset, x + offset, inf);
        return x;
    }
    inline int merge(int a, int b, int l, int r) {
        if(!a || !b) return a + b;
        int x = newnode(); 
//        cout << x + offset << " - [" << l << "," << r <<"]\n"; 
        if(l == r) {
            MF :: adde(a + offset, x + offset, inf);
            MF :: adde(b + offset, x + offset, inf);
            return x;
        }
        int mid = l+r>>1;
        ch[x][0] = merge(ch[a][0], ch[b][0], l, mid); 
        ch[x][1] = merge(ch[a][1], ch[b][1], mid+1, r);        
        if(ch[x][0]) MF :: adde(ch[x][0] + offset, x + offset, inf);
        if(ch[x][1]) MF :: adde(ch[x][1] + offset, x + offset, inf);
        return x;
    }
    inline void link(int x, int l, int r, int L, int R, int p) {
        if(!x) return ;
        if(L <= l && r <= R) {
            MF :: adde(x + offset, p, inf);
            return ;
        }
        int mid = l+r>>1;
        if(L <= mid) link(ch[x][0], l, mid, L, R, p);
        if(R > mid) link(ch[x][1], mid+1, r, L, R, p);
    }
}t;

inline void dfs(int x) {
    rt[x] = t.build(1, n, w[x], x);
    for (int i=head[x]; i; i=nxt[i]) {
        dfs(to[i]);
        rt[x] = t.merge(rt[x], rt[to[i]], 1, n);
    }
}

int main() {
    cin >> n >> m; t.set(); 
    S = n+m+1, offset = T = n+m+2;
    for (int i=2; i<=n; ++i) {
        scanf("%d", par+i);
        add(par[i], i);
    }
    for (int i=1; i<=n; ++i) {
        MF :: adde(S, i, 1);
        scanf("%d", w+i);
    }
    dfs(1);
    for (int i=1, L, R, D, times; i<=m; ++i) {
        scanf("%d%d%d%d", &L, &R, &D, &times); 
         t.link(rt[D], 1, n, L, R, i+n);
         MF :: adde(i+n, T, times); 
    }
    all = offset + t.siz;
    cout << MF::main(); 
    return 0;
}
/*
5 2
1 1 2 2
5 3 2 4 1
1 3 2 1
3 5 1 4
*/