标签：double namespace code ++ name mod geo 复杂度 复杂
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, ×); 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 */
标签：double namespace code ++ name mod geo 复杂度 复杂
原文地址：http://www.cnblogs.com/galaxies/p/bzoj3681.html
