4538: [Hnoi2016]网络

4538: [Hnoi2016]网络

链接

分析：

　　整体二分。

　　对于一次操作，可以二分一个答案mid，判断权值大于mid的路径是否全部经过这个点。如果是 ，那么这次询问的答案在[l,mid-1]之间，否则在[mid,r]之间。

　　判断是否所有的路径经过一个点：等价于数经过这个点的路径条数，对于一条路径(u->v)，可以在u,v处+1，在lca处-1，在fa[lca]处-1，然后询问一个点的子树和即可。

　　多次询问，整体二分即可。

代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<cctype>
#include<set>
#include<queue>
#include<vector>
#include<map>
using namespace std;
typedef long long LL;

inline int read() {
    int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
}

const int N = 200005;
struct Data { int ty, a, b, v, c; } A[N], B[N], C[N];
struct Edge { int to, nxt; } e[N << 1];
int head[N], f[N][20], dfn[N], dep[N], siz[N], ans[N], En, Index;

inline void add_edge(int u,int v) {
    ++En; e[En].to = v, e[En].nxt = head[u]; head[u] = En;
    ++En; e[En].to = u, e[En].nxt = head[v]; head[v] = En;
}
int LCA(int u,int v) {
    if (dep[u] < dep[v]) swap(u, v);
    int d = dep[u] - dep[v];
    for (int i = 18; ~i; --i) if ((d >> i) & 1) u = f[u][i];
    if (u == v) return u;
    for (int i = 18; ~i; --i) if (f[u][i] != f[v][i]) u = f[u][i], v = f[v][i];
    return f[u][0];
}
void dfs(int u) {
    for (int j = 1; j <= 18; ++j) f[u][j] = f[f[u][j - 1]][j - 1];
    dep[u] = dep[f[u][0]] + 1;
    dfn[u] = ++Index; siz[u] = 1;
    for (int i = head[u]; i; i = e[i].nxt) {
        int v = e[i].to;
        if (v == f[u][0]) continue;
        f[v][0] = u;
        dfs(v);
        siz[u] += siz[v];
    }
}
struct Bit{
    int sum[N << 1], n;
    void clear() { memset(sum, 0, sizeof(sum)); }
    void update(int p,int v) { if (p == 0) return ; for (; p <= n; p += (p & (-p))) sum[p] += v; }
    int query(int p) {
        int ans = 0;
        for (; p; p -= (p & (-p))) ans += sum[p];
        return ans;
    }
    int Ask(int l,int r) { return query(r) - query(l - 1); }
}bit;
void add(Data &A,int ty) {
    ty = ty ? -1 : 1;
    bit.update(dfn[A.a], ty); 
    bit.update(dfn[A.b], ty);
    bit.update(dfn[A.c], -ty); 
    bit.update(dfn[f[A.c][0]], -ty);
}
void solve(int l,int r,int H,int T) {
    if (H > T) return ;
    if (l == r) {
        for (int i = H; i <= T; ++i) if (A[i].ty == 2) ans[A[i].b] = l; 
        return ;
    }
    int mid = (l + r + 1) >> 1, cl = H, cr = T, now = 0;
    for (int i = H; i <= T; ++i) {
        if (A[i].ty <= 1) {
            if (A[i].v < mid) B[cl ++] = A[i];
            else add(A[i], A[i].ty), B[cr --] = A[i], now += (A[i].ty ? -1 : 1);
        }
        else {
            int t = bit.Ask(dfn[A[i].a], dfn[A[i].a] + siz[A[i].a] - 1);
            if (t == now) B[cl ++] = A[i];
            else B[cr --] = A[i];
        }
    }
    for (int i = H; i <= T; ++i) if (A[i].ty <= 1 && A[i].v >= mid) add(A[i], !A[i].ty);
    for (int i = H; i < cl; ++i) A[i] = B[i];
    for (int i = cl; i <= T; ++i) A[i] = B[T - i + cl];
    solve(l, mid - 1, H, cl - 1);
    solve(mid, r, cl, T);
}
int main() {
    int n = read(), m = read(), mx = 0, id = 0, tot = 0, now = 0; bit.n = n;
    for (int i = 1; i < n; ++i) {
        int u = read(), v = read();
        add_edge(u, v);
    }
    dfs(1);
    for (int i = 1; i <= m; ++i) {
        int opt = read();
        if (opt == 0) {
            int u = read(), v = read(), w = read(), z = LCA(u, v);
            A[i] = (Data){opt, u, v, w, z};
            mx = max(mx, w);
        } else if (opt == 1) {
            int t = read();
            A[i] = A[t]; A[i].ty = 1;
        } else if (opt == 2) {
            int x = read(); ++id;
            A[i] = (Data){2, x, id, 0, 0};
        }
    }
    for (int i = 1; i <= m; ++i) {
        if (A[i].ty <= 1) add(A[i], A[i].ty), now += (A[i].ty ? -1 : 1), B[++tot] = A[i];
        else {
            if (bit.Ask(dfn[A[i].a], dfn[A[i].a] + siz[A[i].a] - 1) == now) ans[A[i].b] = -1;
            else B[++tot] = A[i];
        }
    }
    for (int i = 1; i <= tot; ++i) A[i] = B[i];
    bit.clear();
    solve(-1, mx, 1, tot);
    for (int i = 1; i <= id; ++i) printf("%d\n", ans[i]);
    return 0;
}