SPOJ-QTREE3 Query on a tree again!

Query on a tree again!

树链剖分 + 二分

通过树链剖分查找，判断一下路径上，最后一个黑点出现在哪一条链上，然后在链上进行二分 dfn 查找第一个黑点所在位置

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn = 1e5 + 10;
int dep[maxn], siz[maxn], hson[maxn], fa[maxn];
int dfn[maxn], rnk[maxn], tr[maxn << 2], top[maxn];
vector<int>gra[maxn];

void dfs1(int now, int pre, int d)
{
    dep[now] = d;
    hson[now] = -1;
    siz[now] = 1;
    fa[now] = pre;
    for(int i=0; i<gra[now].size(); i++)
    {
        int nex = gra[now][i];
        if(nex == fa[now]) continue;
        dfs1(nex, now, d + 1);
        siz[now] += siz[nex];
        if(hson[now] == -1 || siz[hson[now]] < siz[nex])
            hson[now] = nex;
    }
}

int tp = 0;
void dfs2(int now, int t)
{
    tp++;
    top[now] = t;
    dfn[now] = tp;
    rnk[tp] = now;
    if(hson[now] != -1)
    {
        dfs2(hson[now], t);
        for(int i=0; i<gra[now].size(); i++)
        {
            int nex = gra[now][i];
            if(nex == fa[now] || nex == hson[now]) continue;
            dfs2(nex, nex);
        }
    }
}

void init(int n, int rt = 1)
{
    tp = 0;
    dfs1(rt, rt, 1);
    dfs2(rt, rt);
    for(int i=0; i<=n; i++) gra[i].clear();
}

void update(int now, int l, int r, int x)
{
    if(l == r)
    {
        tr[now] ^= 1;
        return;
    }
    int mid = l + r >> 1;
    if(x <= mid)
        update(now << 1, l, mid, x);
    else
        update(now << 1 | 1, mid + 1, r, x);
    tr[now] = tr[now << 1] + tr[now << 1 | 1];
}

int query(int now, int l, int r, int L, int R)
{
    if(L <= l && r <= R)
        return tr[now];
    int mid = l + r >> 1;
    int ans = 0;
    if(L <= mid)
        ans += query(now << 1, l, mid, L, R);
    if(R > mid)
        ans += query(now << 1 | 1, mid + 1, r, L, R);
    return ans;
}

int solve(int v, int n)
{
    int node = 0;
    while(top[v] != 1)
    {
        if(query(1, 1, n, dfn[top[v]], dfn[v])) node = v;
        v = fa[top[v]];
    }
    if(query(1, 1, n, 1, dfn[v])) node = v;
    if(node == 0) return -1;
    
    int l = dfn[top[v]], r = dfn[v];
    while(l < r)
    {
        int mid = l + r >> 1;
        if(query(1, 1, n, l, mid)) r = mid;
        else l = mid + 1;
    }
    return rnk[r];
}

int main()
{
    int n, m;
    scanf("%d%d", &n, &m);
    for(int i=1; i<n; i++)
    {
        int x, y;
        scanf("%d%d", &x, &y);
        gra[x].push_back(y);
        gra[y].push_back(x);
    }
    init(n, 1);
    while(m--)
    {
        int t;
        scanf("%d", &t);
        if(t == 0)
        {
            int i;
            scanf("%d", &i);
            update(1, 1, n, dfn[i]);
        }
        else
        {
            int v;
            scanf("%d", &v);
            printf("%d\n", solve(v, n));
        }
    }
    return 0;
}

标签：hson,include,int,tree,mid,SPOJ,maxn,Query,now
来源： https://www.cnblogs.com/dgsvygd/p/16456739.html