HDU3887 DFS序+ 线段树

查询树上某个节点的子节点的标号小于其标号的数目. 一个trick是建立线段树之后,从标号小的向标号大的来做更新.

#include <cstdio>
#include <cstring>
#include <cctype>
#include <algorithm>
#include <vector>
#include <iostream>
using namespace std;
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define LL(a)   a<<1
#define RR(a)   a<<1|1

const int MaxL = 200030;

int pre[MaxL];  // first travel
int nxt[MaxL];  // nxt travel
int f[MaxL];
bool vis[MaxL];
int N;
vector<vector<int> > E(MaxL);

struct Seg_tree
{
    int left, right;
    int sum;
} tt[MaxL<<2];


void PushUp(int idx)
{
    tt[idx].sum = tt[LL(idx)].sum + tt[RR(idx)].sum;
}

void build(int l,int r,int idx)
{
    tt[idx].left = l, tt[idx].right = r;
    tt[idx].sum = 0;
    if (l == r) return ;
    int m = (l + r) >> 1;
    build(l,m, LL(idx));
    build(m+1, r, RR(idx));
}

void update(int p, int idx = 1)
{
    if(p == tt[idx].left && p == tt[idx].right)
    {
        tt[idx].sum =1;
        return ;
    }
    int mid = (tt[idx].left + tt[idx].right)>>1;
    if(p <= mid) update(p, LL(idx));
    else update(p, RR(idx));
    PushUp(idx);
}

int query(int l, int r, int idx = 1)
{
    if(l == tt[idx].left && r ==tt[idx].right)
        return tt[idx].sum;
    int mid = (tt[idx].left + tt[idx].right)>>1;
    if(r <= mid) return query(l,r, LL(idx));
    else if(l> mid) return query(l,r, RR(idx));
    else
        return query(l, mid, LL(idx))+ query(mid+1, r, RR(idx));
}

int mark = 1;
void dfs( int a)
{
    vis[a] = 1;
    pre[a] = mark++;
    for(int i=0; i<E[a].size(); i++)
    {
        if(!vis[E[a][i]])
            dfs(E[a][i]);
    }
    nxt[a] = mark++;
}

int main()
{
//    freopen("1.txt","r",stdin);
    int p;
    while( scanf("%d%d", &N, &p) && N!=0 && p!=0)
    {
        memset(pre, 0, sizeof(pre));
        memset(nxt, 0 ,sizeof(nxt));
        memset(vis, 0 ,sizeof(vis));
        for(int i=1; i<=N; i++) E[i].clear();

        for(int i=1; i<N; i++)
        {
            int a,b;
            scanf("%d%d", &a,&b);
            E[a].push_back(b);
            E[b].push_back(a);
        }
        mark = 1;
        dfs(p);
        build(1, 2*N, 1);
        for(int i=1; i<=N; i++)
        {
            f[i] =query(pre[i], nxt[i],1);
            update(pre[i],1);
        }
        for(int i=1; i<N; i++)
            printf("%d ",f[i]);
        printf("%d\n",f[N]);
    }
    return 0;
}