POJ3321 - Apple Tree DFS序 + 线段树或树状数组

Apple Tree：http://poj.org/problem?id=3321

题意：

　　告诉你一棵树，每棵树开始每个点上都有一个苹果，有两种操作，一种是计算以x为根的树上有几个苹果，一种是转换x这个点上的苹果，就是有就去掉，没有就加上。

思路：

　　先对树求一遍dfs序，每个点保存一个l，r。l是最早到这个点的时间戳，r是这个点子树中的最大时间戳，这样就转化为区间问题，可以用树状数组，或线段树维护区间的和。

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>
using namespace std;
//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
// #pragma GCC diagnostic error "-std=c++11"
// #pragma comment(linker, "/stack:200000000")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
// #pragma GCC optimize("-fdelete-null-pointer-checks,inline-functions-called-once,-funsafe-loop-optimizations,-fexpensive-optimizations,-foptimize-sibling-calls,-ftree-switch-conversion,-finline-small-functions,inline-small-functions,-frerun-cse-after-loop,-fhoist-adjacent-loads,-findirect-inlining,-freorder-functions,no-stack-protector,-fpartial-inlining,-fsched-interblock,-fcse-follow-jumps,-fcse-skip-blocks,-falign-functions,-fstrict-overflow,-fstrict-aliasing,-fschedule-insns2,-ftree-tail-merge,inline-functions,-fschedule-insns,-freorder-blocks,-fwhole-program,-funroll-loops,-fthread-jumps,-fcrossjumping,-fcaller-saves,-fdevirtualize,-falign-labels,-falign-loops,-falign-jumps,unroll-loops,-fsched-spec,-ffast-math,Ofast,inline,-fgcse,-fgcse-lm,-fipa-sra,-ftree-pre,-ftree-vrp,-fpeephole2",3)

#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue
#define max3(a,b,c) max(max(a,b),c)

typedef long long ll;
typedef unsigned long long ull;

typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl ‘\n‘

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9+7;
const double esp = 1e-8;
const double PI=acos(-1.0);

template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<‘0‘||ch>‘9‘) f|=(ch==‘-‘),ch=getchar();
    while (ch>=‘0‘&&ch<=‘9‘) x=x*10+ch-‘0‘,ch=getchar();
    return x=f?-x:x;
}

/*-----------------------showtime----------------------*/
            const int maxn =  300009;
            struct edge
            {
                int to,nx;
            }e[maxn];
            int h[maxn],all;
            void addedge(int u,int v){
                e[all].to = v;
                e[all].nx = h[u];
                h[u] = all++;
            }
            int sum[maxn],vis[maxn];
            char op[20];
            struct node
            {
                int l,r;
            }a[maxn];
            int tot = 1;
            void dfs(int x,int fa){
                a[x].l = tot;
                for(int i=h[x]; ~i; i = e[i].nx){
                    int v = e[i].to;
                    tot++;
                    if(v!=fa){
                        dfs(v,x);
                    }
                }
                a[x].r = ++tot;
            }
            int lowbit(int x){
                return x&(-x);
            }
            void add(int x,int c){
                while(x < maxn){
                    sum[x] += c;
                    x += lowbit(x);
                }
            }
            int getsum(int x){
                int res = 0;
                while(x>0){
                    res += sum[x];
                    x -= lowbit(x);
                }
                return res;
            }
int main(){
            int n,m,x;
            scanf("%d", &n);
            memset(h,-1,sizeof(h));
            for(int i=1; i<n; i++){
                int u,v;
                scanf("%d%d",&u, &v);
                addedge(u,v);
                addedge(v,u);
            }
            dfs(1,-1);
            for(int i=1; i<=n; i++){
                vis[i] = 1;
                add(a[i].l , 1);
            }
            scanf("%d", &m);
            while(m--){
                scanf("%s%d", op,&x);
                if(op[0] == ‘Q‘){
                    printf("%d\n", getsum(a[x].r) - getsum(a[x].l-1));
                }
                else {
                    if(vis[x] == 1){
                        add(a[x].l,-1);
                        vis[x] = 0;
                    }
                    else {
                        add(a[x].l,1);
                        vis[x] = 1;
                    }
                }
            }
            return 0;
}

POJ 3321

原文地址：https://www.cnblogs.com/ckxkexing/p/9637102.html