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; }