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; //
const ll nmos = 0x80000000; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //
const int mod = 1e9+;
const double esp = 1e-;
const double PI=acos(-1.0); template<typename T>
inline T read(T&x){
x=;int f=;char ch=getchar();
while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();
while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();
return x=f?-x:x;
} /*-----------------------showtime----------------------*/
const int maxn = ;
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[];
struct node
{
int l,r;
}a[maxn];
int tot = ;
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 = ;
while(x>){
res += sum[x];
x -= lowbit(x);
}
return res;
}
int main(){
int n,m,x;
scanf("%d", &n);
memset(h,-,sizeof(h));
for(int i=; i<n; i++){
int u,v;
scanf("%d%d",&u, &v);
addedge(u,v);
addedge(v,u);
}
dfs(,-);
for(int i=; i<=n; i++){
vis[i] = ;
add(a[i].l , );
}
scanf("%d", &m);
while(m--){
scanf("%s%d", op,&x);
if(op[] == 'Q'){
printf("%d\n", getsum(a[x].r) - getsum(a[x].l-));
}
else {
if(vis[x] == ){
add(a[x].l,-);
vis[x] = ;
}
else {
add(a[x].l,);
vis[x] = ;
}
}
}
return ;
}
POJ 3321