ZOJ-3686 A Simple Tree Problem 线段树

　　题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3686

　　题意：给定一颗有根树，每个节点有0和1两种值。有两种操作：o a操作，把以a为根节点的子树的权值全部取反；q a操作，求以a为根节点的子树权值为1的节点个数。

　　先求出树的先序遍历结果，并且记录每颗子树的节点个数，然后就可以用线段树维护了。。

//STATUS:C++_AC_240MS_6524KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,102400000")
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef long long LL;
typedef unsigned long long ULL;
//const
const int N=100010;
const int INF=0x3f3f3f3f;
const int MOD=1e+7,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End

int first
,next[N*2],e[N*2],ra
,id
,sum
,one[N<<2],rev[N<<2];
int n,m,mt,cnt,ans;

void adde(int a,int b)
{
e[mt]=b;
next[mt]=first[a],first[a]=mt++;
e[mt]=a;
next[mt]=first[b],first[b]=mt++;
}

int dfs(int u,int fa)
{
ra[cnt++]=u;
int i,j;
sum[u]=1;
for(i=first[u];i!=-1;i=next[i]){
if(e[i]==fa)continue;
sum[u]+=dfs(e[i],u);
}
return sum[u];
}

void pushdown(int rt,int llen,int rlen)
{
if(rev[rt]){
rev[rt]=0;
one[rt<<1]=llen-one[rt<<1];
one[rt<<1|1]=rlen-one[rt<<1|1];
rev[rt<<1]^=1,rev[rt<<1|1]^=1;
}
}

void update(int l,int r,int rt,int L,int R)
{
if(L<=l && r<=R){
rev[rt]^=1;
one[rt]=r-l+1-one[rt];
return ;
}
int mid=(l+r)>>1;
pushdown(rt,mid-l+1,r-mid);
if(L<=mid)update(lson,L,R);
if(R>mid)update(rson,L,R);
one[rt]=one[rt<<1]+one[rt<<1|1];
}

void query(int l,int r,int rt,int L,int R)
{
if(L<=l && r<=R){
ans+=one[rt];
return ;
}
int mid=(l+r)>>1;
pushdown(rt,mid-l+1,r-mid);
if(L<=mid)query(lson,L,R);
if(R>mid)query(rson,L,R);
one[rt]=one[rt<<1]+one[rt<<1|1];
}

int main()
{
//  freopen("in.txt","r",stdin);
int i,j,t;
char op[2];
while(~scanf("%d%d",&n,&m))
{
mem(first,-1);mt=0;
for(i=2;i<=n;i++){
scanf("%d",&t);
adde(t,i);
}
cnt=1;
dfs(1,-1);
for(i=1;i<=n;i++)id[ra[i]]=i;

mem(one,0);mem(rev,0);
while(m--){
scanf("%s%d",op,&t);
if(op[0]=='o'){
update(1,n,1,id[t],id[t]+sum[t]-1);
}
else {
ans=0;
query(1,n,1,id[t],id[t]+sum[t]-1);
printf("%d\n",ans);
}
}
putchar('\n');
}
return 0;
}