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=;
 const int INF=0x3f3f3f3f;
 const int MOD=1e+,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //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[N],next[N*],e[N*],ra[N],id[N],sum[N],one[N<<],rev[N<<];
 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]=;
     for(i=first[u];i!=-;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]=;
         one[rt<<]=llen-one[rt<<];
         one[rt<<|]=rlen-one[rt<<|];
         rev[rt<<]^=,rev[rt<<|]^=;
     }
 }

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

 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)>>;
     pushdown(rt,mid-l+,r-mid);
     if(L<=mid)query(lson,L,R);
     if(R>mid)query(rson,L,R);
     one[rt]=one[rt<<]+one[rt<<|];
 }

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

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