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[N],next[N*2],e[N*2],ra[N],id[N],sum[N],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
",ans);
            }
        }
        putchar('
');
    }
    return 0;
}