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。。

  1 //STATUS:C++_AC_240MS_6524KB
  2 #include <functional>
  3 #include <algorithm>
  4 #include <iostream>
  5 //#include <ext/rope>
  6 #include <fstream>
  7 #include <sstream>
  8 #include <iomanip>
  9 #include <numeric>
 10 #include <cstring>
 11 #include <cassert>
 12 #include <cstdio>
 13 #include <string>
 14 #include <vector>
 15 #include <bitset>
 16 #include <queue>
 17 #include <stack>
 18 #include <cmath>
 19 #include <ctime>
 20 #include <list>
 21 #include <set>
 22 #include <map>
 23 using namespace std;
 24 //#pragma comment(linker,"/STACK:102400000,102400000")
 25 //using namespace __gnu_cxx;
 26 //define
 27 #define pii pair<int,int>
 28 #define mem(a,b) memset(a,b,sizeof(a))
 29 #define lson l,mid,rt<<1
 30 #define rson mid+1,r,rt<<1|1
 31 #define PI acos(-1.0)
 32 //typedef
 33 typedef long long LL;
 34 typedef unsigned long long ULL;
 35 //const
 36 const int N=100010;
 37 const int INF=0x3f3f3f3f;
 38 const int MOD=1e+7,STA=8000010;
 39 const LL LNF=1LL<<60;
 40 const double EPS=1e-8;
 41 const double OO=1e15;
 42 const int dx[4]={-1,0,1,0};
 43 const int dy[4]={0,1,0,-1};
 44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
 45 //Daily Use ...
 46 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 50 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 51 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 56 //End
 57 
 58 int first[N],next[N*2],e[N*2],ra[N],id[N],sum[N],one[N<<2],rev[N<<2];
 59 int n,m,mt,cnt,ans;
 60 
 61 void adde(int a,int b)
 62 {
 63     e[mt]=b;
 64     next[mt]=first[a],first[a]=mt++;
 65     e[mt]=a;
 66     next[mt]=first[b],first[b]=mt++;
 67 }
 68 
 69 int dfs(int u,int fa)
 70 {
 71     ra[cnt++]=u;
 72     int i,j;
 73     sum[u]=1;
 74     for(i=first[u];i!=-1;i=next[i]){
 75         if(e[i]==fa)continue;
 76         sum[u]+=dfs(e[i],u);
 77     }
 78     return sum[u];
 79 }
 80 
 81 void pushdown(int rt,int llen,int rlen)
 82 {
 83     if(rev[rt]){
 84         rev[rt]=0;
 85         one[rt<<1]=llen-one[rt<<1];
 86         one[rt<<1|1]=rlen-one[rt<<1|1];
 87         rev[rt<<1]^=1,rev[rt<<1|1]^=1;
 88     }
 89 }
 90 
 91 void update(int l,int r,int rt,int L,int R)
 92 {
 93     if(L<=l && r<=R){
 94         rev[rt]^=1;
 95         one[rt]=r-l+1-one[rt];
 96         return ;
 97     }
 98     int mid=(l+r)>>1;
 99     pushdown(rt,mid-l+1,r-mid);
100     if(L<=mid)update(lson,L,R);
101     if(R>mid)update(rson,L,R);
102     one[rt]=one[rt<<1]+one[rt<<1|1];
103 }
104 
105 void query(int l,int r,int rt,int L,int R)
106 {
107     if(L<=l && r<=R){
108         ans+=one[rt];
109         return ;
110     }
111     int mid=(l+r)>>1;
112     pushdown(rt,mid-l+1,r-mid);
113     if(L<=mid)query(lson,L,R);
114     if(R>mid)query(rson,L,R);
115     one[rt]=one[rt<<1]+one[rt<<1|1];
116 }
117 
118 int main()
119 {
120   //  freopen("in.txt","r",stdin);
121     int i,j,t;
122     char op[2];
123     while(~scanf("%d%d",&n,&m))
124     {
125         mem(first,-1);mt=0;
126         for(i=2;i<=n;i++){
127             scanf("%d",&t);
128             adde(t,i);
129         }
130         cnt=1;
131         dfs(1,-1);
132         for(i=1;i<=n;i++)id[ra[i]]=i;
133 
134         mem(one,0);mem(rev,0);
135         while(m--){
136             scanf("%s%d",op,&t);
137             if(op[0]=='o'){
138                 update(1,n,1,id[t],id[t]+sum[t]-1);
139             }
140             else {
141                 ans=0;
142                 query(1,n,1,id[t],id[t]+sum[t]-1);
143                 printf("%d\n",ans);
144             }
145         }
146         putchar('\n');
147     }
148     return 0;
149 }