ZOJ 3686 后序遍历线段树

A Simple Tree Problem

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Time Limit: 3 Seconds      Memory Limit: 65536 KB

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Given a rooted tree, each node has a boolean (0 or 1) labeled on it. Initially, all the labels are 0.

We define this kind of operation: given a subtree, negate all its labels.

And we want to query the numbers of 1's of a subtree.

Input

Multiple test cases.

First line, two integer N and M, denoting the numbers of nodes and numbers of operations and queries.(1<=N<=100000, 1<=M<=10000)

Then a line with N-1 integers, denoting the parent of node 2..N. Root is node 1.

Then M lines, each line are in the format "o node" or "q node", denoting we want to operate or query on the subtree with root of a certain node.

Output

For each query, output an integer in a line.

Output a blank line after each test case.

Sample Input

3 21 1o 2q 1

Sample Output

1

题意容易理解不说了。

思路：刚开始不知道怎么把平常的树转化为线段二叉树，唉……弄了几天，自己以前的线段树模板写了又写，对lazy标记自己的结构体线段树模板不好，所以看了别人的线段树代码是用数组写的，就有点晕了……然后对比了自己的线段树模板，发现用数组做处理lazy标记比较直观，用起来也比较爽……哈哈，然后用了一晚上弄这个线段树模板，相当于把以前的线段树常用的结构体给弄翻了，又理解了好久才会……呵呵……

这道题如果用图画一下，就会发现这棵树每个结点会有超过两个的子树，但是又是边修改边查询，所以肯定是用线段树做，所以就得把这棵不平常的树转化为线段树了。

如果用后序遍历这棵树的话，就可以得到线性区间的线性序列，然后就可以用线段树表示了。

#include <iostream>#include <cstdio>#include <fstream>#include <algorithm>#include <cmath>#include <deque>#include <vector>#include <list>#include <queue>#include <string>#include <cstring>#include <map>#include <stack>#include <set>#define PI acos(-1.0)#define mem(a,b) memset(a,b,sizeof(a))#define sca(a) scanf("%d",&a)#define pri(a) printf("%d\n",a)#define lson i<<1,l,mid#define rson i<<1|1,mid+1,r#define MM 100005#define MN 105#define INF 10000007using namespace std;int head[MM],num,cnt,l[MM],r[MM],sum[MM*4],val[MM*4];struct node{    int v,next;}e[MM];void add(int u,int v){    e[cnt].v=v; e[cnt].next=head[u]; head[u]=cnt++;}void dfs(int u){    l[u]=++num;    for(int i=head[u]; i!=-1; i=e[i].next)        dfs(e[i].v);    r[u]=num;}void push_up(int i){    sum[i]=sum[i<<1]+sum[i<<1|1];}void push_down(int i,int m) //处理lazy标记{    if(val[i])    {        val[i<<1]^=1; val[i<<1|1]^=1;        sum[i<<1]=(m-(m>>1))-sum[i<<1]; //因为长度为奇数的线段树左子树多1，比如l=1,r=7,左子树为1->4,而右子树为5->7,故左子树多1        sum[i<<1|1]=(m>>1)-sum[i<<1|1];        val[i]=0;    }}void build(int i,int l,int r){    val[i]=0;    if(l==r) { sum[i]=0; return ; }    int mid=(l+r)>>1;    build(lson); build(rson);    push_up(i);}void update(int i,int l,int r,int L,int R){    if(L<=l&&r<=R)    {        val[i]^=1;        sum[i]=r-l+1-sum[i]; //长度减去总值即为更新后的值，因为值只有0,1        return ;    }    int mid=(l+r)>>1;    push_down(i,r-l+1);    if(L<=mid) update(lson,L,R);    if(R>mid) update(rson,L,R);    push_up(i);}int query(int i,int l,int r,int L,int R){    if(L<=l&&r<=R) return sum[i];    int ans=0,mid=(l+r)>>1;    push_down(i,r-l+1);    if(L<=mid) ans+=query(lson,L,R);    if(R>mid) ans+=query(rson,L,R);    return ans;}int main(){    int n,m,u;    char s[3];    while(~scanf("%d%d",&n,&m))    {        build(1,1,n);        mem(head,-1),cnt=0,num=0;        for(int i=2;i<=n;i++)            sca(u),add(u,i);        dfs(1);        while(m--)        {            scanf("%s%d",s,&u);            if(s[0]=='o') update(1,1,n,l[u],r[u]);            else pri(query(1,1,n,l[u],r[u]));        }        puts("");    }    return 0;}