ZOJ 3686 A Simple Tree Problem

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码代码的猿猿的AC之路

最新推荐文章于 2021-08-04 15:15:41 发布

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A Simple Tree Problem

Time Limit: 3 Seconds Memory Limit: 65536 KB

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

Sample Output

Author: CUI, Tianyi
Contest: ZOJ Monthly, March 2013

先DFS搜索出每个结点所控制的范围，然后就可以用线段树了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>

#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

using namespace std;

vector<int> g[200010];
int n,m,id;

const int maxn=200010;

struct Interval
{
    int from,to;
}I[200010];

void dfs(int node)
{
    I[node].from=id;
    id++;
    int t=g[node].size();
    for(int i=0;i<t;i++)
    {
        dfs(g[node][i]);
    }
    I[node].to=id-1;
}

int m0[maxn<<2],m1[maxn<<2],xxo[maxn<<2];

void push_up(int rt)
{
    m0[rt]=m0[rt<<1]+m0[rt<<1|1];
    m1[rt]=m1[rt<<1]+m1[rt<<1|1];
}

void push_down(int rt)
{
    if(xxo[rt])
    {
        xxo[rt<<1]^=1; xxo[rt<<1|1]^=1;
        swap(m0[rt<<1|1],m1[rt<<1|1]);swap(m0[rt<<1],m1[rt<<1]);
        xxo[rt]=0;
    }
}

void build(int l,int r,int rt)
{
    xxo[rt]=0;m0[rt]=0;m1[rt]=0;
    if(l==r)
    {
        m0[rt]=1; m1[rt]=0;
        return ;
    }
    int m=(l+r)>>1;
    push_down(rt);
    build(lson); build(rson);
    push_up(rt);
}

void update(int L,int R,int l,int r,int rt)
{
    if(L<=l&&r<=R)
    {
        xxo[rt]^=1;
        swap(m0[rt],m1[rt]);
        return ;
    }
    int m=(l+r)>>1;
    push_down(rt);
    if(L<=m) update(L,R,lson);
    if(R>m) update(L,R,rson);
    push_up(rt);
}

int query(int L,int R,int l,int r,int rt)
{
    if(L<=l&&r<=R)
    {
        push_down(rt);
        return m1[rt];
    }
    int m=(l+r)>>1,ret=0;
    push_down(rt);
    if(L<=m) ret+=query(L,R,lson);
    if(R>m) ret+=query(L,R,rson);
    push_up(rt);
    return ret;
}

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(int i=0;i<=n+10;i++) g[i].clear();
        memset(m0,0,sizeof(m0));
        memset(m1,0,sizeof(m1));
        memset(xxo,0,sizeof(xxo));
        for(int i=2;i<=n;i++)
        {
            int a;
            scanf("%d",&a);
            g[a].push_back(i);
        }
        id=1;
        dfs(1);
        build(1,n,1);
        while(m--)
        {
            char cmd[5]; int a;
            scanf("%s%d",cmd,&a);
            if(cmd[0]=='o')   update(I[a].from,I[a].to,1,n,1);
            else if(cmd[0]=='q')   printf("%d\n",query(I[a].from,I[a].to,1,n,1));
        }
        putchar(10);
    }
    return 0;
}