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ZOJ 3686 A Simple Tree Problem(线段树)

2024-10-05 11:22:02   217人阅读

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

3 2
1 1
o 2
q 1

Sample Output

1

题意：给你一颗树。而且有两种操作，一种是将以节点i为root的子树每一个节点上的值取反，另一种操作时
       查询以节点i为root的子树全部节点上的值得和。节点上的值不是1就是0，初始皆为0。

题解：把子树化为区间的就是裸的线段树了。我们能够先序遍历树，把以x为根节点的子树的用L[x],R[x]区间表示
         然后就是线段树区间更新了。

#include<cstring>
#include<cstdio>
#include<iostream>
#include<cmath>
#include<vector>
#include<algorithm>
#define lc idx<<1
#define rc idx<<1|1
#define lson l,mid,lc
#define rson mid+1,r,rc
#define N 100010

using namespace std;

int n,m;
vector<int>G[N];
int L[N],R[N],id;

struct Tree {
    int st;
    int one;
} tree[N<<2];

void init() {
    for(int i=0; i<=n; i++)
        G[i].clear();
    id=1;
}

void dfs(int fa) {
    L[fa]=id++;
    for(int i=0; i<G[fa].size(); i++) {
        dfs(G[fa][i]);
    }
    R[fa]=id-1;
}

void push_up(int idx) {
    tree[idx].one=tree[lc].one+tree[rc].one;
}

void push_down(int idx,int m) {
    if(tree[idx].st) {
        tree[lc].st^=1,tree[rc].st^=1;
        tree[idx].st=0;
        tree[lc].one=m-m/2-tree[lc].one;
        tree[rc].one=m/2-tree[rc].one;
    }
}

void build(int l,int r,int idx) {
    tree[idx].one=0;
    tree[idx].st=0;
    if(l==r)return;
    int mid=(l+r)>>1;
    build(lson);
    build(rson);
}

void update(int l,int r,int idx,int x,int y) {
    if(x<=l&&r<=y) {
        tree[idx].st^=1;
        tree[idx].one=r-l+1-tree[idx].one;
        return;
    }
    push_down(idx,r-l+1);
    int mid=(l+r)>>1;
    if(x<=mid)update(lson,x,y);
    if(y>mid) update(rson,x,y);
    push_up(idx);
}

int query(int l,int r,int idx,int x,int y) {
    if(x<=l&&r<=y)
        return tree[idx].one;
    push_down(idx,r-l+1);
    int mid=(l+r)>>1;
    int ans=0;
    if(x<=mid) ans+=query(lson,x,y);
    if(y>mid)  ans+=query(rson,x,y);
    return ans;
}

int main() {
    //freopen("test.in","r",stdin);
    while(~scanf("%d%d",&n,&m)) {
        init();
        int fa;
        for(int i=2; i<=n; i++) {
            scanf("%d",&fa);
            G[fa].push_back(i);
        }
        dfs(1);
        build(1,n,1);
        char c[3];
        int rt;
        while(m--) {
            scanf("%s%d",c,&rt);
            if(c[0]==‘o‘) {
                update(1,n,1,L[rt],R[rt]);
                continue;
            }
            printf("%d\n",query(1,n,1,L[rt],R[rt]));
        }
        printf("\n");
    }
    return 0;
}


ZOJ 3686 A Simple Tree Problem(线段树)


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