ZOJ 3686 A Simple Tree Problem(树转线段树+线段树区间更新)
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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.
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.
For each query, output an integer in a line.
Output a blank line after each test case.
3 21 1o 2q 1
1
题解:
题意:
给你一堆节点和节点的父节点,初始全部节点为1,后来又m组操作,输入o x就是将节点x的所有子节点值反转,q x就是询问节点x子节点中为1的个数
题目的难点在于将数转化为线段树,我们先按照他给的父子关系用动态数组建好一棵树,然后将计数器id置为0,用dfs从1开始向下先序遍历,每进入一个节点,id++,然后先序遍历完该节点的所有子节点,可以发现此时id就为该节点控制的所有子节点的右边界,所以刚进入该节点的时候也是该子节点的左边界,这样就获得了每个节点控制的子节点的区间,然后剩下的就是线段树的区间更新了
代码:
#include<algorithm>#include<iostream>#include<cstring>#include<stdio.h>#include<math.h>#include<string>#include<stdio.h>#include<queue>#include<stack>#include<map>#include<deque>#define M (t[k].l+t[k].r)/2#define lson k*2#define rson k*2+1#define ll long longusing namespace std;vector<int>p[100005];//存i的子节点的动态数组int id;struct dev//存每个节点管理的区间左右边界{ int l,r;}a[100005];void dfs(int x)//先序遍历树{ a[x].l=++id;//记录左边界 for(int i=0;i<p[x].size();i++)//遍历每一个子节点 { dfs(p[x][i]); } a[x].r=id;//记录右边界}struct node//线段树部分{ int l,r; int sum; int tag;//反转标记,如果tag%2==1,就是要反转,也可以写成异或 int len() { return r-l+1; }}t[100005*4];void Build(int l,int r,int k){ t[k].l=l; t[k].r=r; t[k].sum=0; t[k].tag=0; if(l==r) return; int mid=M; Build(l,mid,lson); Build(mid+1,r,rson);}void pushup(int k)//向上更新区间{ t[k].sum=t[lson].sum+t[rson].sum;}void pushdown(int k)//向下传递状态,更新子区间{ if(t[k].tag%2) { t[lson].tag+=t[k].tag; t[lson].sum=t[lson].len()-t[lson].sum;//区间长度-当前1的个数就是反转以后1的个数 t[rson].tag+=t[k].tag; t[rson].sum=t[rson].len()-t[rson].sum; t[k].tag=0; }}void update(int l,int r,int k)//日常更新{ if(t[k].l==l&&t[k].r==r) { t[k].tag++; t[k].sum=t[k].len()-t[k].sum; return; } pushdown(k); int mid=M; if(r<=mid) update(l,r,lson); else if(l>mid) update(l,r,rson); else { update(l,mid,lson); update(mid+1,r,rson); } pushup(k);}int query(int l,int r,int k)//日常询问{ if(t[k].l==l&&t[k].r==r) { return t[k].sum; } pushdown(k); int mid=M; if(r<=mid) return query(l,r,lson); else if(l>mid) return query(l,r,rson); else { return query(l,mid,lson)+query(mid+1,r,rson); }}int main(){ int i,j,n,m,x; char s[10]; while(scanf("%d%d",&n,&m)!=EOF) { for(i=1;i<=n;i++) p[i].clear(); for(i=2;i<=n;i++) { scanf("%d",&x); p[x].push_back(i); } id=0; dfs(1); Build(1,n,1); for(i=0;i<m;i++) { scanf("%s%d",s,&x); if(s[0]=='o') { update(a[x].l,a[x].r,1);//更新的是子节点的左右边界 } else { printf("%d\n",query(a[x].l,a[x].r,1)); } } printf("\n"); } return 0;}
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