BNU A Simple Tree Problem 线段树

最新推荐文章于 2018-09-14 22:44:08 发布

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本文介绍了一种基于DFS和线段树的数据结构处理方法，用于解决在一个有根树上进行节点值翻转操作及查询指定子树中特定节点数量的问题。通过递归遍历与区间更新策略，实现了高效的操作响应。

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A Simple Tree Problem

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：有几组操作，s[0]=='o'时改变该子树的所有值，使之和原来的值相反，s[0]=='q'时问该子树上有几个点为1

思路：dfs+线段树，这题跟poj3321 差不多

一直WA了一个早上，只因为把初始化放错位置了

代码：

#include <iostream>
#include <stdio.h>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <map>
#include <queue>
#define lson l,mid,num<<1
#define rson mid+1,r,num<<1|1
using namespace std;
const int M=100050;
struct node
{
    int v;
    int next;
} edge[M*2];
int sum[M*4],cnt[M*4],head[M];
int low[M],high[M],vis[M];
int cc,e,n,m;
void init()
{
    memset(vis,0,sizeof(vis));
    memset(head,-1,sizeof(head));
    memset(low,0,sizeof(low));
    memset(high,0,sizeof(high));
    cc=e=0;
}
void addEdge(int u,int v)
{
    edge[e].v=v;
    edge[e].next=head[u];
    head[u]=e++;
}
void dfs(int u)
{
    low[u]=++cc;
    vis[u]=1;
    for(int k =head[u]; k!=-1; k=edge[k].next)
    {
        if(!vis[edge[k].v])
            dfs(edge[k].v);
    }
    high[u]=cc;
}
void down(int num,int mm)
{
    if(cnt[num]%2!=0)
    {
        cnt[num<<1]+= cnt[num];
        cnt[num<<1|1]+=cnt[num];
        sum[num<<1] = (mm-(mm>>1))-sum[num<<1];
        sum[num<<1|1] =(mm>>1)-sum[num<<1|1];
        cnt[num]=0;
    }
}
void up(int num)
{
    sum[num] = sum[num<<1] + sum[num<<1|1];
}
void build(int l,int r,int num)
{
    //必须写外面，真是，WA了无数次
    sum[num]=0;
    cnt[num]=0;
    if(l==r)
    {　　　	//sum[num]=0;
    	//cnt[num]=0;

return; } int mid = (l+r)>>1; build(lson); build(rson);}void update(int L,int R ,int l, int r ,int num){ if(L<=l && r <=R ) { cnt[num]++; sum[num]=r-l+1-sum[num]; return; } down(num,r-l+1); int mid = (l+r)>>1; if(L <= mid)update(L,R,lson); if(R > mid)update(L,R,rson); up(num);}int query(int L,int R,int l,int r,int num){ if(L <= l && r <= R) { return sum[num]; } down(num,r-l+1); int ret=0; int mid = (l+r)>>1; if(L <= mid ) ret += query(L,R,lson); if(R > mid)ret += query(L,R,rson); return ret;}int main(){ while(scanf("%d%d",&n,&m)!=EOF) { init(); int u; for(int i=2; i<=n; i++) { scanf("%d",&u); addEdge(u,i); addEdge(i,u); } dfs(1); build(1,n,1); char s[2]; int nod; for(int i=1; i<=m; i++) { scanf("%s%d",s,&nod); if(s[0]=='o') { update(low[nod],high[nod],1,n,1); } else if(s[0]=='q') { printf("%d\n",query(low[nod],high[nod],1,n,1)); } } printf("\n"); } return 0;}