CF 题目集锦 PART 5 #266 div 2 E_html/css_WEB-ITnose-html教程-PHP中文网

CF 题目集锦 PART 5 #266 div 2 E_html/css_WEB-ITnose

php中文网

发布： 2016-06-24 11:57:54

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【原题】

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E. Information Graph

time limit per test

1 second

memory limit per test

512 megabytes

input

standard input

output

standard output

There are n employees working in company "X" (let's number them from 1 to n for convenience). Initially the employees didn't have any relationships among each other. On each of m next days one of the following events took place:

either employee y became the boss of employee x (at that, employee x didn't have a boss before);

or employee x gets a packet of documents and signs them; then he gives the packet to his boss. The boss signs the documents and gives them to his boss and so on (the last person to sign the documents sends them to the archive);

or comes a request of type "determine whether employee x signs certain documents".

Your task is to write a program that will, given the events, answer the queries of the described type. At that, it is guaranteed that throughout the whole working time the company didn't have cyclic dependencies.

Input

The first line contains two integers n and m (1?≤?n,?m?≤?105) ? the number of employees and the number of events.

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Each of the next m lines contains the description of one event (the events are given in the chronological order). The first number of the line determines the type of event t (1?≤?t?≤?3).

If t?=?1, then next follow two integers x and y (1?≤?x,?y?≤?n) ? numbers of the company employees. It is guaranteed that employee x doesn't have the boss currently.

If t?=?2, then next follow integer x (1?≤?x?≤?n) ? the number of the employee who got a document packet.

If t?=?3, then next follow two integers x and i (1?≤?x?≤?n; 1?≤?i?≤?[number of packets that have already been given]) ? the employee and the number of the document packet for which you need to find out information. The document packets are numbered started from 1 in the chronological order.

It is guaranteed that the input has at least one query of the third type.

Output

For each query of the third type print "YES" if the employee signed the document package and "NO" otherwise. Print all the words without the quotes.

Sample test(s)

input

4 91 4 32 43 3 11 2 32 23 1 21 3 12 23 1 3

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output

YESNOYES

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【题意】意思看了半天~就是有N个人，M个操作。每次操作有3种。

①把x的父亲置为y。

②从x开始发新的一份文件（是从1开始标号的）。从x开始，一直传递到最远的祖先。

③询问x号文件有没有落到y的手里。

【分析】即要简单地判断y是否是x的祖先。一开始有思维定势，觉得要求一遍LCA。但是后来想了想，可以直接用DFS序搞出同一棵树的深度，再用并查集维护是否在同一棵树上。代码很简单。

【代码】

#include<cstdio>#include<vector>#define N 200005#define pb push_backusing namespace std;vector<int>son[N],ask[N];struct arr{int x,y,id,opt;}a[N];int f[N],L[N],R[N],fa[N],ans[N];int i,n,m,now,num,id,tot,P,j;inline int get(int u){return f[u]==u?f[u]:f[u]=get(f[u]);}void dfs(int k){  L[k]=++tot;  for (int i=0;i<son[k].size();i++)    dfs(son[k][i]);  R[k]=tot;}int main(){  scanf("%d%d",&n,&m);  for (i=1;i<=m;i++)  {    scanf("%d%d",&a[i].opt,&a[i].x);a[i].id=i;    if (a[i].opt!=2) scanf("%d",&a[i].y);    if (a[i].opt==1) son[a[i].y].pb(a[i].x),fa[a[i].x]=1;    if (a[i].opt==3) ask[a[i].y].pb(a[i].id);  }  for (i=1;i<=n;i++) if (!fa[i]) dfs(i);  for (i=1;i<=n;i++) f[i]=i;now=0;  for (i=1;i<=m;i++)  {    if ((P=a[i].opt)==1) {f[get(a[i].x)]=get(a[i].y);continue;}    if (P==3) {puts(ans[i]?"YES":"NO");continue;}now++;    for (j=0;j<ask[now].size();j++)    {      id=ask[now][j];num=a[id].x;      if (get(num)==get(a[i].x)&&L[num]<=L[a[i].x]&&R[num]>=R[a[i].x]) ans[id]=1;    }  }}

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