## Sunday, January 4, 2015

### Codeforces 466E - Information Graph

Problem Statement:
466E - Information Graph

Solution:
The main idea is to build the whole relationship graph before answering the queries of type 3. We can make use of the fact that if we run DFS on the root of a graph and maintain the time information when we first enter and leave a certain node, we can check whether u is a parent of v on the DFS tree in O(1) by comparing the time information. While building the information graph, we also use a disjoint set union structure to maintain information about the parents of each nodes in the connected components currently being built. Finally, a node x receives a packet with index i if and only if x is in between the vertex u that received that packet and the vertex v that was the parent of u when it received the packet. After we build the graph and the disjoint set union structures, we can serve each queries of type 3 in O(1).

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cassert>
#include <vector>
using namespace std;

int col[100005][2], par[100005];
int t[100005][2];
int vis[100005];
vector<vector<int> > adj;
vector<int> P, Q;
int N, M, K;
int ctr;

void init(){
for(int i=0;i<=N;++i) par[i] = i;
}

int find(int u){
return (par[u] == u ? u : (par[u] = find(par[u])));
}

void dfs(int u){
vis[u] = 1;
t[u][0] = ctr++;
for(int i=0;i<adj[u].size();++i){
int v = adj[u][i];
if(vis[v])continue;
dfs(v);
}
t[u][1] = ctr++;
}

int main(){
int type,u,v;
K = 1;
scanf("%d %d", &N, &M);
adj = vector<vector<int> > (N+3);
init();
for(int i=0;i<M;++i){
scanf("%d", &type);
if(type == 1){
scanf("%d %d", &u, &v);
par[find(u)] = v;
adj[u].push_back(v);
adj[v].push_back(u);
} else if(type==2){
scanf("%d", &u);
col[K][0] = u;
col[K][1] = find(u);
++K;
} else if(type==3){
scanf("%d %d", &u,&v);
P.push_back(u);
Q.push_back(v);
}
}
ctr = 0;
for(int i=1;i<=N;++i){
if(par[i]==i){
dfs(i);
}
}
for(int i=0;i<P.size();++i){
u = col[Q[i]][0];
v = col[Q[i]][1];
if(t[u][0] >= t[P[i]][0] && t[P[i]][1] >= t[u][1] &&
t[v][0] <= t[P[i]][0] && t[P[i]][1] <= t[v][1]){
printf("YES\n");
} else {
printf("NO\n");
}
}
return 0;
}