UVALive 6910 Cutting Tree 并查集

盛夏的果实

Cutting Tree

题目连接：
https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4922

Description
Tree in graph theory refers to any connected graph (of nodes and edges) which has no simple cycle,
while forest corresponds to a collection of one or more trees. In this problem, you are given a forest of
N nodes (of rooted trees) and K queries. Each query is in the form of:
· C x : remove the edge connecting node and its parent. If node has no parent, then ignore this
query.
· Q a b : output ‘YES’ if there is a path from node to node in the forest; otherwise, ‘NO’.
For example, let the initial forest is shown by Figure 1.
Figure 1. Figure 2.
Let’s consider the following queries (in order):

• Q 5 7 : output YES.
  
• C 2 : remove edge (2, 1) — the resulting forest is shown in Figure 2.
  
• Q 5 7 : output NO, as there is no path from node 5 to node 7 in Figure 2.
  
• Q 4 6 : output YES.
  

Input
The first line of input contains an integer T (T ≤ 50) denoting the number of cases. Each case begins
with two integers: N and K (1 ≤ N ≤ 20, 000; 1 ≤ K ≤ 5, 000) denoting the number of nodes in the
forest and the number of queries respectively. The nodes are numbered from 1 to N. The next line
contains N integers Pi (0 ≤ Pi ≤ N) denoting the parent of i-th node respectively. Pi = 0 means that
node i does not have any parent (i.e. it’s a root of a tree). You are guaranteed that the given input
corresponds to a valid forest. The next K lines represent the queries. Each query is in the form of ‘C
x’ or ‘Q a b’ (1 ≤ x, a, b ≤ N), as described in the problem statement above

Output
For each case, output ‘Case #X:’ in a line, where X is the case number starts from 1. For each ‘Q
a b’ query in the input, output either ‘YES’ or ‘NO’ (without quotes) in a line whether there is a path
from node a to node b in the forest.
Explanation for 2nd sample case:
The initial forest is shown in Figure 3 below.

• C 3 : remove edge (3, 2) — the resulting forest is shown in Figure 4.
  
• Q 1 2 : output YES.
  
• C 1 : remove edge (1, 2) — the resulting forest is shown in Figure 5.
  
• Q 1 2 : output NO as there is no path from node 1 to node 2 in Figure 5
  

Sample Input
4
7 4
0 1 1 2 2 2 3
Q 5 7
C 2
Q 5 7
Q 4 6
4 4
2 0 2 3
C 3
Q 1 2
C 1
Q 1 2
3 5
0 3 0
C 1
Q 1 2
C 3
C 1
Q 2 3
1 1
0
Q 1 1

Sample Output
Case #1:
YES
NO
YES
Case #2:
YES
NO
Case #3:
NO
YES
Case #4:
YES

Hint

题意
给你个森林，俩操作，1是砍掉与他父亲的连边，2是查询xy是否在同一个连通块里面

题解：
倒着做，砍边就变成连边了，然后并茶几莽一波就好了

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e4+7;
int cas = 0;
int fa[maxn];
int e[maxn];
int flag[maxn];
int a[maxn],b[maxn],c[maxn];;
int fi(int x){
  if(x==fa[x])return x;
  return fa[x]=fi(fa[x]);
}
void init(){
  memset(flag,0,sizeof(flag));
}
void solve(){
  init();
  vector<int>ans;
  int n,m;
  scanf("%d%d",&n,&m);
  for(int i=1;i<=n;i++)
    fa[i]=i;
  for(int i=1;i<=n;i++)
    scanf("%d",&e[i]);
  for(int i=1;i<=m;i++){
    string s;cin>>s;
    if(s[0]=='C'){
      a[i]=1;
      scanf("%d",&b[i]);
      flag[b[i]]++;
    }else{
      a[i]=0;
      scanf("%d%d",&b[i],&c[i]);
    }
  }
  for(int i=1;i<=n;i++){
    if(flag[i]==0&&e[i]!=0){
      fa[fi(i)]=fi(e[i]);
    }
  }
  for(int i=m;i>=1;i--){
    if(a[i]==1){
      flag[b[i]]--;
      if(flag[b[i]]==0&&e[b[i]]!=0)
        fa[fi(b[i])]=fi(e[b[i]]);
    }else{
      if(fi(b[i])==fi(c[i]))ans.push_back(1);
      else ans.push_back(0);
    }
  }
  for(int i=ans.size()-1;i>=0;i--){
    if(ans[i])printf("YES\n");
    else printf("NO\n");
  }
}
int main(){
  //freopen("1.txt","r",stdin);
  int t;
  scanf("%d",&t);
  while(t--){
    printf("Case #%d:\n",++cas);
    solve();
  }
  return 0;
}