Reachability Query(Union-Find)

Time Limit: 3 sec / Memory Limit: 1024 MiB

Score : 450 points

Problem Statement

You are given an undirected graph with N vertices and zero edges.

The vertices are numbered 1,2,...,N, and initially all vertices are white.

Process a total of Q queries of the following three types:

Type 1 : Add an undirected edge connecting vertices u and v.
Type 2 : If vertex v is white, change it to black; if it is black, change it to white.
Type 3 : Determine whether a black vertex can be reached from vertex v by traversing zero or more edges; report Yes if reachable, and No otherwise.

Constraints

All input values are integers.
1≤N≤
1≤Q≤
Type 1 queries satisfy the following constraints:
- 1≤u<v≤N
- For each query, no edge connecting u and v has been added before that query.
Type 2,3 queries satisfy the following constraints:
- 1≤v≤N

Input

The input is given from Standard Input in the following format:

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N 𝑄
Query1
Query2
⋮
QueryQ

where Queryi represents the i-th query.

Type 1 queries are given in the following format:

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1 u v

Type 2 queries are given in the following format:

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2 v

Type 3 queries are given in the following format:

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3 v

Output

For each type 3 query, output the answer as follows:

Yes if a black vertex can be reached from vertex v by traversing zero or more edges;
No if no black vertex can be reached from vertex v by traversing zero or more edges.

Sample Input 1

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Sample Output 1

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No
Yes
No
Yes
Yes
No

In this input, the graph initially has five vertices and zero edges.

This input contains 12 queries.

The 1st query is 3 2.
- At this point, no black vertex can be reached from vertex 2 by traversing zero or more edges, so report No.
The 2nd query is 2 2.
- Vertex 22 is white, so change it to black.
The 3rd query is 3 2.
- At this point, black vertex 22 can be reached from vertex 2 by traversing zero or more edges. Therefore, report Yes.
The 4th query is 1 2 5.
- Add an edge connecting vertices 2,5.
The 5th query is 1 3 4.
- Add an edge connecting vertices 3,4.
The 6th query is 3 4.
- At this point, no black vertex can be reached from vertex 4 by traversing zero or more edges, so report No.
The 7th query is 3 5.
- At this point, black vertex 2 can be reached from vertex 5 by traversing zero or more edges. Therefore, report Yes.
The 8th query is 1 4 5.
- Add an edge connecting vertices 4,5.
The 9th query is 1 1 3.
- Add an edge connecting vertices 1,3.
The 10th query is 3 1.
- At this point, black vertex 2 can be reached from vertex 1 by traversing zero or more edges. Therefore, report Yes.
The 11th query is 2 2.
- Vertex 2 is black, so change it to white.
The 12th query is 3 1.
- At this point, no black vertex can be reached from vertex 1 by traversing zero or more edges, so report No.

思路分析

初始化：使用并查集（Union-Find）数据结构来管理连通分量。每个顶点初始时独立为一个分量，颜色均为白色，黑色计数为零。

处理查询：

类型1（添加边）：合并两个顶点所在的连通分量，合并时黑色计数相加。

类型2（切换颜色）：改变顶点颜色，并更新所在分量的黑色计数。白色变黑色时计数增加，黑色变白色时计数减少。

类型3（检查可达性）：检查顶点所在分量的黑色计数是否大于零，若是则输出"Yes"，否则输出"No"。

效率优化：并查集通过路径压缩和按秩合并优化，确保每次操作的时间复杂度接近常数，适用于大规模输入。

代码

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#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N=2e5+5;
vector<int>fa(N),rankk(N,0),color(N,0),cnt_black(N,0);
int n,q,op,u,v;
int find(int x){
    if(fa[x]!=x){
        fa[x]=find(fa[x]);
    }
    return fa[x];
}
void unionset(int x,int y){
    int rx=find(x),ry=find(y);
    if(rx==ry)return;
    if(rankk[rx]>rankk[ry]){
        fa[ry]=rx;
        cnt_black[rx]+=cnt_black[ry];
    }
    else{
        if(rankk[rx]==rankk[ry])rankk[ry]++;
        fa[rx]=ry;
        cnt_black[ry]+=cnt_black[rx];
    }
}
vector<int>black(N,0);
int main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    cin>>n>>q;
    for(int i=1;i<=n;i++){
        fa[i]=i;
    }
    while(q--){
        cin>>op;
        if(op==1){
            cin>>u>>v;
            unionset(u,v);
        }
        else if(op==2){
            cin>>v;
            int c=color[v];
            if(c==0){
                color[v]=1;
                int root=find(v);
                cnt_black[root]++;
            }
            else{
                color[v]=0;
                int root=find(v);
                cnt_black[root]--;
            }
        }
        else{
            cin>>v;
            int root=find(v);
            if(cnt_black[root]>0){
                cout<<"Yes\n";
            }
            else{
                cout<<"No\n";
            }
        }
    }
    return 0;
}