并查集 atcoder D - 連結 / Connectivity

Time limit : 2sec / Memory limit : 256MB

Score : 400 points

Problem Statement

There are N cities. There are also K roads and L railways, extending between the cities. The i-th road bidirectionally connects the pi-th and qi-th cities, and the i-th railway bidirectionally connects the ri-th and si-th cities. No two roads connect the same pair of cities. Similarly, no two railways connect the same pair of cities.

We will say city A and B are connected by roads if city B is reachable from city A by traversing some number of roads. Here, any city is considered to be connected to itself by roads. We will also define connectivity by railways similarly.

For each city, find the number of the cities connected to that city by both roads and railways.

Constraints

• 2≦N≦2*105
• 1≦K,L≦105
• 1≦pi,qi,ri,si≦N
• pi<qi
• ri<si
• When i≠j, (pi,qi)≠(pj,qj)
• When i≠j, (ri,si)≠(rj,sj)

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Input

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

N K L
p1 q1
:
pK qK
r1 s1
:
rL sL

Output

Print N integers. The i-th of them should represent the number of the cities connected to the i-th city by both roads and railways.

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Sample Input 1

Copy

4 3 1
1 2
2 3
3 4
2 3

Sample Output 1

Copy

1 2 2 1

All the four cities are connected to each other by roads.

By railways, only the second and third cities are connected. Thus, the answers for the cities are 1,2,2and 1, respectively.

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Sample Input 2

Copy

4 2 2
1 2
2 3
1 4
2 3

Sample Output 2

Copy

1 2 2 1

---

Sample Input 3

Copy

7 4 4
1 2
2 3
2 5
6 7
3 5
4 5
3 4
6 7

Sample Output 3

Copy

1 1 2 1 2 2 2

原题链接

大意：有公路和铁路两种路，分别给出两种路链接了哪些城市，然后问对于每个城市，可以通过公路铁路两种路都能到达的城市有哪些，包括自身。

解析：被可以包括自身这一点坑到了，然后去查了题解。其实是一道挺简单的并查集，就是在得到公路铁路分别的并查集之后求交集就可以，但是查题解的时候看到了一个不错的操作，就是利用pair和make_pair来写map[代码53行往后]

代码如下

#include<iostream>
#include<cstdio>
#include<map>
#include<sstream>
#include<cstring>
#include<algorithm>
#include<string>
using namespace std;
const int maxn = 2e5+7;
int a[maxn];
int p[maxn],q[maxn],r[maxn];
int found1(int x)
{
    return x==p[x]?x : p[x] = found1(p[x]);
}
int found2(int x)
{
    return x==q[x]?x : q[x] = found2(q[x]);
}
int main()
{
    int n,l,k;
    cin >> n >> k >> l;
    int a,b;
    for(int i = 0; i < n; i ++)
    {
        p[i] = i;
        q[i] = i;
    }
    for(int i = 0; i < k; i ++)
    {
        scanf("%d%d",&a,&b);
        int x = found1(a),y = found1(b);
        if(x!=y)
        {
            p[x] = y;
        }
    }
    for(int i = 0; i < l; i ++)
    {
        scanf("%d%d",&a,&b);
        int x = found2(a),y = found2(b);
        if(x!=y)
        {
            q[x] = y;
        }
    }
    for(int i = 1; i <= n; i ++)
    {
        found1(i);
        found2(i);
    }
    map<pair<int,int>,int>mp;
    for(int i = 1; i <= n; i ++)
    {
        mp[make_pair(p[i],q[i])]++;
    }
    for(int i = 1; i <= n; i ++)
    {
        cout << mp[make_pair(p[i],q[i])] << " ";
    }
    cout <<endl;
    return 0;
}