并查集 File Transfer

We have a network of computers and a list of bi-directional connections. Each of these connections allows a file transfer from one computer to another. Is it possible to send a file from any computer on the network to any other?

Input Specification:

Each input file contains one test case. For each test case, the first line contains $N$ (2≤N≤1042\le N\le 10^42≤N≤10​4​​), the total number of computers in a network. Each computer in the network is then represented by a positive integer between 1 and $N$. Then in the following lines, the input is given in the format:

I c1 c2

where I stands for inputting a connection between c1 and c2; or

C c1 c2

where C stands for checking if it is possible to transfer files between c1and c2; or

S

where S stands for stopping this case.

Output Specification:

For each C case, print in one line the word "yes" or "no" if it is possible or impossible to transfer files between c1 and c2, respectively. At the end of each case, print in one line "The network is connected." if there is a path between any pair of computers; or "There are k components." where kis the number of connected components in this network.

Sample Input 1:

5
C 3 2
I 3 2
C 1 5
I 4 5
I 2 4
C 3 5
S

Sample Output 1:

no
no
yes
There are 2 components.

Sample Input 2:

5
C 3 2
I 3 2
C 1 5
I 4 5
I 2 4
C 3 5
I 1 3
C 1 5
S

Sample Output 2:

no
no
yes
yes
The network is connected.

思路：经典的并查集，但是需要进行优化才能AC

1. 未优化版

#include <iostream>

using namespace std;

int adj[10002];

int myfind(int i) {
    while(adj[i] != 0)
        i = adj[i];
    return i;
}

void myunion(int i, int j) {
    int f1 = myfind(i);
    int f2 = myfind(j);
    
    if(adj[f2] > adj[f1])
        adj[f1] = f2;
}

int main()
{
    
    int n;
    cin >> n;

    while(true) {
        char c;
        cin >> c;
        if(c == 'S')    break;

        int i, j;
        cin >>i >> j;
        if(c == 'I') {
            myunion(i, j);
        } else {
            int f1 = myfind(i);
            int f2 = myfind(j);
            if(f1 == f2)
                cout << "yes" << endl;
            else
                cout << "no" << endl;
        }
    }

    int cnt = 0;
    for(int i=1; i<=n; i++)
        if(adj[i] == 0)	cnt ++;
    if(cnt == 1)
        cout << "The network is connected." << endl;
    else
        cout << "There are " << cnt << " components." << endl;

    return 0;
}

2. 优化版

#include <iostream>

using namespace std;

int adj[10002];

// 路径压缩，用g++编译器报 段错误 好像没有对尾递归进行优化
int myfind(int i) {
//    if(adj[i] < 0)
//        return i;
//    else
//        return adj[i] = myfind(adj[i]);

    while(adj[i] > 0)
        i = adj[i];
    return i;
}

void myunion(int i, int j) {
    int f1 = myfind(i);
    int f2 = myfind(j);

    // 这里用的是按规模归并，每次归并都要注意更新规模
    // 更新规模要在归并前！！
//    if(adj[f2] < adj[f1]) {
//        adj[f2] += adj[f1];
//        adj[f1] = f2;
//    } else {
//        adj[f1] += adj[f2];
//        adj[f2] = f1;
//    }

    // 按高度归并
    if(adj[f1] < adj[f2])
        adj[f2] = f1;
    else {
        if(adj[f1] == adj[f2])  // 只有当树一样高的时候，才要树高加1
            adj[f2]--;
        adj[f1] = f2;
    }
}

// 考察优化：
// 1. 按秩归并：按高度或者按规模选择谁加到谁
// 2. 路径压缩：每次find了自后，都把遍历过的节点的adj都更新为最顶端的父节点

int main()
{

    int n;
    cin >> n;

    while(true) {
        char c;
        cin >> c;
        if(c == 'S')    break;

        int i, j;
        cin >>i >> j;
        if(c == 'I') {
            myunion(i, j);
        } else {
            int f1 = myfind(i);
            int f2 = myfind(j);
            if(f1 == f2)
                cout << "yes" << endl;
            else
                cout << "no" << endl;
        }
    }

    int cnt = 0;
    for(int i=1; i<=n; i++)
        if(adj[i] <= 0)	cnt ++;
    if(cnt == 1)
        cout << "The network is connected." << endl;
    else
        cout << "There are " << cnt << " components." << endl;

    return 0;
}