牛客多校(2020第八场)I - Interesting Computer Game

Apollo is playing an interesting computer game. There are N rounds in the game.
At each round, the computer will give Apollo two integers ( aia_iai​ and bib_ibi​), and Apollo can do exactly one of the following three actions.

•  Apollo can do nothing.
•  If integer aia_iai​ has not been selected in all previous rounds, Apollo can select integer aia_iai​.
•  If integer bib_ibi​ has not been selected in all previous rounds, Apollo can select integer bib_ibi​.

Apollo has cracked the game, and he has known all the candidate numbers of each round before the game starts. Now he wants to know the maximum number of integers he can select with the optimal strategy.
I believe it would be very simple question for you, please help Apollo solve this question.

输入描述:

The first line is an integer T\mathbf{T}T (1≤T≤101 \leq \mathbf{T} \leq 101≤T≤10), which is the number of test cases.

Each test case begins with a line containing a positive integer N (1≤N≤1051 \le N \le 10^51≤N≤105), indicating the number of rounds in this game.

Then N lines follow. The i-th line contains two integers aia_iai​ and bib_ibi​ (1≤ai≤1091 \le a_i \le 10^91≤ai​≤109, 1≤bi≤1091 \le b_i \le 10^91≤bi​≤109), indicating that two integers of the i-th round.

输出描述:

For each test case, output one line containing ‘‘Case #x: y′′``Case\ \#x:\ y''‘‘Case #x: y′′, where x is the test case number and y is the answer.

示例1

输入

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

输出

Case #1: 4
Case #2: 4

题意：

• 给了俩个数组a, b
• 第i步可从ai和bi种选择一个数
• 求最后选择的数的个数，不同的数要多

题解：

　首先要想到可以用并查集做题，然后将ai, bi看成俩个点，[ai, bi]看成一条边，所以题目会给出 n 条边，然后根据并查集的合成，会形成一个或多个连通块。

　连通块有俩种形式

• 有环：n个点，n条边，如果选数，每条边可以选择一个点， 则最多可以选择 n 个数
• 无环：n个点，n-1条边，如果选数，则只能选择n-1个数

　最后，因为ai, bi的数据有点大，所以需要对数据进行离散化，可以选择使用map数组或者unordered_map(当然unordered_map效率会更高一些)。

 

 

 

 1 /* 
 2 并查集，离散化
 3 合并的集合n个顶点有环，则可选择n个，无环则只能选择n-1个
 4  */
 5 #include
 6 #include
 7 #include
 8 #include
 9 
10 using namespace std;
11 
12 typedef long long ll;
13 const int N = 1e5+5;
14 
15 int par[2 * N];//并查集数组,记录每个结点的父节点
16 bool circle[2 * N]; //记录有没有环
17 unordered_map<int, int> mp; //离散化,第一个是原来的值，第二个是离散化后的值
18 
19 inline int read() {
20     int x = 0, f = 1;
21     char ch = getchar();
22     while(ch<'0'||ch>'9'){
23         if(ch=='-')
24             f=-1;
25         ch=getchar();
26     }
27     while(ch>='0'&&ch<='9'){
28         x = x * 10 + ch - '0';
29         ch = getchar();
30     }
31     return x * f;
32 }
33 
34 int find(int x) {
35     return par[x] == x ? x : par[x] = find(par[x]); //如果自己是自己的父节点则返回自己，否则寻找父节点的父节点，并更新
36 }
37 
38 void merge(int x, int y) {
39     int par_x = find(x), par_y = find(y); //寻找x, y的父节点
40     if (par_x != par_y) { //父节点不同，则合并
41         par[par_x] = par_y; //将x的父节点的父节点设置为y
42         circle[par_y] |= circle[par_x]; //若x的父节点有环，则y的父节点也该变为有环(需位的或操作)
43     }
44     else {
45         circle[par_x] = true; //若父节点相同，则其有环
46     }
47 }
48 
49 void init() {
50     mp.clear();
51     for (int i = 0; i < 2 * N; i++) {
52         par[i] = i;
53         circle[i] = false;
54     }
55 }
56 
57 int main() {
58     int t;
59     t = read();
60     for (int k = 1; k <= t; k++) {
61         init();
62         int n = read();
63 
64         int cnt = 0; //用于离散化
65         for (int i = 0; i < n; i++) {
66             int a, b;
67             a = read();
68             b = read();
69             //离散化
70             if (!mp[a]) mp[a] = ++cnt;
71             if (!mp[b]) mp[b] = ++cnt;
72             merge(mp[a], mp[b]);
73         }
74 
75         int ans = cnt;
76         
77         for (int i = 1; i <= cnt; i++) {
78             if (!circle[i] && par[i] == i)  ans--; //对于每个并查集，如果无环则需减1
79         }
80         printf ("Case #%d: %d\n", k, ans);
81     }
82 }