1 /*
 2 6
 3 1 1
 4 2 1
 5 5 1
 6 6 6 
 7 4 6
 8 7 4
 9 */
10 #define _CRT_SECURE_NO_WARNINGS
11 #include <iostream>
12 #include <cstring>
13 #include <cstdio>
14 #include <cstdlib>
15 using namespace std;
16 
17 const int maxn = 1000 + 100;
18 int par[maxn];     //父亲,  当par[x] = x时,x是所在的树的根
19 int Rank[maxn];    //树的高度
20 
21 //初始化n个元素
22 void init(int n)
23 {
24     for (int i = 0; i < n; i++) {
25         par[i] = i;
26         Rank[i] = 0;
27     }
28 }
29 
30 //查询树的根
31 int find(int x) {
32     if (par[x] == x) {
33         return x;
34     }
35     else {
36         return par[x] = find(par[x]);
37     }
38 }
39 
40 //合并x和y所属集合
41 void unite(int x, int y) {
42     x = find(x);
43     y = find(y);
44     if (x == y) return;
45     
46     if (Rank[x] < Rank[y]) {
47         par[x] = y;
48     } else {
49         par[y] = x;
50         if (Rank[x] == Rank[y]) Rank[x]++;    //如果x,y的树高相同,就让x的树高+1
51     }
52 }
53 
54 //判断x和y是否属于同一个集合
55 bool same(int x, int y) {
56     return find(x) == find(y);
57 }
58 
59 int main()
60 {
61     int n;
62     scanf("%d", &n);
63     init(n);
64 
65     int data, p;
66     cout << "输入数据: 
";
67     for (int i = 0; i < n; i++) {
68         scanf("%d%d", &data, &p);
69         par[data] = p;
70         Rank[p]++;
71     }
72 
73     cout << "输入合并集合: 
";
74     int p1, p2;
75     cin >> p1 >> p2;
76     unite(p1, p2);
77     cout << "查询是否属于一个集合: 
";
78     cin >> p1 >> p2;
79 
80     if (same(p1, p2)) {
81         puts("same");
82     }
83     else {
84         puts("diff");
85     }
86 
87     return 0;
88 }

  1 #define _CRT_SECURE_NO_WARNINGS
  2 #include <iostream>
  3 #include <cstring>
  4 #include <cstdio>
  5 using namespace std;
  6 
  7 const int maxn = 100000 * 2 + 100000;
  8 //输入(T是信息的类型)
  9 int N, K;           //N-N种类, K-K条信息
 10 int T[maxn], X[maxn], Y[maxn];
 11 
 12 //在这里省略了并查集部分代码
 13 int par[maxn];
 14 int Rank[maxn];
 15 void solve();
 16 void init(int n);
 17 int find(int x);
 18 void unite(int x, int y);
 19 bool same(int x, int y);
 20 
 21 void init(int n) {
 22     for (int i = 0; i < n; i++) {
 23         par[i] = i;
 24         Rank[i] = 0;
 25     }
 26 }
 27 
 28 //查询树根
 29 int find(int x)
 30 {
 31     if (par[x] == x) {
 32         return x;
 33     }
 34     else {
 35         return par[x] = find(par[x]);
 36     }
 37 }
 38 
 39 //合并x和y所属的集合
 40 void unite(int x, int y)
 41 {
 42     x = find(x);
 43     y = find(y);
 44     if (x == y) return;
 45 
 46     if (Rank[x] < Rank[y]) {
 47         par[x] = y;
 48     }
 49     else {
 50         par[y] = x;
 51         if (Rank[x] == Rank[y]) Rank[x]++;
 52     }
 53 }
 54 
 55 bool same(int x, int y) {
 56     return find(x) == find(y);
 57 }
 58 void solve()
 59 {
 60     //初始化并查集
 61     //元素x, x + N, x + 2 * N 分别代码 x-A, x-B, x-C
 62     init(N * 3);
 63 
 64     int ans = 0;
 65     for (int i = 0; i < K; i++) {
 66         int t = T[i];
 67         int x = X[i] - 1, y = Y[i] - 1; //把输入变成0,...N-1的范围
 68 
 69         //不正确的编号
 70         if (x < 0 || N <= x || y < 0 || N <= y) {
 71             ans++; continue;
 72         }
 73         
 74         if (t == 1) {
 75             // "x和y属于同一类"的信息
 76             if (same(x, y + N) || same(x, y + 2 * N)) {
 77                 ans++;
 78             }
 79             else {
 80                 unite(x, y);
 81                 unite(x + N, y + N);
 82                 unite(x + N * 2, y + N * 2);
 83             }
 84         }
 85         else {
 86             //"x吃y"的信息
 87             if (same(x, y) || same(x, y + 2 * N)) {  //A和A，A和C不能相等
 88                 ans++; 
 89             }
 90             else {
 91                 unite(x, y + N);         
 92                 unite(x + N, y + 2 * N);
 93                 unite(x + 2 * N, y);
 94             }
 95         }
 96     }
 97     printf("%d
", ans);
 98 }
 99 int main()
100 {
101     cin >> N >> K;
102     for (int i = 0; i < K; i++)
103     {
104         cin >> T[i] >> X[i] >> Y[i];
105     }
106     solve();
107     return 0;
108 }