PAT_A1118#Birds in Forest

Source:

PAT A1118 Birds in Forest (25 分)

Description:

Some scientists took pictures of thousands of birds in a forest. Assume that all the birds appear in the same picture belong to the same tree. You are supposed to help the scientists to count the maximum number of trees in the forest, and for any pair of birds, tell if they are on the same tree.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive number N (≤) which is the number of pictures. Then N lines follow, each describes a picture in the format:

K B​1​​ B​2​​ ... B​K​​

where K is the number of birds in this picture, and B​i​​'s are the indices of birds. It is guaranteed that the birds in all the pictures are numbered continuously from 1 to some number that is no more than 1.

After the pictures there is a positive number Q (≤) which is the number of queries. Then Q lines follow, each contains the indices of two birds.

Output Specification:

For each test case, first output in a line the maximum possible number of trees and the number of birds. Then for each query, print in a line Yes if the two birds belong to the same tree, or No if not.

Sample Input:

4
3 10 1 2
2 3 4
4 1 5 7 8
3 9 6 4
2
10 5
3 7

Sample Output:

2 10
Yes
No

Keys:

• 并查集（Union-Find）

Code:

 1 /*
 2 Data: 2019-06-23 14:01:57
 3 Problem: PAT_A1118#Birds in Forest
 4 AC: 19:34
 5 
 6 题目大意：
 7 一张照片中的所有鸟都在同一棵树上，现在给出多张照片，
 8 找出一共有多少棵树，并判断给定的两只鸟是否在同一棵树上。
 9 
10 基本思路：
11 并查集
12 */
13 #include<cstdio>
14 #include<set>
15 using namespace std;
16 const int M=1e4+10;
17 int mp[M];
18 
19 int Father(int v)
20 {
21     int x=v,s;
22     while(mp[v] != v)
23         v = mp[v];
24     while(mp[x] != x){
25         s = mp[x];
26         mp[x] = v;
27         x = s;
28     }
29     return v;
30 }
31 
32 void Union(int v1, int v2)
33 {
34     int f1 = Father(v1);
35     int f2 = Father(v2);
36     mp[f2] = f1;
37     Father(v2);
38 }
39 
40 int main()
41 {
42 #ifdef ONLINE_JUDGE
43 #else
44     freopen("Test.txt", "r", stdin);
45 #endif // ONLINE_JUDGE
46 
47     for(int i=0; i<M; i++)
48         mp[i]=i;
49 
50     int n,m,f,b1,b2;
51     set<int> tree,bird;
52     scanf("%d", &n);
53     while(n--)
54     {
55         scanf("%d", &m);
56         scanf("%d", &b1);
57         bird.insert(b1);
58         for(int i=1; i<m; i++)
59         {
60             scanf("%d", &b2);
61             bird.insert(b2);
62             Union(b1,b2);
63             b1=b2;
64         }
65     }
66     scanf("%d", &n);
67     for(auto it=bird.begin(); it!=bird.end(); it++)
68         tree.insert(Father(*it));
69     printf("%d %d\n", tree.size(), bird.size());
70     for(int i=0; i<n; i++)
71     {
72         scanf("%d%d", &b1,&b2);
73         b1 = Father(b1);
74         b2 = Father(b2);
75         if(b1 == b2)
76             printf("Yes\n");
77         else
78             printf("No\n");
79     }
80 
81     return 0;
82 }

 

转载于:https://www.cnblogs.com/blue-lin/p/11005772.html