1118 Birds in Forest (25分)—

题目

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(≤104)N(\le10^4)N(≤104) which is the number of pictures. Then NNN lines follow, each describes a picture in the format:

KKK B1B_1B1 B2B_2B2… BKB_KBK

where KKK is the number of birds in this picture, and BiB_iBi'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 10410^4104.

After the pictures there is a positive number Q(≤104)Q(\le10^4)Q(≤104) which is the number of queries. Then QQQ 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

题目大意

有很多张图片，图片上面有树有鸟，规定每张图片上的鸟都属于同一棵树（鸟的编号连续），要求求出有多少棵树，有多少只鸟，并且有K次查询，判断两只鸟是否在同一棵树上。

思路

我的第一天想法是图的连通区域问题，每遍历一个连同区域时，对这个连通区域编号，即对这棵树编号，那么访问连同区域的没一只鸟时，标记这只鸟属于那棵树，后边遍历直接比较即可；按这种方法可以解题，但仔细一想，这与经典的求连通区域问题有点区别，似乎更像是并查集的问题，但是目前还不会并查集，后序会跟新并查集的做法；(更新并查集的解法，连通分量解法见另一篇博客）

代码

#include <iostream>
#include <cstdio>
#include <vector>
using namespace std;int parent[10001];
bool exist[10001];int find(int x){if(parent[x] == x)return x;elsereturn parent[x] = find(parent[x]);
}void merge(int x, int y){x = find(x);y = find(y);if(x != y)parent[x] = y;
}bool check(int x, int y){return find(x) == find(y);
}int main(){int n;for(int i=1; i<=10001; i++)parent[i] = i;fill(exist, exist+10001, false);scanf("%d", &n);for(int i=0; i<n; i++){int k, root;scanf("%d%d", &k, &root);exist[root] = true;for(int j=1, t; j<k; j++){scanf("%d", &t);exist[t] = true;merge(root, t);}}int tree = 0, bird = 0;for(int i=1; i<=10001; i++){if(exist[i] == false)break;if(parent[i] == i)tree++;bird++;}printf("%d %d\n", tree, bird);int k;scanf("%d", &k);for(int i=0, a, b; i<k; i++){scanf("%d %d", &a, &b);if(check(a, b))printf("Yes\n");elseprintf("No\n");}return 0;
}