1118 Birds in Forest (25 分) 并查集

朴灿烈づ我的快乐病毒、 2021-11-01 12:26 226阅读 0赞

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) which is the number of pictures. Then N lines follow, each describes a picture in the format:

K B1 B2 ... BK

where K is the number of birds in this picture, and Bi'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 104.

After the pictures there is a positive number Q (≤104) 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

**题意：**有n张照片，每张照片上的鸟默认在一棵树上，问有几棵树，几只鸟(鸟的编号从1开始，连续)?

给k个询问，问a鸟和b鸟是否在一棵树上。

**思路：**并查集把每张照片上的鸟combine，几棵树=连通块个数，几只鸟=最大编号maxx

问a鸟和b鸟是否在一棵树上用find祖先是否相同即可解决。

#include <stdio.h>
    #include <string.h>
    #include <string>
    #include <map>
    #include <vector>
    #include <algorithm>
    #include <iostream>
    #define INF 0x3f3f3f3f
    using namespace std;
    int ma[10005][10005];
    int fa[10005];
    int find(int x)
    {
    	int r=x,i=x,j;
    	while(fa[r]!=r)
    	r=fa[r];
    	while(fa[i]!=i)
    	j=fa[i],fa[i]=r,i=j;
    	return r;
    }
    void combine(int x,int y)
    {
    	int xx=find(x);
    	int yy=find(y);
    	if(xx<yy)
    	fa[yy]=xx;
    	else
    	fa[xx]=yy;
    }
    int main()
    {
    	int n,m,k,i,j,l,p,maxx=-1;
    	scanf("%d",&n);
    	for(i=0;i<10005;i++)
    	fa[i]=i;
    	for(i=1;i<=n;i++)
    	{
    		scanf("%d%d",&m,&p);
    		if(p>maxx)maxx=p;
    		for(j=1;j<m;j++)
    		{
    			scanf("%d",&k);
    			if(k>maxx)maxx=k;
    			combine(p,k);
    		}
    	}
    	int num[10005]={0};
    	for(i=1;i<=maxx;i++)
    	{
    		num[find(i)]++;
    		//printf("%d : %d\n",i,find(i));
    	}
    	int sum=0;
    	for(i=1;i<=maxx;i++)
    	{
    		if(num[i])
    		sum++;
    	}
    	printf("%d %d\n",sum,maxx);
    	scanf("%d",&k);
    	while(k--)
    	{
    		int a,b;
    		scanf("%d%d",&a,&b);
    		if(find(a)==find(b))
    		printf("Yes\n");
    		else
    		printf("No\n");
    	}
    }