浙江财经大学
信息管理与工程学院

HDU 3172 Virtual Friends

本文由 Ocrosoft 于 2016-08-13 17:22:27 发表

Virtual Friends

Problem Description

These days, you can do all sorts of things online. For example, you can use various websites to make virtual friends. For some people, growing their social network (their friends, their friends’ friends, their friends’ friends’ friends, and so on), has become an addictive hobby. Just as some people collect stamps, other people collect virtual friends. 
Your task is to observe the interactions on such a website and keep track of the size of each person’s network. 
Assume that every friendship is mutual. If Fred is Barney’s friend, then Barney is also Fred’s friend.

Input

Input file contains multiple test cases. 
The first line of each case indicates the number of test friendship nest.
each friendship nest begins with a line containing an integer F, the number of friendships formed in this frindship nest, which is no more than 100 000. Each of the following F lines contains the names of two people who have just become friends, separated by a space. A name is a string of 1 to 20 letters (uppercase or lowercase).

Output

Whenever a friendship is formed, print a line containing one integer, the number of people in the social network of the two people who have just become friends.

Sample Input

1
3
Fred Barney
Barney Betty
Betty Wilma

Sample Output

2
3
4

Solution

题意:每次给出两个人的关系,例如说a和b交朋友,那么输出b所在的朋友圈(其实也是a)的人数。
思路:并查集合并的时候加一个re数组,表示这个根所在的朋友圈的人数。

//#pragma comment(linker, "/STACK:1024000000,1024000000")
//#pragma GCC optimize("Og")
#include <set>
#include <map>
#include <list>
#include <cmath>
#include <stack>
#include <queue>
#include <ctime>
#include <string>
#include <cstdio>
#include <vector>
#include <cctype>
#include <climits>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define ms(a) memset(a,0,sizeof(a))
typedef long long ll;
const int INF = INT_MAX;
const int MAXN = 100000 + 10;
using namespace std;
int n,m,pre[MAXN], re[MAXN], mp_size;
map<string, int>mp;
int find(int x)
{
	if (pre[x] != x)pre[x] = find(pre[x]);
	return pre[x];
}
void Union(int x, int y)
{
	int fx = find(x), fy = find(y);
	if (fx != fy)
	{
		pre[fx] = fy;
		re[fy] += re[fx];
	}
	printf("%d\n", re[fy]);
}
int mp_find(string s)
{
	if (mp.count(s))return mp[s];
	else
	{
		re[mp_size] = 1;
		pre[mp_size] = mp_size;
		mp[s] = mp_size;
		return mp_size++;
	}
}
int main()
{
	int T;
	while (cin >> T)
	{
		while (T--)
		{
			mp_size = 1;
			mp.clear();
			cin >> n;
			while (n--)
			{
				string a, b;
				cin >> a >> b;
				Union(mp_find(a), mp_find(b));
			}
		}
	}
	//system("pause");
	return 0;
}

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