【codeforces】Misha and Changing Handles（映射+并查集）

向右看齐 2022-09-26 11:59 7阅读 0赞

Misha and Changing Handles

**Time Limit:**1000MS **Memory Limit:**262144KB **64bit IO Format:**%I64d & %I64u

Submit [Status][]

Description

Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.

Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.

Input

The first line contains integer *q* (1 ≤ *q* ≤ 1000), the number of handle change requests.

Next *q* lines contain the descriptions of the requests, one per line.

Each query consists of two non-empty strings *old* and *new*, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings *old* and *new* are distinct. The lengths of the strings do not exceed 20.

The requests are given chronologically. In other words, by the moment of a query there is a single person with handle *old*, and handle*new* is not used and has not been used by anyone.

Output

In the first line output the integer *n* — the number of users that changed their handles at least once.

In the next *n* lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, *old* and*new*, separated by a space, meaning that before the user had handle *old*, and after all the requests are completed, his handle is *new*. You may output lines in any order.

Each user who changes the handle must occur exactly once in this description.

Sample Input

Input

5
    Misha ILoveCodeforces
    Vasya Petrov
    Petrov VasyaPetrov123
    ILoveCodeforces MikeMirzayanov
    Petya Ivanov

Output

3
    Petya Ivanov
    Misha MikeMirzayanov
    Vasya VasyaPetrov123

题意：让求两个最终关联的id名称并输出，中间id略过输出。

写法：先map映射一下，并用一个字符数组记录当前数字对应的整数值；然后对数字用并查集进行关联；把结果保存在另一个字符数组中输出即可。

#include<cstdio>
    #include<iostream>
    #include<cstring>
    #include<map>
    #include<cmath>
    #include<algorithm>
    using namespace std;
    const int N = 2020;
    char c[N][22],d[N][22];
    bool vis[N];
    int pre[N];
    int find(int x) {
    	int r=x;
    	while(r!=pre[r]) {
    		vis[pre[r]]=1;
    		r=pre[r];
    	}
    	return r;
    }
    //void Union(int x,int y) {
    //	int fx=find(x);
    //	int fy=find(y);
    //	if(fx!=fy) {
    //		pre[fx]=fy;
    //	}
    //}
    int main() {
    	int n;
    	while(scanf("%d",&n)!=EOF) {
    		memset(vis,0,sizeof(vis));
    		map<string,int>mp;
    		mp.clear();
    		char a[22],b[22];
    		int pos=0;
    		for(int i=0; i<N; i++) {
    			pre[i]=i;
    		}
    		for(int i=0; i<n; i++) {
    			scanf("%s%s",a,b);
    			if(!mp[a]) mp[a]=++pos,strcpy(c[pos],a);
    			if(!mp[b]) mp[b]=++pos,strcpy(c[pos],b);
    			pre[mp[a]]=mp[b];
    		}
    		int temp=0,num=0;
    		for(int i=1; i<=pos; i++) {
    			if(!vis[i]) {
    				vis[i]=1;
    				int cnt=find(i);
    				strcpy(d[num++],c[i]);
    				strcpy(d[num++],c[cnt]);
    				temp++;
    			}
    		}
    		printf("%d\n",temp);
    		for(int i=0; i<num; i++) {
    			printf("%s%c",d[i],i&1?'\n':' ');
    		}
    	}
    	return 0;
    }

[Status]: http://vjudge.net/contest/132750#status//E/0