3172 Virtual Friends【并查集】
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Virtual Friends
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6971 Accepted Submission(s): 1991
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.
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).
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
13Fred BarneyBarney BettyBetty Wilma
Sample Output
234
题目说的是,网上交友,朋友的朋友也是自己的朋友,问每次两个交了朋友之后,他们朋友圈里有多少人,
并查集处理比较顺,用上map 编号,比较简单就处理了.....但是这个题目的输入比较坑,很少见到这样的输入..........无限次指定次数的输入.........
#include<stdio.h>#include<cstring>#include<string>#include<map>#define max(a,b) (a>b?a:b)using namespace std;int per[100005],num[100005],maxn,k,cnt,n,x[100005];void init(){maxn=k=cnt=0;for(int i=0;i<100005;++i){per[i]=i;num[i]=1;//全部赋值为 1 }}int find(int x){int r=x;while(r!=per[r]){r=per[r];}int i=x,j;while(i!=r){j=per[i];per[i]=r;i=j;}return r;}void slove(){init();map<string,int> map;char a[105],b[105];for(int i=0;i<n;++i){scanf("%s%s",a,b);if(!map.count(a)){map[a]=k++;}if(!map.count(b)){map[b]=k++;}int u=map[a],v=map[b];int fx=find(u),fy=find(v);if(fx!=fy){per[fy]=fx;num[fx]+=num[fy];//累加元素数量x[cnt++]=num[fx];//记录}else{x[cnt++]=num[fx];}}for(int i=0;i<cnt;++i){printf("%d\n",x[i]);//输出}}int main(){int t;while(~scanf("%d",&t))//!!!!注意输入格式!!!!!{while(t--){scanf("%d",&n);slove();}}return 0;}
也是今天突然发现...原来不需要保存下来,直接就输出就可以了...
优化了一下,不过发现时间也没省多少啊.........
和上面的思路一样,也就不注释了....
#include<stdio.h>#include<cstring>#include<string>#include<map>#define max(a,b) (a>b?a:b)using namespace std;int per[100005],num[100005],maxn,k,cnt,n;void init(){maxn=k=cnt=0;for(int i=0;i<100005;++i){per[i]=i;num[i]=1;}}int find(int x){int r=x;while(r!=per[r]){r=per[r];}int i=x,j;while(i!=r){j=per[i];per[i]=r;i=j;}return r;}void slove(){init();map<string,int> map;char a[105],b[105];for(int i=0;i<n;++i){scanf("%s%s",a,b);if(!map.count(a)){map[a]=k++;}if(!map.count(b)){map[b]=k++;}int u=map[a],v=map[b];int fx=find(u),fy=find(v);if(fx!=fy){per[fy]=fx;num[fx]+=num[fy];//数值累加}printf("%d\n",num[fx]);//直接输出就可以了.....}}int main(){int t;while(~scanf("%d",&t)){while(t--){scanf("%d",&n);slove();}}return 0;}
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