UVa 11503:virtual Friends

Topic Link:

Uva:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&itemid=8&category=24&page=show_ problem&problem=2498

hdu:http://acm.hdu.edu.cn/showproblem.php?pid=3172

Type: And look-up set, hash

Original title:

This is the sorts of things online. For example, your 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), the 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 of the size of all person ' s track.

Assume that every friendship is mutual. If Fred is Barney's friend, then Barney is also Fred ' s friend.

Input specification

The "a" of input contains one integer specifying the number of test cases to follow. Each test case begins with a line containing an integer F, the number of friendships formed, which are 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. A name is a string of 1 to letters (uppercase or lowercase).

Sample Input

1
3
Fred Barney
Barney Betty
Betty Wilma

Output specification

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

Output for Sample Input

2
3
4

Meaning

If A and B are friends, C is a friend of B, then A and B are friends. In this way, a network of Friends is formed, and only one person in this network is the one who knows all the people in the net. Enter some friends relationship, each input a relationship, output two people in the net of the total number of friends

Analysis and Summary:

It is obvious that we are looking at the topic of the collection. The key to this question is to convert people's names into numbers.

This is a very big problem. A place for dad: The input T is also multiple groups, to use while (scanf ("%d", &t)!=eof) {while (t--) {...} }

1. Directly with the STL map to deal with. But the speed is poor: 2.124s (UVa), 625MS (HDU)

* * and check set + map * time:2.124s (UVa), 625MS (HDU) */#include <iostream> #include <map> #include <str  
     
ing> #include <cstdio> #include <cstring> #define N 100005 using namespace std;  
map<string,int>mp;  
     
int father[n], num[n];  
void Init () {for (int i=0; i<n; ++i) father[i] = i;  
    int find (int x) {int I, j = x;  
    while (j!=father[j]) j = father[j];  
        while (x!=j) {i =father[x];  
        Father[x] = j;  
    x = i;  
} return J;  
    } void Union (int x, int y) {int a=find (x);  
    int B=find (y);  
        if (a!=b) {Father[a] = b;  
    NUM[B] + = Num[a];  
the int main () {#ifdef local freopen ("Input.txt", "R", stdin);  
    #endif int T, n, K;  
    String Person1, Person2;  
            while (scanf ("%d", &t)!=eof) {while (t--) {scanf ("%d", &n);  
           int cnt=1; Init ();  
            Mp.clear ();  
                while (n--) {cin >> person1 >> Person2;  
                int x1, x2; if (!) (  
                X1=mp[person1]) {x1 = Mp[person1] = cnt++; num[cnt-1]=1;} if (!) (  
                    
                X2=mp[person2]) {x2 = Mp[person2] = cnt++; num[cnt-1]=1;}  
                     
                Union (x1, x2);  
                int x = FIND (x1);  
            printf ("%d\n", num[x]);  
}} return 0; }