hdu 5.1.7 virtual friend

hdu终于正常了……之前的题，都有些记不清了（什么记性啊喂！）

Virtual Friends

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 54 Accepted Submission(s): 27

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

朴素的并查集，唯一的亮点是要用map建立字符串到数字的映射。map<string,int>p p.find(a) p.end() p.clear,就这些函数。用多了还挺顺手的。

Input file contains multiple test cases. ……害我wa了一次。

#include <iostream>
#include <map>
#include <string>
using namespace std;

int root[100002];
int cnt[100002];//how many friend does (the root) have
int num;
map<string ,int> p;//map, cast string to int

void clear()
{
     int i;
     for(i=0;i<100002;i++)
     {
         root[i]=i;
         cnt[i]=1;
     }
     p.clear();
     num=0;
}

int find(int x)
{
    if(root[x]!=x)
    {
        root[x]=find(root[x]);//update root[]
    }
    return root[x];
}

void merge(int a,int b)
{
     int fa=find(a);
     int fb=find(b);
     if(fa!=fb)
     {
         root[fb]=fa;
         cnt[fa]+=cnt[fb];
     }
     cout<<cnt[fa]<<endl;
}
     
     

int main()
{
    int cas;
    while(cin>>cas)//Input file contains multiple test cases.
    {
    while(cas--)
    {
        int n;
        
        
        clear();
        cin>>n;
        while(n--)
        {
            string a,b;
            cin>>a>>b;
            if(p.find(a)==p.end())//no a in map
            {
                p[a]=++num;//cast a to ++num
            }
            if(p.find(b)==p.end())//no a in map
            {
                p[b]=++num;//
            }
            merge(p[a],p[b]);
        }
    }
    }
    system("pause");
    return 0;
}
            
            
            

二专这周结课，可以周末和米娜（当然主要是队长大人哈哈）一起做题了~可惜本周的两个还是参加不了……残念……
恩,发现上面这种做法现在超时了,感谢   littlesheep_shaun~~

不过,网上找找不超时的方法只有hash表,这,看不懂...

唉...ctrl-c&v

#include<iostream>
#include<cstring>
#include<cstdio>
#define MAXN 200003
using namespace std;
typedef char Word[21];
Word que[MAXN];

const int HashSize = 1000003;
int head[HashSize], next[MAXN], rear;
int father[MAXN], num[MAXN]; 

inline void init_lookup_table(){
    rear=1; 
    memset(head, 0, sizeof(head));
}

inline int hash(Word &s){
    int seed = 131, v=0;
    for(int i=0; i<strlen(s); ++i){
        v = (v*seed + s[i]);
    }
    return (v & 0x7FFFFFFF)%HashSize;
}

bool try_to_insert(int s){
    int h = hash(que[s]);
    int u = head[h];
    while(u){
        if(strcmp(que[u], que[s])==0) return false;
        u = next[u];
    }
    next[s] = head[h];
    head[h] = s;
    num[s] = 1; // 加入成功后， 该位置的人数变为1
    return true;
}

int search(Word &s){
    int h = hash(s);
    int u = head[h];
    while(u){
        if( strcmp(que[u], s)==0 ) return u;
        u = next[u];
    }
}


void init_unionset(){
    for(int i=0; i<MAXN; i++){
        father[i]=i; num[i] = 0;
    }
}

int find(int x){  
    int i, j=x;
    while(j!=father[j]) j = father[j]; // 找树根
    while(x != j){ // 路径压缩
        i = father[x];
        father[x] = j;
        x = i;
    }
    return j;
}  

void Union(int x,int y){  
    x = find(x);  
    y = find(y);  
    if(x!=y){
        father[y] = x;
        num[x] += num[y];
    }
}


int main(){

    int T, n;
    Word name1, name2;
    while(scanf("%d", &T)!=EOF){

		while(T--){

		    init_lookup_table();
		    init_unionset();

		    scanf("%d",&n);

		    if(n==0) {
		        printf("0\n");
		        continue;
		    }
		    for(int i=0; i<n; ++i){
		        scanf("%s %s",name1, name2);
		        
		        strcpy(que[rear], name1);
		        if(try_to_insert(rear)) ++rear;

		        strcpy(que[rear], name2);
		        if(try_to_insert(rear)) ++rear;

		        int a = search(name1),  b = search(name2);

		        Union(a, b);

		        printf("%d\n",num[find(a)]); 
		    }
		}
    }
    return 0;
}

咕噜咕噜滚走