HDU 4409 Family Name List (LCA & pitfall)

Family Name List Time Limit: 2000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)
Total submission (s): 882 accepted submission (s): 271


Problem descriptionkong belongs to a huge family. Recently he got a family name list which lists all men (no women) in his family over many generations.

The list shows that the whole family has a common ancestor, let's call him Mr. x. of course, everybody does t mr. X in the list is Mr. x's descendant. everybody's father is shown in the list doesn't that Mr. x's father is not recorded. we define that Mr. x's generation number is 0. his son's generation number is 1.his grandson's generation number is 2, and so on. in a word, everybody's generation number is 1 smaller than his son's generation number. everybody's generation number is marked in some way in the list.

Now Kong is willing to pay a lot of money for a program which can re-arrange the list as he requires, and answer his questions such as how many brothers does a certain man have, etc. please write this program for him.
 
Inputthere are no more than 15 test cases.
For each test case:
The first line is an integer N (1 <= n <= 30,000), indicating the number of names in the list.
The second line is the name of Mr. X.
In the next N-1 lines, there is a man's name in each line. and if the man's generation number is K, there are k dots ('. ') before his name.

Please note that:
1) A name consists of only letters or digits ('0'-'9 ').
2) All names are unique.
3) Every line's length is no more than 60 characters.
4) in the list, a man M's father is the closest one above M Whose generation number is 1 less than M.
5) for any 2 adjacent lines in the list, if the above line's generation number is G1 and the lower line's generation number is G2, than G2 <= G1 + 1 is guaranteed.

After the name list, a line containing an integer Q (1 <= q <= 30,000) follows, meaning that there are Q queries or operations below.

In the next Q lines, each line indicates a query or operation. It can be in the following 3 formats:
1) L
Print the family list in the same format as the input, but in a sorted way. the sorted way means that: if a and B are brothers (cousins don't count), and a's name is alphabetically smaller than B's name, then a must appear earlier than B.
2) B Name
Print out how many brothers does "name" have, including "name" himself.
3) c name1 name2
Print out the closest common ancestor of "name1" and "name2 ". "closest" means the generation number is the largest. since Mr. X has no ancestor in the list, so it's guaranteed that there is no question asking about mr. x's ancestor.

The input ends with n = 0.
 
Outputalready mentioned in the input.
 
Sample Input

9Kongs.son1..son1son2..son1son1...sonkson2son1...son1son2son2..son1son3...son1son3son1.son07Lb son1son3son1b son1son2b sonkson2son1b son1c sonkson2son1 son1son2son2c son1son3son1 son1son20 

 
Sample output

Kongs.son0.son1..son1son1...son1son2son2...sonkson2son1..son1son2..son1son3...son1son3son11322son1son1son1

 
Source2012 ACM/ICPC Asia Regional Jinhua online
Recommendzhoujiaqi2010 | we have carefully selected several similar problems for you: 5053 5052 5051 5050 question: give you a tree. There are three operations. 1. DFS sequence of the output tree. Small lexicographic orders are output first. 2. Output how many sons a node's father has. Include yourself. 3. output the LCA of U and V. Idea: This question is a pain point. I built it with a stack. I don't know if there are any other advanced methods. Then for operation 1, because the lexicographically small first DFS. So we have to use a vector edge that is not very familiar. The edges are ordered by name lexicographically. Then DFS saves the answer at a time. 2. Just record who is the father of the next node. For 3.tar Jan offline processing. There is a pitfall in this question, that is, the LCA cannot be its own. Over. For details, see the code:

#include<cstdio>#include<vector>#include<algorithm>#include<string>#include<cstring>#include<iostream>#include<map>using namespace std;const int maxn=30010;int cnt,ptr,pp,vis[maxn],ty[maxn],aans[maxn];int st[maxn],rk[maxn],fa[maxn],pa[maxn],uu[maxn],vv[maxn];char na[100];vector<int> G[maxn];string name[maxn],ans[maxn];map<string,int> mp;struct node{    int v,id;    node *next;} ed[maxn<<1],*head[maxn];void adde(int u,int v,int id){    ed[ptr].v=v;    ed[ptr].id=id;    ed[ptr].next=head[u];    head[u]=&ed[ptr++];}bool cmp(int a,int b){    return name[a]<name[b];}void dfs(int u){    string op=".";    ans[pp]="";    for(int i=0;i<rk[u];i++)        ans[pp]+=op;    ans[pp++]+=name[u];    for(int i=0;i<G[u].size();i++)    {        pa[G[u][i]]=u;        dfs(G[u][i]);    }}int getfa(int x){    if(fa[x]==x)        return x;    return fa[x]=getfa(fa[x]);}void tarjan(int u){    vis[u]=1,fa[u]=u;    for(node *p=head[u];p!=NULL;p=p->next)    {        if(vis[p->v])            aans[p->id]=getfa(p->v);    }    for(int i=0;i<G[u].size();i++)    {        tarjan(G[u][i]);        fa[G[u][i]]=u;    }}int main(){    int i,j,n,m,tp,ct,id,u,v;    string aa,bb;    char cmd[20];    while(scanf("%d",&n),n)    {        for(i=0;i<=n;i++)            G[i].clear();        mp.clear();        tp=cnt=1;        st[0]=0,rk[0]=-1;        for(i=0;i<n;i++)        {            scanf("%s",na);            ct=0;            for(j=0;na[j];j++)                if(na[j]=='.')                    ct++;                else                    break;            string nna(na+ct);            //cout<<nna<<endl;            if(!mp.count(nna))            {                name[cnt]=nna;                rk[cnt]=ct;                mp[nna]=cnt++;            }            id=mp[nna];            while(rk[st[tp-1]]>=rk[id])                tp--;            G[st[tp-1]].push_back(id);            st[tp++]=id;        }        for(i=1;i<=n;i++)            sort(G[i].begin(),G[i].end(),cmp);        pp=0;        dfs(1);        ptr=0;        memset(head,0,sizeof head);        memset(vis,0,sizeof vis);        scanf("%d",&m);        for(i=0;i<m;i++)        {            scanf("%s",cmd);            if(cmd[0]=='L')                ty[i]=0;            else if(cmd[0]=='b')            {                ty[i]=1;                cin>>aa;                id=mp[aa];                aans[i]=G[pa[id]].size();            }            else            {                ty[i]=2;                cin>>aa>>bb;                u=mp[aa],v=mp[bb];                uu[i]=u,vv[i]=v;                adde(u,v,i);                adde(v,u,i);            }        }        tarjan(1);        for(i=0;i<m;i++)        {            if(ty[i]==0)            {                for(j=0;j<n;j++)                    cout<<ans[j]<<endl;            }            else if(ty[i]==1)                printf("%d\n",aans[i]);            else            {                if(aans[i]==uu[i]||aans[i]==vv[i])                    aans[i]=pa[aans[i]];                cout<<name[aans[i]]<<endl;            }        }    }    return 0;}

HDU 4409 Family Name List (LCA & pitfall)