hdu 4409 Family Name List(LCA&有坑点)

Posted 2020-09-18 zsychanpin

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 Description

Kong 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 except Mr.X in the list is Mr. X‘s descendant. Everybody‘s father is shown in the list except 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.

 

Input

There 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.

 

Output

Already mentioned in the input.

 

Sample Input

9
Kongs
.son1
..son1son2
..son1son1
...sonkson2son1
...son1son2son2
..son1son3
...son1son3son1
.son0	
7
L
b son1son3son1
b son1son2
b sonkson2son1
b son1
c sonkson2son1 son1son2son2
c son1son3son1 son1son2
0 

 

Sample Output

Kongs
.son0
.son1
..son1son1
...son1son2son2
...sonkson2son1
..son1son2
..son1son3
...son1son3son1
1
3
2
2
son1son1
son1

 

Source

2012 ACM/ICPC Asia Regional Jinhua Online

 

Recommend

zhoujiaqi2010   |   We have carefully selected several similar problems for you:  5053 5052 5051 5050 5049 

 

题意：

给你一棵树。

有三种操作。

1.输出树的dfs序。

字典序小的先输出。

2.输出一个结点的父亲有多少儿子。包含自己。

3.输出u，v的LCA。

思路：

这题输入比較蛋疼。我是用一个栈来建树的。

不知道有其他什么高级方法没。

然后对于操作1.因为要字典序小的的先dfs。那么仅仅好用 不是非常熟悉的vector存边了。然后对边按名字字典序排序。

然后dfs一次把答案存起来。对于2记录下一个结点的父亲是谁即可了。

对于3.tarjan离线处理。这题有个坑点就是LCA不能是自己。over。

具体见代码：

#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;
}