(LCA在线算法)Closest Common Ancestors POJ - 1470

Description

Write a program that takes as input a rooted tree and a list of pairs of vertices. For each pair (u,v) the program determines the closest common ancestor of u and v in the tree. The closest common ancestor of two nodes u and v is the node w that is an ancestor of both u and v and has the greatest depth in the tree. A node can be its own ancestor (for example in Figure 1 the ancestors of node 2 are 2 and 5)

Input

The data set, which is read from a the std input, starts with the tree description, in the form: 

nr_of_vertices 
vertex:(nr_of_successors) successor1 successor2 ... successorn 
...
where vertices are represented as integers from 1 to n ( n <= 900 ). The tree description is followed by a list of pairs of vertices, in the form: 
nr_of_pairs 
(u v) (x y) ... 

The input file contents several data sets (at least one). 
Note that white-spaces (tabs, spaces and line breaks) can be used freely in the input.

Output

For each common ancestor the program prints the ancestor and the number of pair for which it is an ancestor. The results are printed on the standard output on separate lines, in to the ascending order of the vertices, in the format: ancestor:times 
For example, for the following tree: 

Sample Input

5
5:(3) 1 4 2
1:(0)
4:(0)
2:(1) 3
3:(0)
6
(1 5) (1 4) (4 2)
      (2 3)
(1 3) (4 3)

Sample Output

2:1
5:5

Hint

Huge input, scanf is recommended.

Source

Southeastern Europe 2000

 

1：如果使用离线LCA算法做这道题的话，查询数的最大值要定义为50W！
2：多组输入！
3：需要对根节点进行特判，因为第一个点不一定是根节点！
4：然后一直看大家都说输入有问题，输入直接在括号前面加上一个空格不就直接吸收缓冲区里面的回车了吗。
scanf(" (%d %d)",&x,&y);
5：别忘了清空数组，因为如果你一开始是用单组输入的话，改成组输入的时候确实很容易忘。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<vector>
#include<queue>
#include<cmath>
#include<map>
#include<string.h>
#include<string>
using namespace std;
const int maxn=20000;
const int DEG=20;
int head[maxn],tot,top;
int fa[maxn][DEG],depth[maxn];
int number;
bool flag[maxn];

struct Edge
{
    int to,next;
} edge[maxn*2];

struct xiao
{
    int u,w;
} mapp[maxn];

bool cmp(xiao a,xiao b)
{
    return a.u<b.u;
}

void addedge(int u,int v)
{
    edge[tot].to=v;
    edge[tot].next=head[u];
    head[u]=tot++;
}

void init()
{
    number=0;
    tot=0;
    memset(head,-1,sizeof(head));
    memset(flag,false,sizeof(flag));
    memset(depth,0,sizeof(depth));
    memset(fa,0,sizeof(fa));
}

void debug()
{
    cout<<"____5_____"<<endl;
}

void BFS(int root)
{
    queue<int>que;
    depth[root]=0;
    fa[root][0]=root;
    que.push(root);
    while(!que.empty())
    {
        int tmp=que.front();
        que.pop();
        for(int i=1; i<DEG; ++i)
        {
            fa[tmp][i]=fa[fa[tmp][i-1]][i-1];
        }
        for(int i=head[tmp]; i!=-1; i=edge[i].next)
        {
            int v=edge[i].to;
            if(v==fa[tmp][0])
                continue;
            depth[v]=depth[tmp]+1;
            fa[v][0]=tmp;
            que.push(v);
        }

    }
}

int LCA(int u,int v)
{
    if(depth[u]>depth[v])
    {
        int x=u;
        u=v;
        v=x;
    }
    int hu=depth[u],hv=depth[v];
    int tu=u,tv=v;
    for(int det=hv-hu,i=0; det; det>>=1,i++)
        if(det&1)
            tv=fa[tv][i];

    if(tu==tv)
        return tu;
    for(int i=DEG-1; i>=0; --i)
    {
        if(fa[tu][i]==fa[tv][i])
            continue;
        tu=fa[tu][i];
        tv=fa[tv][i];
    }

    return fa[tu][0];
}

int main()
{

    int n;
    while(cin>>n)
    {
        init();
        for(int i=1; i<=n; ++i)
        {
            int u,x,v;
            scanf("%d:(%d)",&u,&x);
            for(int k=1; k<=x; ++k)
            {
                scanf("%d",&v);
                addedge(u,v);
                addedge(v,u);
                flag[v]=true;
            }
        }
        int root;
        for(int i=1; i <=n ; ++i)
        {
            if(!flag[i])
            {
                root=i;
                break;
            }
        }
        BFS(root);
        map<int,int>mmp;
        int m,x,y;
        cin>>m;
        while(m--)
        {
            scanf(" (%d %d)",&x,&y);
            int mm=LCA(x,y);
            if(mmp[mm]==0)
            {
                mapp[number++].u=mm;
            }

            mmp[mm]++;

        }
        sort(mapp,mapp+number,cmp);
        for(int i=0; i<number; ++i)
        {
            printf("%d:%d\n",mapp[i].u,mmp[mapp[i].u]);
        }
    }
    return 0;
}