poj 1470 Closest Common Ancestors (LCA)

Closest Common Ancestors

Time Limit: 2000MS		Memory Limit: 10000K
Total Submissions: 14314		Accepted: 4607

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

模板题：

给出在线做法和离线做法：

在线的DFS+RMQ做法：

//444K    485MS    C++    2082B    2014-04-15 19:41:38
#include<stdio.h>
#include<string.h>
#include<math.h>
#define N 1005
struct node{
    int u,v;
    int next;
}edge[2*N];
int n,num,to,head[N];
int set[2*N],dep[2*N];
int rank[N],vis[N];
int dp[2*N][30];
void addedge(int u,int v)
{
    edge[num].u=u;
    edge[num].v=v;
    edge[num].next=head[u];
    head[u]=num++;
}
int Min(int a,int b)
{
    return dep[a]<dep[b]?a:b;
}
void dfs(int u,int deep)
{
    vis[u]=1;
    set[++num]=u;
    rank[u]=num;
    dep[num]=deep;
    for(int i=head[u];i!=-1;i=edge[i].next){
        int v=edge[i].v;
        if(vis[v]) continue;
        dfs(v,deep+1);
        set[++num]=u;
        dep[num]=deep;
    }
}
void init()
{
    int nn=2*n-1;
    int m=(int)(log(nn*1.0)/log(2.0));
    for(int i=1;i<=nn;i++)
        dp[i][0]=i;
    for(int j=1;j<=m;j++)
        for(int i=1;i+(1<<j)-1<=nn;i++)
            dp[i][j]=Min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
}
int RMQ(int l,int r)
{
    int m=(int)(log((r-l+1)*1.0)/log(2.0));
    return Min(dp[l][m],dp[r-(1<<m)+1][m]);
}
int LCA(int a,int b)
{
    int x=rank[a];
    int y=rank[b];
    if(x>y) return set[RMQ(y,x)];
    else return set[RMQ(x,y)];
}
int main(void)
{
    int m,u,v,mm;
    int in[N];
    while(scanf("%d",&n)!=EOF)
    {
        memset(in,0,sizeof(in));
        memset(vis,0,sizeof(vis));
        memset(head,-1,sizeof(head));
        num=0;
        for(int i=0;i<n;i++){
            scanf("%d:(%d)",&u,&mm);
            while(mm--){
                scanf("%d",&v);
                in[v]++;
                addedge(u,v);
                addedge(v,u);
            }
        }
        int id=0;
        num=0;
        while(in[++id]);
        dfs(id,1);
        init();
        memset(in,0,sizeof(in));
        scanf("%d",&m);
        //printf("%%%d
",m);
        while(m--){
            scanf("%(%d %d%)",&u,&v); //这里蛋疼了一会 
            //printf("*%d %d
",u,v);
            in[LCA(u,v)]++;
        }
        for(int i=1;i<=n;i++)
            if(in[i]) printf("%d:%d
",i,in[i]);
    }
    return 0;
}

离线的Tarjan做法：

//3024K    563MS    C++    1575B    2014-04-15 20:32:44
//Runtime Error 了好几次 T T 
#include<iostream>
#include<queue>
#include<vector>
#define N 1005
using namespace std;
vector<int>child[N],V[N];
int set[N],vis[N];
int in[N];
int n;
void init()
{
    for(int i=0;i<=n;i++){
        child[i].clear();
        V[i].clear();
    }
    memset(vis,0,sizeof(vis));
    memset(in,0,sizeof(in));
}
int find(int x)
{
    if(set[x]!=x) set[x]=find(set[x]);
    return set[x];
}
void merge(int a,int b)
{
    int x=find(a);
    int y=find(b);
    set[y]=x;
}
void LCA(int u)
{
    set[u]=u;
    vis[u]=true;
    for(int i=0;i<V[u].size();i++)
        if(vis[V[u][i]])
            in[find(V[u][i])]++;
    for(int i=0;i<child[u].size();i++){
        if(!vis[child[u][i]]){
            LCA(child[u][i]);
            merge(u,child[u][i]);
            set[child[u][i]]=u;
        }
    }
}
int main(void)
{
    int m,u,v,mm;
    
    while(scanf("%d",&n)!=EOF)
    {
        init();
        for(int i=0;i<n;i++){
            scanf("%d:(%d)",&u,&mm);
            while(mm--){
                scanf("%d",&v);
                child[u].push_back(v);
                child[v].push_back(u);
                in[v]++;
            }
        }
        scanf("%d",&m);
        for(int i=1;i<=m;i++){
            scanf("%(%d %d%)",&u,&v);
            V[u].push_back(v);
            V[v].push_back(u);
        }
        int id=0;
        while(in[++id]);
        memset(in,0,sizeof(in));
        LCA(id);
        for(int i=1;i<=n;i++)
            if(in[i]) printf("%d:%d
",i,in[i]);
    }    
    return 0;
}