POJ1330Nearest Common Ancestors——近期公共祖先(离线Tarjan)

http://poj.org/problem?

id=1330

给一个有根树,一个查询节点(u,v)的近期公共祖先

836K 16MS

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<string>
#include<set>
#include<map>
const int maxn=10005;
using namespace std;
int n,m;//结点数,查询数
int f[maxn];//并查集
bool vis[maxn];//訪问标记
int ancestor[maxn];//祖先
int answer[2];//存储最后的查询结果
int in[maxn];//入度
vector<int> que[maxn];
struct Edge{
    int to;
    int next;
}edge[maxn<<1];int head[maxn],tot;

void addedge(int u,int v){
    edge[tot].to=v;
    edge[tot].next=head[u];
    head[u]=tot++;
}
void Init(){
    tot=0;
    memset(in,0,sizeof(in));
    memset(f,-1,sizeof(f));
    memset(head,-1,sizeof(head));
    memset(vis,false,sizeof(vis));
    memset(ancestor,0,sizeof(ancestor));
    for(int i=1;i<=n;++i){
        que[i].clear();
    }
}
int find(int x){
    return f[x]==-1 ?

x:f[x]=find(f[x]); } void Union(int u,int v){ int t1=find(u); int t2=find(v); if(t1!=t2) f[t2]=t1; } void LCA(int u){ ancestor[u]=u; for(int i=head[u];i!=-1;i=edge[i].next){ int v=edge[i].to; if(vis[v]) continue; LCA(v); Union(u,v); ancestor[find(v)]=u;//可省略 } vis[u]=true; int sz=que[u].size(); for(int i=0;i<sz;++i){ int v=que[u][i]; if(vis[v]){ printf("%d ",ancestor[find(v)]);//可替换find(v) } } } int main() { //freopen("in.txt","r",stdin); int T,u,v,w; cin>>T; while(T--){ scanf("%d",&n); Init(); for(int i=0;i<n-1;++i){ scanf("%d%d",&u,&v); if(u!=v){ addedge(u,v); in[v]++; } } scanf("%d%d",&u,&v); que[u].push_back(v); que[v].push_back(u); for(int i=1;i<=n;++i){ if(in[i]==0){ LCA(i); break; } } } return 0; }

原文地址:https://www.cnblogs.com/bhlsheji/p/5344559.html