POJ 1470 LCA tarjan 离线算法

这里有个讲解非常明白，虽然没证明。点击打开链接
还有个数据结构的学习
点击打开链接 伸展树/Treap/划分树/ 归并树

步骤：

tarjan算法的步骤是(当dfs到节点u时):

1 在并查集中建立仅有u的集合,设置该集合的祖先为u

1 对u的每个孩子v:

1.1 tarjan之

1.2 合并v到父节点u的集合,确保集合的祖先是u

2 设置u为已遍历

3 处理关于u的查询,若查询(u,v)中的v已遍历过,则LCA(u,v)=v所在的集合的祖先

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
/*
* POJ 1470
* 给出一颗有向树，Q个查询
* 输出查询结果中每个点出现次数
*/
/*
* LCA离线算法，Tarjan
* 复杂度O(n+Q);
*/
const int MAXN = 1010;
const int MAXQ = 500010;//查询数的最大值

//并查集部分
int F[MAXN];//需要初始化为-1
int find(int x)
{
if(F[x] == -1)return x;
return F[x] = find(F[x]);
}
void bing(int u,int v)
{
int t1 = find(u);
int t2 = find(v);
if(t1 != t2)
F[t1] = t2;
}
//************************
bool vis[MAXN];//访问标记
int ancestor[MAXN];//祖先
struct Edge
{
int to,next;
}edge[MAXN*2];
int head[MAXN],tot;
void addedge(int u,int v)
{
edge[tot].to = v;
edge[tot].next = head[u];
head[u] = tot++;
}

struct Query
{
int q,next;
int index;//查询编号
}query[MAXQ*2];
int answer[MAXQ];//存储最后的查询结果，下标0~Q-1
int h[MAXQ];
int tt;
int Q;

void add_query(int u,int v,int index)
{
query[tt].q = v;
query[tt].next = h[u];
query[tt].index = index;
h[u] = tt++;
query[tt].q = u;
query[tt].next = h[v];
query[tt].index = index;
h[v] = tt++;
}

void init()
{
tot = 0;
memset(head,-1,sizeof(head));
tt = 0;
memset(h,-1,sizeof(h));
memset(vis,false,sizeof(vis));
memset(F,-1,sizeof(F));
memset(ancestor,0,sizeof(ancestor));
}

void LCA(int u)
{
ancestor[u] = u;
vis[u] = true;
for(int i = head[u];i != -1;i = edge[i].next)
{
int v = edge[i].to;
if(vis[v])continue;
LCA(v);
bing(u,v);
ancestor[find(u)] = u;
}
for(int i = h[u];i != -1;i = query[i].next)
{
int v = query[i].q;
if(vis[v])
{
answer[query[i].index] = ancestor[find(v)];
}
}
}

bool flag[MAXN];
int Count_num[MAXN];
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int n;
int u,v,k;
while(scanf("%d",&n) == 1)
{
init();
memset(flag,false,sizeof(flag));
for(int i = 1;i <= n;i++)
{
scanf("%d:(%d)",&u,&k);
while(k--)
{
scanf("%d",&v);
flag[v] = true;
addedge(u,v);
addedge(v,u);
}
}
scanf("%d",&Q);
for(int i = 0;i < Q;i++)
{
char ch;
cin>>ch;
scanf("%d %d)",&u,&v);
add_query(u,v,i);
}
int root;
for(int i = 1;i <= n;i++)
if(!flag[i])
{
root = i;
break;
}
LCA(root);
memset(Count_num,0,sizeof(Count_num));
for(int i = 0;i < Q;i++)
Count_num[answer[i]]++;
for(int i = 1;i <= n;i++)
if(Count_num[i] > 0)
printf("%d:%d\n",i,Count_num[i]);
}
return 0;
}