【POJ 1470】 Closest Common Ancestors

【題目鏈接】

          點擊打開鏈接

【算法】

          離線tarjan求最近公共祖先

【代碼】

         

#include <algorithm>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <clocale>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <limits>
#include <list>
#include <map>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <utility>
#include <vector>
#include <cwchar>
#include <cwctype>
#include <stack>
#include <limits.h>
using namespace std;
#define MAXN 910

int i,n,m,u,v,s,root,tot;
int head[MAXN],fa[MAXN],f[MAXN],ans[MAXN];
bool visited[MAXN];
vector<int> query[MAXN];

struct Edge
{
		int to,nxt;
} e[MAXN];

inline int find(int x)
{
		if (f[x] == x) return x;
		return f[x] = find(f[x]);
}
inline void tarjan(int u)
{
		int i,v;
		visited[u] = true;
		f[u] = u;
		for (i = 0; i < query[u].size(); i++)
		{
				v = query[u][i];
				if (visited[v]) ans[find(v)]++;
		}		
		for (i = head[u]; i; i = e[i].nxt)
		{
				v = e[i].to;
				tarjan(v);
				f[v] = u;
		}
}
inline void add(int u,int v)
{
		tot++;
		e[tot] = (Edge){v,head[u]};
		head[u] = tot;
}

int main() 
{
		
		while (scanf("%d",&n) != EOF)
		{
				tot = 0;
				for (i = 1; i <= n; i++) 
				{
						head[i] = 0;
						fa[i] = 0;
						ans[i] = 0;
						visited[i] = false;
						query[i].clear();
				}
				for (i = 1; i <= n; i++)
				{
						scanf("%d : (%d)",&u,&s);
						while (s--)
						{
								scanf("%d",&v);
								add(u,v);
								fa[v] = u;	
						}	
				}	
				for (i = 1; i <= n; i++) 
				{
						if (!fa[i]) 
								root = i;
				}
				scanf("%d",&m);
				for (i = 1; i <= m; i++)
				{
						scanf(" (%d %d)",&u,&v);
						query[u].push_back(v);
						query[v].push_back(u);
				}
				tarjan(root);
				for (i = 1; i <= n; i++)
				{
						if (ans[i])
								printf("%d:%d\n",i,ans[i]);
				}
		}
		
		return 0;
	
}