POJ 1330 Nearest Common Ancestors LCA

题目：http://poj.org/problem?id=1330

题意：一棵n个点的数，输入n - 1条边，最后输入一组查询，求它们的最近公共祖先

思路：tarjan算法求LCA第一题，模板题

总结：tarjan求强连通，tarjan求双联通，tarjan求LCA，粗糙的学完了。。。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int N = 10010;
typedef long long ll;
struct edge
{
int to, cost, next;
} G1
, G2
;
int head1
, head2
, par
;
int cnt1, cnt2;
bool vis
, deg
;
int n, m;
void init()
{
for(int i = 1; i <= n; i++)
par[i] = i;
memset(head1, -1, sizeof head1);
memset(head2, -1, sizeof head2);
memset(deg, 0, sizeof deg);
memset(vis, 0, sizeof vis);
cnt1 = cnt2 = 0;
}
void add_edge1(int v, int u)
{
G1[cnt1].to = u;
G1[cnt1].next = head1[v];
head1[v] = cnt1++;
}
void add_edge2(int v, int u)
{
G2[cnt2].to = u;
G2[cnt2].next = head2[v];
head2[v] = cnt2++;
}
int ser(int x)
{
int r = x, i = x, j;
while(r != par[r]) r = par[r];
while(i != r) j = par[i], par[i] = r, i = j;
return r;
}
void tarjan_lca(int v)
{
int u;
for(int i = head1[v]; i != -1; i = G1[i].next)
{
u = G1[i].to;
tarjan_lca(u);
par[u] = v;
}
vis[v] = true; /*这个标记的位置很巧妙，求LCA时不能放在函数首部，否则会重复输出某些答案*/
for(int i = head2[v]; i != -1; i = G2[i].next)
if(vis[u = G2[i].to])
{
printf("%d\n", ser(u));
return;
}
}
int main()
{
int t, a, b;
scanf("%d", &t);
while(t--)
{
scanf("%d", &n);
init();
for(int i = 0; i < n - 1; i++)
{
scanf("%d%d", &a, &b);
add_edge1(a, b);
deg[b] = true;
}
scanf("%d%d", &a, &b);
add_edge2(a, b);
add_edge2(b, a);
for(int i = 1; i <= n; i++)
if(! deg[i]) /*树根*/
{
tarjan_lca(i);
break;
}
}
return 0;
}