ping ping ping

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 207    Accepted Submission(s): 47

Problem Description

The structure of the computer room in Northeastern University is pretty miraculous. There are n servers, some servers connect to the gateway whose IP address is 0 directly. All servers are connected with each other by n netting twines. It is said that this structure is favorable for maintaining physical problem of servers.
But because of an unexpected rainstorm, the computer room was destroyed by a terrible thunderclap!
Our maintainer Bittersweet found that many servers were not able to be visited, so he hurried to the computer room to lookup the reason. After several hours, Bittersweet realized that some net gape of servers were broken by thunderclap. However, there were too many servers to find out all the broken net gapes quickly. So he came up with an idea to assess the damaged condition roughly. Bittersweet decided to turn on some servers and ping other servers randomly, then record the unsuccessful pairs of servers.
Now he need a program to analyze the record to confirm what is the minimum number of servers whose net gape was destroyed by thunderclap. Can you help him to complete this work?

Input

There are at most 20 test cases.
In each test case, the first line is an integer n (3≤n≤104), denoting the number of servers. The IP address of these servers is 1…n.
Then follows n lines, each line contains two integers u and v (0≤u,v≤n), denoting that the server whose IP address is u is connected with the server whose IP address is v by netting twine initially.
After those, there is one line contains only an integer p (p≤50000), denoting the number that Bittersweet uses ping.
Then follows p lines, each line contains two integers U and V , denoting when using server U to ping server V, it returned unsuccessful.

Output
A single integer x in a line, denoting at least x servers whose net gape were broken.

Sample Input
4
1 0
4 2
2 0
3 2
2
1 3
2 1

Sample Output
1

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <string>
#include <stack>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define fin(name) freopen(name,"r",stdin)
#define fout(name) freopen(name,"w",stdout)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int, int> pii;
typedef long long LL;
const double PI = acos(-1.0);
const int N = 10010;
const int M = 50010;
struct edge
{
int to, nxt;
edge() {}
edge(int _to, int _nxt): to(_to), nxt(_nxt) {}
} E[N << 1];
struct info
{
int u, v, lca, dep;
bool operator<(const info &rhs)const
{
return dep > rhs.dep;
}
} node[M];
int D[N << 1], F[N], ver[N << 1], dp[N << 1][19], ts, dep[N];
int T[N], L[N], R[N], idx;

void init()
{
tot = 0;
ts = 0;
CLR(T, 0);
idx = 0;
}
inline void add(int s, int t)
{
}
void dfs(int u, int f, int d)
{
ver[++ts] = u;
D[ts] = d;
F[u] = ts;

L[u] = ++idx;
dep[u] = d;
for (int i = head[u]; ~i; i = E[i].nxt)
{
int v = E[i].to;
if (v != f)
{
dfs(v, u, d + 1);
ver[++ts] = u;
D[ts] = d;
}
}
R[u] = idx;
}
void RMQ_init(int l, int r)
{
int i, j;
for (i = l; i <= r; ++i)
dp[i][0] = i;
for (j = 1; l + (1 << j) - 1 <= r; ++j)
{
for (i = l; i + (1 << j) - 1 <= r; ++i)
{
int a = dp[i][j - 1], b = dp[i + (1 << (j - 1))][j - 1];
dp[i][j] = D[a] < D[b] ? a : b;
}
}
}
int LCA(int u, int v)
{
int l = F[u], r = F[v];
if (l > r)
swap(l, r);
int len = r - l + 1;
int k = 0;
while (1 << (k + 1) <= len)
++k;
int a = dp[l][k], b = dp[r - (1 << k) + 1][k];
return D[a] < D[b] ? ver[a] : ver[b];
}
namespace BIT
{
{
while (k < N)
{
T[k] += v;
k += (k & -k);
}
}
int sum(int k)
{
int ret = 0;
while (k)
{
ret += T[k];
k -= (k & -k);
}
return ret;
}
}
int main(void)
{
int n, u, v, i, q;
while (~scanf("%d", &n))
{
init();
for (i = 0; i < n; ++i)
{
scanf("%d%d", &u, &v);
}
dfs(0, -1, 0);
RMQ_init(1, ts);
scanf("%d", &q);
for (i = 0; i < q; ++i)
{
scanf("%d%d", &node[i].u, &node[i].v);
node[i].lca = LCA(node[i].u, node[i].v);
node[i].dep = dep[node[i].lca];
}
sort(node, node + q);
int ans = 0;
for (i = 0; i < q; ++i)
{
if (BIT::sum(L[node[i].u]) || BIT::sum(L[node[i].v]))
continue;
}