POJ 1470 Closest Common Ancestors(LCA&RMQ)

题意比较费劲：输入看起来很麻烦。处理括号冒号的时候是用%1s就可以。还有就是注意它有根节点。。。Q次查询，我是用在线st做的。

/*************************************************************************
    > File Name:            3.cpp
    > Author:               Howe_Young
    > Mail:                 [email protected]
    > Created Time:         2015年10月08日 星期四 19时03分30秒
 ************************************************************************/

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>

using namespace std;
typedef long long ll;
const int maxn = 10000;
struct Edge {
    int to, next;
}edge[maxn<<1];
int tot, head[maxn];
int ans[maxn];
int Euler[maxn<<1];
int R[maxn];
int dep[maxn];
int dp[maxn<<1][20]; // RMQ
bool in[maxn];
int cnt;
void init()
{
    cnt = 0;
    tot = 0;
    memset(head, -1, sizeof(head));
    memset(ans, 0, sizeof(ans));
    memset(in, false, sizeof(in));
}
void addedge(int u, int v)
{
    edge[tot].to = v;
    edge[tot].next = head[u];
    head[u] = tot++;
}
void dfs(int u, int depth)
{
    Euler[++cnt] = u;
    R[u] = cnt;
    dep[cnt] = depth;
    for (int i = head[u]; i != -1; i = edge[i].next)
    {
        int v = edge[i].to;
        dfs(v, depth + 1);
        Euler[++cnt] = u;
        dep[cnt] = depth;
    }
}

void RMQ(int n)
{
    for (int i = 1; i <= n; i++) dp[i][0] = i;
    int m = (int)(log(n) / log(2));
    for (int j = 1; j <= m; j++)
    {
        for (int i = 1; i + (1 << j) - 1 <= n; i++)
            dp[i][j] = dep[dp[i][j - 1]] < dep[dp[i + (1 << (j - 1))][j - 1]] ? dp[i][j - 1] : dp[i + (1 << (j - 1))][j - 1];
    }
}
int query(int u, int v)
{
    int l = R[u], r = R[v];
    if (l > r) swap(l, r);
    int k = (int)(log(r - l + 1) / log(2));
    int lca = dep[dp[l][k]] < dep[dp[r - (1 << k) + 1][k]] ? dp[l][k] : dp[r - (1 << k) + 1][k];
    return Euler[lca];
}
int main()
{
    int n;
    while (~scanf("%d", &n))
    {
        init();
        char s1[3], s2[3];
        int u, v, m;
        for (int i = 0; i < n; i++)
        {
            scanf("%d %1s %1s %d %1s", &u, s1, s1, &m, s2);
            for (int j = 0; j < m; j++)
            {
                scanf("%d", &v);
                addedge(u, v);
                in[v] = true;
            }
        }
        for (int i = 1; i <= n; i++)
            if (!in[i])
            {
                dfs(i, 1);
                break;
            }
        RMQ(cnt);
        int Q;
        scanf("%d", &Q);
        while (Q--)
        {
            scanf("%1s %d %d %1s", s1, &u, &v, s2);
            ans[query(u, v)]++;
        }
        for (int i = 1; i <= n; i++)
            if (ans[i])
                printf("%d:%d\n", i, ans[i]);
    }
    return 0;
}