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

在线st算法

/*************************************************************************
    > File Name:            3.cpp
    > Author:               Howe_Young
    > Mail:                 1013410795@qq.com
    > 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;
}

 离线tarjan算法：

#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 1000;//节点
const int maxm = 1000010;//最大查询数
struct Edge {
    int to, next;
}edge[maxn * 2];
int tot, head[maxn];
struct Query {
    int q, next;
    int index;
}query[maxm * 2];
int cnt, h[maxn];
int fa[maxn];
int r[maxn];
int ancestor[maxn];
int ans[maxm];
int Q;
bool vis[maxn];

void init(int n)
{
    Q = 0;
    tot  = 0;
    cnt = 0;
    memset(head, -1, sizeof(head));
    memset(h, -1, sizeof(h));
    memset(fa, -1, sizeof(fa));
    memset(ancestor, 0, sizeof(ancestor));
    memset(vis, false, sizeof(vis));
    for (int i = 1; i <= n; i++) r[i] = 1;
}
void addedge(int u, int v)
{
    edge[tot].to = v;
    edge[tot].next = head[u];
    head[u] = tot++;
}
void addquery(int u, int v, int index)
{
    query[cnt].q = v;
    query[cnt].index = index;
    query[cnt].next = h[u];
    h[u] = cnt++;
}
int find(int x)
{
    if (fa[x] == -1) return x;
    return fa[x] = find(fa[x]);
}
void Union(int x, int y)
{
    int tx = find(x);
    int ty = find(y);
    if (tx != ty)
    {
        if (tx < ty)
        {
            fa[tx] = ty;
            r[ty] += r[tx];
        }
        else
        {
            fa[ty] = tx;
            r[tx] += r[ty];
        }
    } 
}
void LCA(int u)
{
    vis[u] = true;
    ancestor[u] = u;
    for (int i = head[u]; i != -1; i = edge[i].next)
    {
        int v = edge[i].to;
        if (vis[v]) continue;
        LCA(v);
        Union(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])
        {
            ans[query[i].index] = ancestor[find(v)];
        }
    }
}
int res[maxn];
bool in[maxn];
int main()
{
    int n;
    while (~scanf("%d", &n))
    {
        init(n);
        memset(in, false, sizeof(in));
        int u, v, m;    
        char ch[2];
        for (int i = 1; i <= n; i++)
        {
            scanf("%d %1s %1s %d %1s", &u, ch, ch, &m, ch);
            for (int j = 0; j < m; j++)
            {
                scanf("%d", &v);
                in[v] = true;
                addedge(u, v);
                addedge(v, u);
            }
        }
        int q;
        scanf("%d", &q);
        while (q--)
        {
            scanf("%1s %d %d %1s", ch, &u, &v, ch);
            addquery(u, v, Q);
            addquery(v, u, Q++);
        }
        int root;
        for (int i = 1; i <= n; i++)
        {
            if (!in[i])
            {
                root = i;
                break;
            }
        }
        LCA(root);
        memset(res, 0, sizeof(res));
        for (int i = 0; i < Q; i++)
            res[ans[i]]++;
        for (int i = 1; i <= n; i++)
            if (res[i])
                printf("%d:%d\n", i, res[i]);
    }
    return 0;
}