(好题)树状数组+离散化+DFS序+离线/莫队 HDOJ 4358 Boring counting

题目传送门

题意：给你一棵树，树上的每个节点都有树值，给m个查询，问以每个点u为根的子树下有多少种权值恰好出现k次。

分析：首先要对权值离散化，然后要将树形转换为线形，配上图：。然后按照右端点从小到大排序，离线操作：将每一个深度的权值分组到相同权值的cnt中，当sz == k时，用树状数组更新+1，表示在该深度已经存在k个相同的权值，如果>k，之前k个-2(-1是恢复原样，再-1是为下次做准备？)，然后一个点的子树的答案就是 sum (r) - sum (l-1)。

当然，区间离线问题用莫队算法是很好做的。

收获：1. 离散化　　2. DFS序，树形变线形　　3. 离线操作　　4. 莫队算法

代码(树状数组)：

/************************************************
* Author        :Running_Time
* Created Time  :2015/9/10 星期四 19:15:22
* File Name     :I.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
struct Edge {
    int v, nex;
}edge[N*2];
struct Query    {
    int l, r, id;
    Query (int _l = 0, int _r = 0, int _id = 0) : l (_l), r (_r), id (_id) {}
    bool operator < (const Query &x) const  {
        return r < x.r;
    }
}q[N];
struct BIT  {
    int c[N];
    void init(void) {
        memset (c, 0, sizeof (c));
    }
    void updata(int i, int x)   {
        while (i < N)   {
            c[i] += x;  i += i & (-i);
        }
    }
    int query(int i)    {
        int ret = 0;
        while (i)   {
            ret += c[i];    i -= i & (-i);
        }
        return ret;
    }
}bit;
int head[N], dfn[N], low[N], w[N], p[N], val[N], ans[N];
vector<int> cnt[N];
int e, dep;

void init(void) {
    memset (head, -1, sizeof (head));
    e = 0;  dep = 0;
}

void add_edge(int u, int v) {
    edge[e].v = v;  edge[e].nex = head[u];
    head[u] = e++;
}

bool cmp(int i, int j)  {
    return w[i] < w[j];
}

void compress(int n)    {
    for (int i=1; i<=n; ++i)    p[i] = i;
    sort (p+1, p+1+n, cmp);
    int rk = 0, pre = w[p[1]] - 1;
    for (int i=1; i<=n; ++i)    {
        if (pre != w[p[i]]) {
            pre = w[p[i]];
            w[p[i]] = ++rk;
        }
        else    {
            w[p[i]] = rk;
        }
    }
}

void DFS(int u, int fa) {
    dfn[u] = ++dep; val[dep] = w[u];
    for (int i=head[u]; ~i; i=edge[i].nex)  {
        int v = edge[i].v;
        if (v == fa)    continue;
        DFS (v, u);
    }
    low[u] = dep;
}

int main(void)    {
    int T, cas = 0;  scanf ("%d", &T);
    while (T--) {
        if (cas)    puts ("");
        printf ("Case #%d:
", ++cas);
        int n, k;   scanf ("%d%d", &n, &k);
        init ();

        for (int i=1; i<=n; ++i)    {
            scanf ("%d", &w[i]);
        }
        compress (n);                                           //离散化，升序排序，相同的还是相同的

        for (int u, v, i=1; i<n; ++i)   {
            scanf ("%d%d", &u, &v);
            add_edge (u, v);    add_edge (v, u);
        }
        DFS (1, -1);                                            //先序遍历，得到DFS序，树形变线形

        int m;  scanf ("%d", &m);
        for (int u, i=1; i<=m; ++i)    {
            scanf ("%d", &u);
            q[i] = Query (dfn[u], low[u], i);
        }
        sort (q+1, q+1+m);                                      //按照DFS序排序
        
        for (int i=1; i<=n; ++i)    cnt[i].clear ();
        int qu = 1; bit.init ();
        for (int i=1; i<=n; ++i)    {                           //按照dep深度从小到大
            int v = val[i];
            cnt[v].push_back (i);                               //表示离散后的相同的权值的个数
            int sz = cnt[v].size ();
            if (sz >= k)    {
                if (sz == k)    bit.updata (cnt[v][sz-k], 1);
                else    {
                    bit.updata (cnt[v][sz-k-1], -2);            //?
                    bit.updata (cnt[v][sz-k], 1);
                }
            }
            while (qu <= m && q[qu].r == i) {
                ans[q[qu].id] = bit.query (q[qu].r) - bit.query (q[qu].l - 1);
                qu++;
            }
        }

        for (int i=1; i<=m; ++i)    {
            printf ("%d
", ans[i]);
        }
    }

    return 0;
}

代码(莫队算法)：

/************************************************
* Author        :Running_Time
* Created Time  :2015/9/10 星期四 19:15:22
* File Name     :I.cpp
 ************************************************/
 
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
 
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
struct Edge {
    int v, nex;
}edge[N*2];
struct Data {
    int l, r, id, b;
    Data () {}
    Data (int l, int r, int id, int b) : l (l), r (r), id (id), b (b) {}
    bool operator < (const Data &x) const {
        if (b == x.b)   return r < x.r;
        else    return b < x.b;
    }
}data[N];
int head[N], dfn[N], low[N], w[N], p[N], val[N], ans[N];
int cnt[N];
int n, k, m, e, dep, sum;
 
void init(void) {
    memset (head, -1, sizeof (head));
    e = 0;  dep = 0;
}
 
void add_edge(int u, int v) {
    edge[e].v = v;  edge[e].nex = head[u];
    head[u] = e++;
}
 
bool cmp(int i, int j)  {
    return w[i] < w[j];
}
 
void compress(int n)    {
    for (int i=1; i<=n; ++i)    p[i] = i;
    sort (p+1, p+1+n, cmp);
    int rk = 0, pre = w[p[1]] - 1;
    for (int i=1; i<=n; ++i)    {
        if (pre != w[p[i]]) {
            pre = w[p[i]];
            w[p[i]] = ++rk;
        }
        else    {
            w[p[i]] = rk;
        }
    }
}
 
void DFS(int u, int fa) {
    dfn[u] = ++dep; val[dep] = w[u];
    for (int i=head[u]; ~i; i=edge[i].nex)  {
        int v = edge[i].v;
        if (v == fa)    continue;
        DFS (v, u);
    }
    low[u] = dep;
}
 
inline void updata(int x, int c)    {
    cnt[x] += c;
    if (cnt[x] == k)    sum++;
    else if (c > 0 && cnt[x] == k + 1) sum--;
    else if (c < 0 && cnt[x] == k - 1)  sum--;
}
 
void Modui(void)    {
    memset (cnt, 0, sizeof (cnt));
    sum = 0;
    int l = 1, r = 0;
    for (int i=1; i<=m; ++i)    {
        while (data[i].l < l)   updata (val[--l], 1);
        while (data[i].l > l)   updata (val[l], -1), l++;
        while (data[i].r > r)   updata (val[++r], 1);
        while (data[i].r < r)   updata (val[r], -1), r--;
        ans[data[i].id] = sum;
    }
    for (int i=1; i<=m; ++i)    {
        printf ("%d
", ans[i]);
    }
}
 
int main(void)    {
    int T, cas = 0;  scanf ("%d", &T);
    while (T--) {
        if (cas)    puts ("");
        printf ("Case #%d:
", ++cas);
        scanf ("%d%d", &n, &k);
        init ();
 
        for (int i=1; i<=n; ++i)    {
            scanf ("%d", &w[i]);
        }
        compress (n);                                           //离散化，升序排序，相同的还是相同的
 
        for (int u, v, i=1; i<n; ++i)   {
            scanf ("%d%d", &u, &v);
            add_edge (u, v);    add_edge (v, u);
        }
        DFS (1, -1);                                            //先序遍历，得到DFS序，树形变线形
 
        int block = (int) sqrt (n + 0.5);
        scanf ("%d", &m);
        for (int u, i=1; i<=m; ++i)    {
            scanf ("%d", &u);
            data[i] = Data (dfn[u], low[u], i, dfn[u] / block);
        }
        sort (data+1, data+1+m);                                //按照DFS序排序
        Modui ();
    }
 
    return 0;
}

　　

Codeforces Round #136 (Div. 1) B. Little Elephant and Array

题目传送门

分析：估计这是上面那道题的母题(弱化版)，直接套用就行了，现在稍微理解了树状数组的维护方法，配上图：

代码(树状数组)：

/************************************************
* Author        :Running_Time
* Created Time  :2015/9/11 星期五 18:39:17
* File Name     :B.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int n, m;
struct BIT  {
    int c[N];
    void init(void) {
        memset (c, 0, sizeof (c));
    }
    void updata(int i, int x)   {
        while (i <= n)  {
            c[i] += x;  i += i & (-i);
        }
    }
    int query(int i)    {
        int ret = 0;
        while (i)   {
            ret += c[i];    i -= i & (-i);
        }
        return ret;
    }
}bit;
struct Query    {
    int l, r, id;
    bool operator < (const Query &x) const  {
        return r < x.r;
    }
}q[N];
int a[N], ans[N];
vector<int> cnt[N];

int main(void)    {
    scanf ("%d%d", &n, &m);
    for (int i=1; i<=n; ++i)    scanf ("%d", &a[i]);
    for (int i=1; i<=m; ++i)    {
        scanf ("%d%d", &q[i].l, &q[i].r);
        q[i].id = i;
    }
    sort (q+1, q+1+m);
    int qu = 1;
    for (int i=1; i<=n; ++i)    cnt[i].clear ();
    bit.init ();
    for (int i=1; i<=n; ++i)    {
        if (a[i] > n)   continue;
        cnt[a[i]].push_back (i);
        int sz = cnt[a[i]].size ();
        if (sz >= a[i]) {
            bit.updata (cnt[a[i]][sz-a[i]], 1);
            if (sz > a[i])  bit.updata (cnt[a[i]][sz-a[i]-1], -2);
            if (sz > a[i] + 1)  bit.updata (cnt[a[i]][sz-a[i]-2], 1);
        }
        while (qu <= m && q[qu].r == i) {
            ans[q[qu].id] = bit.query (q[qu].r) - bit.query (q[qu].l - 1);
            qu++;
        }
    }
    for (int i=1; i<=m; ++i)    {
        printf ("%d
", ans[i]);
    }

    return 0;
}

　　

 代码(莫队)：

/************************************************
* Author        :Running_Time
* Created Time  :2015/9/11 星期五 20:05:22
* File Name     :B_Modui.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
struct Data {
    int l, r, id, b;
    bool operator < (const Data &x) const {
        if (b == x.b)   return r < x.r;
        else    return b < x.b;
    }
}data[N];
int a[N], ans[N], cnt[N];
int n, m, sum;

inline void updata(int x, int c) {
    if (x > n)  return ;
    cnt[x] += c;
    if (cnt[x] == x)    sum++;
    else if (c > 0 && cnt[x] == x + 1) sum--;
    else if (c < 0 && cnt[x] == x - 1) sum--;
}

void Modui(void)    {
    sum = 0;
    memset (cnt, 0, sizeof (cnt));
    int l = 1, r = 0;
    for (int i=1; i<=m; ++i)    {
        while (data[i].l < l)   updata (a[--l], 1);
        while (data[i].l > l)   updata (a[l], -1), l++;
        while (data[i].r > r)   updata (a[++r], 1);
        while (data[i].r < r)   updata (a[r], -1), r--;
        ans[data[i].id] = sum;
    }
    for (int i=1; i<=m; ++i)    {
        printf ("%d
", ans[i]);
    }
}

int main(void)    {
    scanf ("%d%d", &n, &m);
    int block = (int) sqrt (n + 0.5);
    for (int i=1; i<=n; ++i)    {
        scanf ("%d", &a[i]);
    }
    for (int i=1; i<=m; ++i)    {
        scanf ("%d%d", &data[i].l, &data[i].r);
        data[i].id = i; data[i].b = data[i].l / block;
    }
    sort (data+1, data+1+m);
    Modui ();

    return 0;
}

　　

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原文地址：https://www.cnblogs.com/Running-Time/p/4799169.html