Problem

查询区间第k大，但保证区间不互相包含（可以相交）

Solution

只需要对每个区间左端点进行排序，那它们的右端点必定单调递增，不然会出现区间包含的情况。

所以我们暴力对下一个区间加上这个区间没有的点，删去下个区间没有的点。

因为每个点最多被加入，删除1次，所以时间复杂度为O（nlogn）

Notice

当相邻两段区间不相交时，那么我们要先加入点，在删去点。

Code

非旋转Treap

  1. #include<cmath>
  2. #include<cstdio>
  3. #include<cstring>
  4. #include<iostream>
  5. #include<algorithm>
  6. using namespace std;
  7. #define sqz main
  8. #define ll long long
  9. #define reg register int
  10. #define rep(i, a, b) for (reg i = a; i <= b; i++)
  11. #define per(i, a, b) for (reg i = a; i >= b; i--)
  12. #define travel(i, u) for (reg i = head[u]; i; i = edge[i].next)
  13. const int INF = 1e9, N = 100000, M = 50000;
  14. const double eps = 1e-6, phi = acos(-1.0);
  15. ll mod(ll a, ll b) {if (a >= b || a < 0) a %= b; if (a < 0) a += b; return a;}
  16. ll read(){ ll x = 0; int zf = 1; char ch; while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
  17. if (ch == '-') zf = -1, ch = getchar(); while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x * zf;}
  18. void write(ll y) { if (y < 0) putchar('-'), y = -y; if (y > 9) write(y / 10); putchar(y % 10 + '0');}
  19. int point = 0, T[N + 5], root, ans[M + 5];
  20. struct Node
  21. {
  22.     int left, right, ask, id;
  23. }Q[M + 5];
  24. struct node
  25. {
  26.     int Size[N + 5], Val[N + 5], Level[N + 5], Son[2][N + 5];
  27.     inline void up(int u)
  28.     {
  29.         Size[u] = Size[Son[0][u]] + Size[Son[1][u]] + 1;
  30.     }
  31.     int Newnode(int v)
  32.     {
  33.         int u = ++point;
  34.         Val[u] = v, Level[u] = rand();
  35.         Size[u] = 1, Son[0][u] = Son[1][u] = 0;
  36.         return u;
  37.     }
  38.     int Merge(int X, int Y)
  39.     {
  40.         if (X * Y == 0) return X + Y;
  41.         if (Level[X] < Level[Y])
  42.         {
  43.             Son[1][X] = Merge(Son[1][X], Y);
  44.             up(X); return X;
  45.         }
  46.         else
  47.         {
  48.             Son[0][Y] = Merge(X, Son[0][Y]);
  49.             up(Y); return Y;
  50.         }
  51.     }
  52.     void Split(int u, int t, int &x, int &y)
  53.     {
  54.         if (!u)
  55.         {
  56.             x = y = 0;
  57.             return;
  58.         }
  59.         if (Val[u] <= t) x = u, Split(Son[1][u], t, Son[1][u], y);
  60.         else y = u, Split(Son[0][u], t, x, Son[0][u]);
  61.         up(u);
  62.     }
  63.     void Build(int l, int r)
  64.     {
  65.         int last, s[N + 5], top = 0;
  66.         rep(i, l, r)
  67.         {
  68.             int u = Newnode(T[i]);
  69.             last = 0;
  70.             while (top && Level[s[top]] > Level[u])
  71.             {
  72.                 up(s[top]);
  73.                 last = s[top--];
  74.             }
  75.             if (top) Son[1][s[top]] = u;
  76.             Son[0][u] = last;
  77.             s[++top] = u;
  78.         }
  79.         while (top) up(s[top--]);
  80.         root = s[1];
  81.     }
  83.     int Find_num(int u, int t)
  84.     {
  85.     	if (t <= Size[Son[0][u]]) return Find_num(Son[0][u], t);
  86.     	else if (t <= Size[Son[0][u]] + 1) return u;
  87.     	else return Find_num(Son[1][u], t - Size[Son[0][u]] - 1);
  88. 	}
  89.     void Insert(int v)
  90.     {
  91.         int t = Newnode(v), x, y;
  92.         Split(root, v, x, y);
  93.         root = Merge(Merge(x, t), y);
  94.     }
  95.     void Delete(int v)
  96.     {
  97.         int x, y, z;
  98.         Split(root, v, x, z), Split(x, v - 1, x, y);
  99.         root = Merge(Merge(x, Merge(Son[0][y], Son[1][y])), z);
  100.     }
  101. }Treap;
  102. int cmp(Node X, Node Y)
  103. {
  104. 	return X.left < Y.left || (X.left == Y.left && X.right < Y.right);
  105. }
  106. int sqz()
  107. {
  108. 	int n = read(), m = read();
  109. 	rep(i, 1, n) T[i] = read();
  110. 	rep(i, 1, m) Q[i].left = read(), Q[i].right = read(), Q[i].ask = read(), Q[i].id = i;
  111. 	sort(Q + 1, Q + m + 1, cmp);
  112.     rep(i, Q[1].left, Q[1].right) Treap.Insert(T[i]);
  113.     ans[Q[1].id] = Treap.Val[Treap.Find_num(root, Q[1].ask)];
  114.     rep(i, 2, m)
  115.     {
  116.         rep(j, Q[i - 1].right + 1, Q[i].right) Treap.Insert(T[j]);
  117.         rep(j, Q[i - 1].left, Q[i].left - 1) Treap.Delete(T[j]);
  118.         ans[Q[i].id] = Treap.Val[Treap.Find_num(root, Q[i].ask)];
  119.     }
  120.     rep(i, 1, m) printf("%d\n", ans[i]);
  121.     return 0;
  122. }

旋转Treap

  1. #include<cmath>
  2. #include<cstdio>
  3. #include<cstring>
  4. #include<iostream>
  5. #include<algorithm>
  6. using namespace std;
  7. #define sqz main
  8. #define ll long long
  9. #define reg register int
  10. #define rep(i, a, b) for (reg i = a; i <= b; i++)
  11. #define per(i, a, b) for (reg i = a; i >= b; i--)
  12. #define travel(i, u) for (reg i = head[u]; i; i = edge[i].next)
  13. const int INF = 1e9, N = 100000, M = 50000;
  14. const double eps = 1e-6, phi = acos(-1.0);
  15. ll mod(ll a, ll b) {if (a >= b || a < 0) a %= b; if (a < 0) a += b; return a;}
  16. ll read(){ ll x = 0; int zf = 1; char ch; while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
  17. if (ch == '-') zf = -1, ch = getchar(); while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x * zf;}
  18. void write(ll y) { if (y < 0) putchar('-'), y = -y; if (y > 9) write(y / 10); putchar(y % 10 + '0');}
  19. int point = 0, T[N + 5], root, ans[M + 5];
  20. struct node
  21. {
  22.     int left, right, ask, id;
  23. }Q[M + 5];
  24. int cmp(node X, node Y)
  25. {
  26. 	return X.left < Y.left || (X.left == Y.left && X.right < Y.right);
  27. }
  28. struct Node
  29. {
  30.     int Val[N + 5], Son[2][N + 5], Level[N + 5], Size[N + 5], Num[N + 5];
  31.     inline void up(int u)
  32.     {
  33.         Size[u] = Size[Son[0][u]] + Size[Son[1][u]] + Num[u];
  34.     }
  35.     inline void Lturn(int &x)
  36.     {
  37.         int y = Son[1][x]; Son[1][x] = Son[0][y]; Son[0][y] = x;
  38.         up(x), up(y); x = y;
  39.     }
  40.     inline void Rturn(int &x)
  41.     {
  42.         int y = Son[0][x]; Son[0][x] = Son[1][y]; Son[1][y] = x;
  43.         up(x), up(y); x = y;
  44.     }
  45.     inline void Newnode(int &u, int v)
  46.     {
  47.         u = ++point;
  48.         Level[u] = rand(), Val[u] = v;
  49.         Num[u] = Size[u] = 1, Son[0][u] = Son[1][u] = 0;
  50.     }
  52.     void Insert(int &u, int t)
  53.     {
  54.         if (!u)
  55.         {
  56.             Newnode(u, t);
  57.             return;
  58.         }
  59.         Size[u]++;
  60.         if (t == Val[u]) Num[u]++;
  61.         else if (t < Val[u])
  62.         {
  63.             Insert(Son[0][u], t);
  64.             if (Level[Son[0][u]] < Level[u]) Rturn(u);
  65.         }
  66.         else
  67.         {
  68.             Insert(Son[1][u], t);
  69.             if (Level[Son[1][u]] < Level[u]) Lturn(u);
  70.         }
  71.     }
  72.     void Delete(int &u, int t)
  73.     {
  74.         if (!u) return;
  75.         if (Val[u] == t)
  76.         {
  77.             if (Num[u] > 1)
  78.             {
  79.                 Size[u]--, Num[u]--;
  80.                 return;
  81.             }
  82.             if (Son[0][u] * Son[1][u] == 0) u = Son[0][u] + Son[1][u];
  83.             else if (Level[Son[0][u]] < Level[Son[1][u]]) Rturn(u), Delete(u, t);
  84.             else Lturn(u), Delete(u, t);
  85.         }
  86.         else if (t < Val[u]) Size[u]--, Delete(Son[0][u], t);
  87.         else Size[u]--, Delete(Son[1][u], t);
  88.     }
  90.     int Find_num(int u, int t)
  91.     {
  92.         if (!u) return 0;
  93.         if (t <= Size[Son[0][u]]) return Find_num(Son[0][u], t);
  94.         else if (t <= Size[Son[0][u]] + Num[u]) return u;
  95.         else return Find_num(Son[1][u], t - Size[Son[0][u]] - Num[u]);
  96.     }
  97. }Treap;
  98. int sqz()
  99. {
  100.     int n = read(), m = read();
  101.     rep(i, 1, n) T[i] = read();
  102.     rep(i, 1, m) Q[i].left = read(), Q[i].right = read(), Q[i].ask = read(), Q[i].id = i;
  103.     sort(Q + 1, Q + m + 1, cmp);
  104.     rep(i, Q[1].left, Q[1].right) Treap.Insert(root, T[i]);
  105.     ans[Q[1].id] = Treap.Val[Treap.Find_num(root, Q[1].ask)];
  106.     rep(i, 2, m)
  107.     {
  108.         if (Q[i].left <= Q[i - 1].right)
  109.         {
  110.             rep(j, Q[i - 1].left, Q[i].left - 1) Treap.Delete(root, T[j]);
  111.             rep(j, Q[i - 1].right + 1, Q[i].right) Treap.Insert(root, T[j]);
  112.         }
  113.         else
  114.         {
  115.             rep(j, Q[i - 1].left, Q[i - 1].right) Treap.Delete(root, T[j]);
  116.             rep(j, Q[i].left, Q[i].right) Treap.Insert(root, T[j]);
  117.         }
  118.         ans[Q[i].id] = Treap.Val[Treap.Find_num(root, Q[i].ask)];
  119.     }
  120.     rep(i, 1, m) printf("%d\n", ans[i]);
  121.     return 0;
  122. }

[POJ2761]Feed the dogs的更多相关文章

  1. 【莫队算法】【权值分块】poj2104 K-th Number / poj2761 Feed the dogs

    先用莫队算法保证在询问之间转移的复杂度,每次转移都需要进行O(sqrt(m))次插入和删除,权值分块的插入/删除是O(1)的. 然后询问的时候用权值分块查询区间k小值,每次是O(sqrt(n))的. ...

  2. [POJ2761] Feed the dogs (Treap)

    题目链接:http://poj.org/problem?id=2761 题目大意:给你n个数,m次查询,m次查询分别是a,b,k,查询下表从a到b的第k小元素是哪个.这m个区间不会互相包含. Trea ...

  3. [Poj2761]Feed the dogs（主席树）

    Desciption 题意:求区间第K小(N<=100000) Solution 主席树模板题 Code #include <cstdio> #include <algorit ...

  4. 【POJ2761】【区间第k大】Feed the dogs(吐槽)

    Description Wind loves pretty dogs very much, and she has n pet dogs. So Jiajia has to feed the dogs ...

  5. poj 2761 Feed the dogs (treap树)

    /************************************************************* 题目: Feed the dogs(poj 2761) 链接: http: ...

  6. 划分树---Feed the dogs

    POJ  2761 Description Wind loves pretty dogs very much, and she has n pet dogs. So Jiajia has to fee ...

  7. poj2761 feed the dog

    题目链接:http://poj.org/problem?id=2761 Description Wind loves pretty dogs very much, and she has n pet ...

  8. POJ 2761 Feed the dogs（平衡树or划分树or主席树）

    Description Wind loves pretty dogs very much, and she has n pet dogs. So Jiajia has to feed the dogs ...

  9. POJ 2761 Feed the dogs （主席树）（K-th 值）

                                                                Feed the dogs Time Limit: 6000MS   Memor ...

随机推荐

  1. ArcFace Android 人脸检测与人脸识别集成分享

    目前我们的应用内使用了 ArcFace 的人脸检测功能,其他的我们并不了解,所以这里就和大家分享一下我们的集成过程和一些使用心得集成ArcFace FD 的集成过程非常简单在 ArcFace FD 的 ...

  2. Codeforces 458C - Elections

    458C - Elections 思路: 三分凹形函数极小值域 代码: #include<bits/stdc++.h> using namespace std; #define ll lo ...

  3. redhat 7.0 配置Bond

    把/etc/sysconfig/network-scripts/目录下的ifcfg-bond*文件手动移动到/tmp目录. 重启网络:systemctl restart NetworkManager ...

  4. Birdman Quotes

    Popularity is the slutty little cousin of prestige (威望,声望) . A man becomes a critic when he cannot b ...

  5. xsd与xml和类(class)对象之间的互相转换

    xsd与xml和类(class)对象之间的互相转换 . 第一:通过现有的已经写好的xsd来生成class(.cs)文件. 在您Visual Studio的安装目录下的SDKv2.0Bin中有个应用程序 ...

  6. WCF利用Stream上传大文件

    WCF利用Stream上传大文件 转自别人的文章,学习这个例子,基本上wcf也算入门了,接口用法.系统配置都有了 本文展示了在asp.net中利用wcf的stream方式传输大文件,解决了大文件上传问 ...

  7. IntelliJ IDEA的这些配置，你值得拥有

    一.自动编译开关 二.忽略大小写开关 IDEA默认是匹配大小写,此开关如果未关.你输入字符一定要符合大小写.比如你敲string是不会出现代码提示或智能补充.但是,如果你开了这个开关,你无论输入Str ...

  8. A strange lift HDU - 1548

    There is a strange lift.The lift can stop can at every floor as you want, and there is a number Ki(0 ...

  9. BZOJ 1833 数字计数 数位DP

    题目链接 做的第一道数位DP题,听说是最基础的模板题,但还是花了好长时间才写出来..... 想深入了解下数位DP的请点这里 先设dp数组dp[i][j][k]表示数位是i,以j开头的数k出现的次数 有 ...

  10. 6月13 ThinkPHP框架基础

    ThinkPHP 一.php框架基础介绍 真实项目开发步骤: 多人同时开发项目,协作开发项目.分工合理.效率有提高(代码风格不一样.分工不好) 测试阶段 上线运行 对项目进行维护.修改.升级(单个人维 ...