3224: Tyvj 1728 普通平衡树
Description
您需要写一种数据结构(可参考题目标题),来维护一些数,其中需要提供以下操作:
1. 插入x数
2. 删除x数(若有多个相同的数,因只删除一个)
3. 查询x数的排名(若有多个相同的数,因输出最小的排名)
4. 查询排名为x的数
5. 求x的前驱(前驱定义为小于x,且最大的数)
6. 求x的后继(后继定义为大于x,且最小的数)
Input
第一行为n,表示操作的个数,下面n行每行有两个数opt和x,opt表示操作的序号(1<=opt<=6)
Output
对于操作3,4,5,6每行输出一个数,表示对应答案
Sample Input
10
1 106465
4 1
1 317721
1 460929
1 644985
1 84185
1 89851
6 81968
1 492737
5 493598
Sample Output
106465
84185
492737
HINT
1.n的数据范围:n<=100000
2.每个数的数据范围:[-1e7,1e7]
Source
题目类型:平衡树(Treap)
算法分析:直接构建一个平衡树来求解即可,注意旋转操作之后一定要更新根节点的size值!!!
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 |
#include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; #define CFF freopen ("aaa.txt", "r", stdin) #define CPPFF ifstream cin ("aaa.txt") #define LL long long #define ULL unsigned long long #define DB(ccc) cout << #ccc << " = " << ccc << endl const int INF = 0x7FFFFFFF; const int MOD = 1e9 + 7; const double EPS = 1e-6; const double PI = 2 * acos (0.0); const int maxn = 1e5 + 66; struct node { int l, r, v, size, rnd, w; }tr[maxn]; int n, size, root, res; void UpDate (int rt)//更新结点信息 { tr[rt].size = tr[tr[rt].l].size + tr[tr[rt].r].size + tr[rt].w; } void rturn (int &rt) { int tt = tr[rt].l; tr[rt].l = tr[tt].r; tr[tt].r = rt; tr[tt].size = tr[rt].size; UpDate(rt); rt = tt; } void lturn (int &rt) { int tt = tr[rt].r; tr[rt].r = tr[tt].l; tr[tt].l = rt; tr[tt].size = tr[rt].size; UpDate(rt); rt = tt; } void Insert (int &rt, int x) { if (rt == 0) { rt = ++size; tr[rt].size = tr[rt].w = 1; tr[rt].v = x; tr[rt].rnd = rand(); return ; } tr[rt].size++; if (tr[rt].v == x) tr[rt].w++;//每个结点顺便记录下与该节点相同值的数的个数 else if (x > tr[rt].v) { Insert (tr[rt].r, x); if (tr[tr[rt].r].rnd < tr[rt].rnd) lturn (rt);//维护堆性质 } else { Insert (tr[rt].l, x); if (tr[tr[rt].l].rnd < tr[rt].rnd) rturn (rt); } } void Del (int &rt, int x) { if (rt == 0) return ; if (tr[rt].v == x) { if (tr[rt].w > 1) { tr[rt].w--; tr[rt].size--; return ;//若不止相同值的个数有多个,删去一个 } if (tr[rt].l * tr[rt].r == 0) rt = tr[rt].l + tr[rt].r;//有一个儿子为空 else if (tr[tr[rt].l].rnd < tr[tr[rt].r].rnd) rturn (rt), Del (rt, x); else lturn (rt), Del (rt, x); } else if (x > tr[rt].v) tr[rt].size--, Del (tr[rt].r, x); else tr[rt].size--, Del (tr[rt].l, x); } int QueryRank (int rt, int x) { if (rt == 0) return 0; if (tr[rt].v == x) return tr[tr[rt].l].size + 1; else if (x > tr[rt].v) return tr[tr[rt].l].size + tr[rt].w + QueryRank (tr[rt].r, x); else return QueryRank (tr[rt].l, x); } int QueryNum (int rt, int x) { if (rt == 0) return 0; if (x <= tr[tr[rt].l].size) return QueryNum (tr[rt].l, x); else if (x > tr[tr[rt].l].size + tr[rt].w) return QueryNum (tr[rt].r, x - tr[tr[rt].l].size - tr[rt].w); else return tr[rt].v; } void QueryPro (int rt, int x) { if (rt == 0) return ; if (tr[rt].v < x) { res = rt; QueryPro (tr[rt].r, x); } else QueryPro (tr[rt].l, x); } void QuerySub (int rt, int x) { if (rt == 0) return ; if (tr[rt].v > x) { res = rt; QuerySub (tr[rt].l, x); } else QuerySub (tr[rt].r, x); } int main() { // CFF; int n; scanf ("%d", &n); for (int i = 1; i <= n; i++) { int oper, val; scanf ("%d%d", &oper, &val); if (oper == 1) Insert (root, val); else if (oper == 2) Del (root, val); else if (oper == 3) printf ("%d\n", QueryRank (root, val)); else if (oper == 4) printf ("%d\n", QueryNum (root, val)); else if (oper == 5) { res = 0; QueryPro (root, val); printf ("%d\n", tr[res].v); } else if (oper == 6) { res = 0; QuerySub (root, val); printf ("%d\n", tr[res].v); } } return 0; } |