BZOJ3224 Tyvj 1728 普通平衡树

就是裸的平衡树的说= =

怎么就是做不出呢。。。

直接写Treap好了，反正Splay至今不会233

 

  1 /**************************************************************
  2     Problem: 3224
  3     User: rausen
  4     Language: C++
  5     Result: Accepted
  6     Time:212 ms
  7     Memory:4032 kb
  8 ****************************************************************/
  9  
 10 #include <cstdio>
 11 #include <cstdlib>
 12 #include <ctime>
 13  
 14 #define lson tr[r].l
 15 #define rson tr[r].r
 16 using namespace std;
 17 const int N = 100001;
 18 const int Maxlen = 900005;
 19  
 20 struct treap_node {
 21     int l, r, v, sz, key, cnt;
 22 }tr[N];
 23  
 24 int n, cnt, root, ans, x;
 25 char buf[Maxlen];
 26 int Left, Len;
 27  
 28 inline int read() {
 29     int x = 0, sgn = 1;
 30     while (buf[Left] < '0' || '9' < buf[Left])
 31         if (buf[Left++] == '-') sgn = -1;
 32     while ('0' <= buf[Left] && buf[Left] <= '9')
 33         x = x * 10 + buf[Left++] - '0';
 34     return sgn * x;
 35 }
 36   
 37 int len = 0, pr[15];
 38 inline void print(int x) {
 39     if (x < 0) putchar('-'), x = -x;
 40     while (x)
 41         pr[++len] = x % 10, x /= 10;
 42     if (!len) putchar('0');
 43     while (len)
 44         putchar(pr[len--] + '0');
 45     puts("");
 46 }
 47  
 48  
 49 inline void update(int r) {
 50     tr[r].sz = tr[lson].sz + tr[rson].sz + tr[r].cnt;
 51 }
 52  
 53 inline void rturn(int &r) {
 54     int t = lson;
 55     lson = tr[t].r;
 56     tr[t].r = r, tr[t].sz = tr[r].sz;
 57     update(r);
 58     r = t;
 59 }
 60  
 61 inline void lturn(int &r) {
 62     int t = rson;
 63     rson = tr[t].l;
 64     tr[t].l = r, tr[t].sz = tr[r].sz;
 65     update(r);
 66     r = t;
 67 }
 68  
 69 void ins(int &r) {
 70     if (r == 0) {
 71         r = ++cnt;
 72         tr[r].sz = tr[r].cnt = 1;
 73         tr[r].v = x;
 74         tr[r].key = rand();
 75         return;
 76     }
 77     ++tr[r].sz;
 78     if (tr[r].v == x)
 79         ++tr[r].cnt;
 80     else if (x > tr[r].v) {
 81         ins(rson);
 82         if (tr[rson].key < tr[r].key)
 83             lturn(r);
 84     } else {
 85         ins(lson);
 86         if (tr[lson].key < tr[r].key)
 87             rturn(r);
 88     }
 89 }
 90  
 91 void del(int &r) {
 92     if (r == 0) return;
 93     if (tr[r].v == x) {
 94         if (tr[r].cnt > 1) {
 95             --tr[r].cnt, --tr[r].sz;
 96             return;
 97         }
 98         if (!lson || !rson)
 99             r = lson + rson;
100         else if (tr[lson].key < tr[rson].key)
101                 rturn(r), del(r);
102         else lturn(r), del(r);              
103     }
104     else if (x > tr[r].v) --tr[r].sz, del(rson);
105     else --tr[r].sz, del(lson);
106 }
107  
108 int query_rank(int r) {
109     if (r == 0) return 0;
110     if (tr[r].v == x) 
111         return tr[lson].sz + 1;
112     else if (x > tr[r].v) 
113         return tr[lson].sz + tr[r].cnt + query_rank(rson);
114     return query_rank(lson);
115 }
116  
117 int query_kth (int r, int x) {
118     if (r == 0) return 0;
119     if (x <= tr[lson].sz) return query_kth(lson, x);
120     else if (x > tr[lson].sz + tr[r].cnt)
121         return query_kth(rson, x - tr[lson].sz - tr[r].cnt);
122     else return tr[r].v;
123 }
124  
125 void query_pre(int r) {
126     if (r == 0) return;
127     if (tr[r].v < x) {
128         ans = r;
129         query_pre(rson);
130     } else query_pre(lson);
131 }
132  
133 void query_next(int r) {
134     if (r == 0) return;
135     if (tr[r].v > x) {
136         ans = r;
137         query_next(lson);
138     } else query_next(rson);
139 }
140  
141 int main() {
142     int i, oper;
143     Len = fread(buf, 1, Maxlen, stdin);
144     buf[Len] = ' ';
145     n = read();
146     for (i = 1; i <= n; ++i) {
147         oper = read(), x = read();
148         if (oper == 1)
149             ins(root); else
150         if (oper == 2)
151             del(root); else
152         if (oper == 3)
153             print(query_rank(root)); else
154         if (oper == 4)
155             print(query_kth(root, x)); else
156         if (oper == 5) {
157             ans = 0, query_pre(root);
158             print(tr[ans].v);
159         } else
160         if (oper == 6) {
161             ans = 0, query_next(root);
162             print(tr[ans].v);
163         }
164     }
165     return 0;
166 }
View Code

 

posted on 2014-11-21 22:15  Xs酱~  阅读(215)  评论(0)  编辑  收藏  举报