BZOJ3224——Tyvj 1728 普通平衡树

来源：互联网发布：apachemysql数据库管理编辑：程序博客网时间：2024/06/06 14:29
1、题目大意：数据结构题，是treap，全都是treap比较基本的操作
2、分析：没啥思考的
#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>using namespace std;struct Node{    Node *ch[2];    int r, v, s, num;    bool operator < (const Node& rhs) const{        return r < rhs.r;    }    int cmp(int x){        if(x < v) return 0;        if(x == v) return -1;        return 1;    }    int cmp1(int x){        int k = num;        if(ch[0]) k += ch[0] -> s;        if(k - num + 1 <= x && x <= k) return -1;        if(x <= k - num) return 0;        return 1;     }    void maintain(){        s = num;        if(ch[0]) s += ch[0] -> s;        if(ch[1]) s += ch[1] -> s;    }};struct treap{    Node ft[5000000];    int tot;    Node *root;    void rotate(Node* &o, int d){        Node* k = o -> ch[d ^ 1];        o -> ch[d ^ 1] = k -> ch[d];        k -> ch[d] = o;        o -> maintain();        k -> maintain();        o = k;     }    void insert(Node* &o, int x){        if(o == NULL){            o = &ft[tot ++];            o -> ch[0] = o -> ch[1] = NULL;            o -> v = x;            o -> r = rand();            o -> num = o -> s = 1;            return;         }        int d = o -> cmp(x);        if(d == -1) o -> num ++;        else {            insert(o -> ch[d], x);            if(o -> ch[d] > o) rotate(o, d ^ 1);        }        o -> maintain();     }    void remove(Node* &o, int x){        int d = o -> cmp(x);        if(d == -1){            if(o -> num > 1) o -> num --;            else if(o -> ch[0] && o -> ch[1]) {                int d2 = 0;                if(o -> ch[0] > o -> ch[1]) d2 = 1;                rotate(o, d2);                remove(o -> ch[d2], x);            }            else {                if(o -> ch[0]){                    o = o -> ch[0];                }                else o = o -> ch[1];            }        }        else remove(o -> ch[d], x);        if(o) o -> maintain();    }    int find(Node* &o, int x){        if(o == NULL) return 0;        int d = o -> cmp(x);        if(d == -1) return o -> num;        return find(o -> ch[d], x);    }    int less_k(Node* &o, int k){        if(o == NULL) return 0;        int d = o -> cmp(k);        int yy = o -> num;        if(o -> ch[0]) yy += o -> ch[0] -> s;          if(d == -1) return yy - o -> num;        else if(d == 0) return less_k(o -> ch[0], k);        else return less_k(o -> ch[1], k) + yy;     }    int kth(Node* &o, int k){        int d = o -> cmp1(k);        int yy = o -> num;        if(o -> ch[0]) yy += o -> ch[0] -> s;        if(d == -1) return o -> v;        if(d == 0) return kth(o -> ch[0], k);        return kth(o -> ch[1], k - yy);    }} wt;int main(){    int n;    scanf("%d", &n);    for(int i = 1; i <= n; i ++){        int op, x;        scanf("%d%d", &op, &x);        if(op == 1){            wt.insert(wt.root, x);        }        else if(op == 2){            wt.remove(wt.root, x);        }        else if(op == 3){            printf("%d\n", wt.less_k(wt.root, x) + 1);        }        else if(op == 4){            printf("%d\n", wt.kth(wt.root, x));        }        else if(op == 5){            int yy = wt.less_k(wt.root, x);            printf("%d\n", wt.kth(wt.root, yy));        }        else{            int yy = wt.less_k(wt.root, x);            yy += wt.find(wt.root, x) + 1;            printf("%d\n", wt.kth(wt.root, yy));        }    }    return 0;}
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