初探treap

treap的基本操作

treap类似二分查找树，只是加了一个堆，用随机值维护平衡，只是期望平衡。小数据下表现并不是特别优秀，但是足够用了。
先水两发，之后再继续搞- -、

poj1338 Ugly Numbers

把质因子只含2,3,5的数叫Ugly Number.通式为:

x=2i×3j×5k

注意到是一个幂次计算，因此大致地有:

0≤i,j,k≤30

那么我们只需要枚举∏(0≤i≤30,x=2,3,5)xi，然后排序即可。当然也可以直接logn地插入，于是可以用STL的set维护，这里不用set，用Treap维护一下，每个节点还要增加一个rank，用来维护子树节点的数量。
代码：

#include <cstdio>#include <cstring>#include <cstdlib>#include <ctime>#include <algorithm>using namespace std;typedef long long type;struct node {    type val;    int fix;    int rank; //增加维护一个rank值,保存以该节点为根子树的节点数目    node* left;    node* right;    node() {        left = right = NULL;    }    node(type x) {        val = x;        fix = rand();        left = right = NULL;    }};inline int rank(node* &T) {    return T == NULL ? 0 : T->rank;}void rotate_left(node* &T) {    //printf("rotate_left %d\n", T->val);    node* tp = T->right;    T->right = tp->left;    T->rank = rank(T->left) + rank(T->right) + 1;    tp->left = T;    tp->rank = rank(tp->left) + rank(tp->right) + 1;    T = tp;}void rotate_right(node* &T) {    //printf("rotate_right %d\n", T->val);    node* tp = T->left;    T->left = tp->right;    T->rank = rank(T->left) + rank(T->right) + 1;    tp->right = T;    tp->rank = rank(tp->left) + rank(tp->right) + 1;    T = tp;}void insert(node* &T, type& val) {    if (T == NULL) {        T = new node(val);        T->rank = 1;    } else if (val == T->val) {        return;    } else if (val < T->val) {        insert(T->left, val);        ++T->rank;        if (T->left->fix < T->fix) {            rotate_right(T);        }    } else {        insert(T->right, val);        ++T->rank;        if (T->right->fix < T->fix) {            rotate_left(T);        }    }}void erase(node* &T, type& val) {    if (val == T->val) {        if (T->left == NULL || T->right == NULL) {            node* t = T;            if (T->left == NULL) {                T = T->right;            } else {                T = T->left;            }            T->rank = rank(T->left) + rank(T->right) + 1;            delete t;        } else {            if (T->left->fix < T->right->fix) {                rotate_right(T);                erase(T->right, val);            } else {                rotate_left(T);                erase(T->left, val);            }        }    } else if (val < T->val) {        erase(T->left, val);        --(T->rank);    } else {        erase(T->right, val);        --(T->rank);    }}bool exist(node* &T, type val) {    if (T == NULL) {        return false;    } else if (T->val == val) {        return true;    } else if (val < T->val) {        return exist(T->left, val);    } else {        return exist(T->right, val);    }}void clear(node* &T) {    if (T == NULL) return;    clear(T->left);    clear(T->right);    delete T;    T = NULL;}const type INF = 0xfffffff;type find_k(node* &T, int k) {    //printf("find_k(%d)\n", k);    if (k <= 0 || k > rank(T)) {        return INF;    } else if (k == rank(T->left) + 1) {        return T->val;    } else if (k <= rank(T->left)) {        return find_k(T->left, k);    } else {        return find_k(T->right, k - 1 - rank(T->left));    }}void visit(node* &T) {    if (T == NULL) return;    visit(T->left);    printf("(%d,%d) ", T->val, T->rank);    visit(T->right);}const int MAX = 32;type p2[MAX], p3[MAX], p5[MAX];int main() {    p2[0] = p3[0] = p5[0] = 1;    for (int i = 1; i < MAX; ++i) {        p2[i] = p2[i - 1] << 1;        p3[i] = p3[i - 1] * 3;        p5[i] = p5[i - 1] * 5;    }    node* T = NULL;    for (int i = 0; i < MAX; ++i) {        if (p2[i] <= 0) break;        for (int j = 0; j < MAX; ++j) {            if (p3[j] <= 0) break;            for (int k = 0; k < MAX; ++k) {                long long t = p2[i] * p3[j] * p5[k];                if (t <= 0) break;                //printf("insert %d\n", t);                insert(T, t);            }        }    }    //visit(T);    //puts("");    int n;    while (~scanf(" %d", &n)) {        if (n == 0) break;        printf("%lld\n", find_k(T, n));    }    clear(T);    return 0;}

poj1552

这题只需要用到查找操作，遍历维护的treap，对任意节点x只需要检查是否存在元素2×x即可。代码：

#include <cstdio>#include <cstring>#include <cstdlib>#include <ctime>#include <algorithm>using namespace std;typedef int type;struct node {    type val;    int fix;    node* left;    node* right;    node() {        left = right = NULL;    }    node(type x) {        val = x;        fix = rand();        left = right = NULL;    }};void rotate_left(node* &T) {    node* tp = T->right;    T->right = tp->left;    tp->left = T;    T = tp;}void rotate_right(node* &T) {    node* tp = T->left;    T->left = tp->right;    tp->right = T;    T = tp;}void insert(node* &T, type& val) {    if (T == NULL) {        T = new node(val);    } else if (val <= T->val) {        insert(T->left, val);        if (T->left->fix < T->fix) {            rotate_right(T);        }    } else {        insert(T->right, val);        if (T->right->fix < T->fix) {            rotate_left(T);        }    }}void erase(node* &T, type& val) {    if (val == T->val) {        if (T->left == NULL || T->right == NULL) {            node* t = T;            if (T->left == NULL) {                T = T->right;            } else {                T = T->left;            }            delete t;        } else {            if (T->left->fix < T->right->fix) {                rotate_right(T);                erase(T->right, val);            } else {                rotate_left(T);                erase(T->left, val);            }        }    } else if (val < T->val) {        erase(T->left, val);    } else {        erase(T->right, val);    }}bool exist(node* &T, type val) {    if (T == NULL) {        return false;    } else if (T->val == val) {        return true;    } else if (val < T->val) {        return exist(T->left, val);    } else {        return exist(T->right, val);    }}void clear(node* &T) {    if (T == NULL) return;    clear(T->left);    clear(T->right);    delete T;    T = NULL;}int query(node* &root, node* &T) {    if (T == NULL) return 0;    return exist(root, (T->val) * 2)        + query(root, T->left) + query(root, T->right);}void visit(node* &T) {    if (T == NULL) return;    visit(T->left);    printf("%d ", T->val);    visit(T->right);}int main() {    node* T = NULL;    int n;    while (~scanf(" %d", &n) && n != - 1) {        while (n != 0) {            insert(T, n);            //visit(T);            //puts("");            scanf(" %d", &n);        }        printf("%d\n", query(T, T));        clear(T);    }    return 0;}

HDU5058 So easy

题意类似判断两个集合是否相等。即是否一个数组中的元素在另一个元素中也一定存在。去重后比较即可。但是，同样地，用treap来练习…
代码：

#include <cstdio>#include <cstring>#include <cstdlib>#include <ctime>#include <algorithm>using namespace std;typedef int type;struct node {    type val;    int fix;    node* left;    node* right;    node() {        left = right = NULL;    }    node(type x) {        val = x;        fix = rand();        left = right = NULL;    }};void rotate_left(node* &T) {    node* tp = T->right;    T->right = tp->left;    tp->left = T;    T = tp;}void rotate_right(node* &T) {    node* tp = T->left;    T->left = tp->right;    tp->right = T;    T = tp;}void insert(node* &T, type& val) {    if (T == NULL) {        T = new node(val);    } else if (val == T->val) {        return;    } else if (val < T->val) {        insert(T->left, val);        if (T->left->fix < T->fix) {            rotate_right(T);        }    } else {        insert(T->right, val);        if (T->right->fix < T->fix) {            rotate_left(T);        }    }}void erase(node* &T, type& val) {    if (val == T->val) {        if (T->left == NULL || T->right == NULL) {            node* t = T;            if (T->left == NULL) {                T = T->right;            } else {                T = T->left;            }            delete t;        } else {            if (T->left->fix < T->right->fix) {                rotate_right(T);                erase(T->right, val);            } else {                rotate_left(T);                erase(T->left, val);            }        }    } else if (val < T->val) {        erase(T->left, val);    } else {        erase(T->right, val);    }}bool exist(node* &T, type val) {    if (T == NULL) {        return false;    } else if (T->val == val) {        return true;    } else if (val < T->val) {        return exist(T->left, val);    } else {        return exist(T->right, val);    }}void clear(node* &T) {    if (T == NULL) return;    clear(T->left);    clear(T->right);    delete T;    T = NULL;}bool query(node* &root, node* &T) {    if (T == NULL) return true;    return exist(root, T->val)        && query(root, T->left) && query(root, T->right);}int main() {    int n;    type x;    node* T1 = NULL, *T2 = NULL;    while (~scanf(" %d", &n)) {        for (int i = 0; i < n; ++i) {            scanf(" %d", &x);            insert(T1, x);        }        for (int i = 0; i < n; ++i) {            scanf(" %d", &x);            insert(T2, x);        }        puts(query(T1, T2) && query(T2, T1) ? "YES" : "NO");        clear(T1);        clear(T2);    }    return 0;}