POJ 3481 Double Queue (数据结构)

题目类型  数据结构

题目意思

给出一系列指令 其中

指令 1  插入一个优先级为 B 值为 A 的人

指令2  去掉优先级最高的人并输出这个人的值

指令3  去掉优先级最低的人并输出这个人的值

解题方法

很多方法都可以做 例如 优先队列 线段树 treap 伸展树等

参考代码 - 有疑问的地方在下方留言 看到会尽快回复的

treap

#include <iostream>#include <cstdio>#include <ctime>#include <cstdlib>using namespace std;const int maxn = 30000 + 10;int a[maxn], b[maxn];struct Node {Node * ch[2];int r;int v;int msg;Node(int v, int msg) : v(v), msg(msg) { ch[0] = ch[1] = NULL; r = rand(); }int cmp(int x) const {if(x == v) return -1;return x < v ? 0 : 1;}};struct Treap {Node * rt;Treap() { rt = NULL; }void rotate(Node* & o, int d) {Node * k = o->ch[d^1]; o->ch[d^1] = k->ch[d]; k->ch[d] = o; o = k;}void insert(Node* & o, int x, int msg) {if(o == NULL) o = new Node(x, msg);else {int d = (x > o->v ? 1 : 0); // 相等放左儿子那边insert(o->ch[d], x, msg);if(o->ch[d]->r > o->r) rotate(o, d ^ 1);}}void remove(Node* & o, int x) {int d = o->cmp(x);if(d == -1) {Node * u = o;if(o->ch[0] != NULL && o->ch[1] != NULL) {int d2 = (o->ch[0] > o->ch[1] ? 1 : 0);rotate(o, d2); remove(o->ch[d2], x);}else {if(o->ch[0] == NULL) o = o->ch[1]; else o = o->ch[0];delete u;}}else remove(o->ch[d], x);}void removetree(Node* & x) {if(x == NULL) return ;if(x->ch[0] != NULL) removetree(x->ch[0]);if(x->ch[1] != NULL) removetree(x->ch[1]);delete x;x = NULL;}};int main() {freopen("in", "r", stdin);  srand(time(0));int n;Treap trp;while(scanf("%d", &n), n) {if(n == 1) {int msg, x;scanf("%d%d", &msg, &x);trp.insert(trp.rt, x, msg);}else if(n == 2) {if(trp.rt == NULL) printf("0\n");else {Node * t = trp.rt;while(t->ch[1] != NULL) t = t->ch[1];printf("%d\n", t->msg);trp.remove(trp.rt, t->v);}}else {if(trp.rt == NULL) printf("0\n");else {Node * t = trp.rt;while(t->ch[0] != NULL) t = t->ch[0];printf("%d\n", t->msg);trp.remove(trp.rt, t->v);}}}trp.removetree(trp.rt);return 0;}

splay

#include <iostream>#include <cstdio>#include <ctime>#include <cstdlib>using namespace std;struct Node {Node * ch[2];int v;int msg;Node(int v, int msg) : v(v), msg(msg) { ch[0] = ch[1] = NULL; }int cmp(int x) const {if(x == v) return -1;return x < v ? 0 : 1;}};struct Splay {Node * rt;Splay() { rt = NULL; }void rotate(Node* & o, int d) {Node * k = o->ch[d^1]; o->ch[d^1] = k->ch[d]; k->ch[d] = o; o = k;}void splay(Node* & o, int x) {int d = o->cmp(x);if(d != -1) {Node * p = o->ch[d];int d2 = p->cmp(x);if(d2 != -1) {splay(p->ch[d2], x);if(d == d2) rotate(o, d^1); else rotate(o->ch[d], d);}rotate(o, d^1);}}void insert(Node* & o, int x, int msg) {if(o == NULL) {o = new Node(x, msg);splay(rt, x);}else {int d = (x < o->v ? 0 : 1); // avoid repetitioninsert(o->ch[d], x, msg);}}int findMax(Node* & o) {Node * t = o;while(t->ch[1] != NULL) {t = t->ch[1];}splay(o, t->v);return o->msg;}int findMin(Node* & o) {Node * t = o;while(t->ch[0] != NULL) {t = t->ch[0];}splay(o, t->v);return o->msg;}void remove(int x) {splay(rt, x);Node * t = rt;if(rt->ch[0] == NULL) {rt = rt->ch[1];delete t;}else {findMax(rt->ch[0]);rt->ch[0]->ch[1] = rt->ch[1];rt = rt->ch[0];delete t;}}};int main() {//freopen("in", "r", stdin);Splay spy;int n;while(scanf("%d", &n), n) {if(n == 1) {int x, msg;scanf("%d%d", &msg, &x);spy.insert(spy.rt, x, msg);}else if(n == 2) {if(spy.rt == NULL) printf("0\n");else {printf("%d\n", spy.findMax(spy.rt));spy.remove(spy.rt->v);}}else {if(spy.rt == NULL) printf("0\n");else {printf("%d\n", spy.findMin(spy.rt));spy.remove(spy.rt->v);}}}return 0;}