SGU 311. Ice-cream Tycoon（平衡树）

题目链接：点击打开链接

思路：

最简单的思路是， 维护一棵平衡树， 以单价作为键值， 维护一个结点个数的信息。  然后类似寻找第K小的方法找到第n个小的价值处， 顺便维护总价值， 判断是否happy。

然后就是删除操作。 我们从新从根结点开始， 沿路删除可以删除的子树就行了。 由于每次删除都是删除的叶子结点处， 所以不需要旋转操作。

细节参见代码：

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <string>#include <vector>#include <stack>#include <bitset>#include <cstdlib>#include <cmath>#include <set>#include <list>#include <deque>#include <map>#include <queue>#define Max(a,b) ((a)>(b)?(a):(b))#define Min(a,b) ((a)<(b)?(a):(b))using namespace std;typedef long long ll;typedef long double ld;const ld eps = 1e-9, PI = 3.1415926535897932384626433832795;const int mod = 1000000000 + 7;const int INF = 0x3f3f3f3f;const int seed = 131;const ll INF64 = ll(1e18);ll n, c;struct node {    node *ch[2];    int r;      // 左右子树    ll s, tt, t, ss;   //ss是该结点的物品数，s是结点总数，tt是单价，t是总价， r是优先级    node(ll tt=0, ll ss=0): tt(tt), ss(ss) {        ch[0] = ch[1] = NULL;        s = ss;        t = tt * ss;        r = rand();    }    ll cmp(ll x) const {        if(x == tt) return -1;        return x < tt ? 0 : 1;    }    void maintain() {        s = ss;        t = tt * ss;        if(ch[0] != NULL) s += ch[0]->s, t += ch[0]->t; //维护这个结点下的所有节点数， 维护一个结点下的总价钱        if(ch[1] != NULL) s += ch[1]->s, t += ch[1]->t;    }} *g;void rotate(node* &o, ll d) {    node* k = o->ch[d^1];  //旋转， 使得优先级满足堆的意义    o->ch[d^1] = k->ch[d];    k->ch[d] = o;    o->maintain();    k->maintain();    o = k;}void insert(node* &o, ll x, ll s) {    if(o == NULL) o = new node(x, s);    else {        int d = (x < o->tt ? 0 : 1);        insert(o->ch[d], x, s);        if(o->ch[d]->r > o->r) rotate(o, d^1);    }    o->maintain();}void removetree(node* &x) {    if(x->ch[0] != NULL) removetree(x->ch[0]);    if(x->ch[1] != NULL) removetree(x->ch[1]);    delete x;    x = NULL;}void _remove(node* &gg, node* &o, ll k, ll sum) {    if(o == NULL || k <= 0 || k > o->s) return ;    ll s = (o->ch[0] != NULL ? o->ch[0]->s : 0); //s是左子树的元素个数    ll hehe = (o->ch[0] != NULL ? o->ch[0]->t : 0);    if(k <= s) {        _remove(o, o->ch[0], k, sum);        o->maintain();    }    else if(k <= s + (o->ss)) {        ll tot = sum; //这k件最便宜的物品的总价        if(o->ch[0] != NULL) tot += o->ch[0]->t;        tot += o->tt * (k - s);            o->s -= k;            o->ss -= (k - s);            o->t -= tot - sum;            if(o->ch[0] != NULL) removetree(o->ch[0]);    }    else {        ll hehe2 = (o->ss * o->tt);        ll hehe3 = o->ss;        node* u = o;        if(o->ch[0] != NULL) removetree(o->ch[0]);        o = o->ch[1];        delete u;        _remove(gg, o, k-s-hehe3, sum + hehe + hehe2);        o->maintain();    }    o->maintain();}bool solve(node* o, ll k, ll sum) {    if(o == NULL || k <= 0 || k > o->s) return false;    ll s = (o->ch[0] != NULL ? o->ch[0]->s : 0); //s是左子树的元素个数    ll hehe = (o->ch[0] != NULL ? o->ch[0]->t : 0);    if(k <= s) {        int ok = solve(o->ch[0], k, sum);        o->maintain();        return ok;    }    else if(k <= s + (o->ss)) {        ll tot = sum; //这k件最便宜的物品的总价        if(o->ch[0] != NULL) tot += o->ch[0]->t;        tot += o->tt * (k - s);        if(tot > c) return false;        else return true;    }    else {        int ok = solve(o->ch[1], k-s-(o->ss), sum + hehe + (o->ss * o->tt));        o->maintain();        return ok;    }    o->maintain();}char s[100];int main() {    g = NULL;    while(~scanf("%s%I64d%I64d",s, &n, &c)) {        if(s[0] == 'A') {            insert(g, c, n);        }        else {            if(solve(g, n, 0)) _remove(g, g, n, 0), printf("HAPPY\n");            else printf("UNHAPPY\n");        }    }    return 0;}