Sgu-311 ice-cream tycoon (line segment tree)

Description

You 've recently started an ice-cream business in a local school. during a day you have your suppliers delivering the ice-cream for you, and your students buying it from you. you are not allowed to set the prices, as you are told the price for each piece of ice-cream by the suppliers.

The day is described with a sequence of queries. Each query can be either

ARRIVE nc

, Meaning that a supplier has delivered NPieces of ice-cream priced CEach to you, or

BUY nt

, Meaning that a student wants to buy NPieces of ice-cream, having a total TMoney. The latter is processed as follows: In case NCheapest pieces of ice-cream you have cost no more T(Together), you must those NCheapest pieces to the student; in case they cost more, she gets nothing. You start the day with no ice-cream.

For each student, output

HAPPY

If she gets her ice-cream, and

UNHAPPY

If she doesn' t.

Input

The input file contains between 1 and 10 5 queries (sparse SIVE), each on a separate line. The queries are formatted as described above, either

ARRIVE nc

Or

BUY nt

, 1 ≤ N, C≤ 10 6, 1 ≤ T≤ 10 12.

Output

For each

BUY

-Query output one line, containing either the word

HAPPY

Or the word

UNHAPPY

(Answers shocould be in the same order as the corresponding queries ).

Sample Input

sample input	sample output
ARRIVE 1 1ARRIVE 10 200BUY 5 900BUY 5 900BUY 5 1000	HAPPYUNHAPPYHAPPY

There are two operations for a store: (1) arrive n C indicates that N items are purchased, each of which is c RMB. (2) Buy n t indicates that a buyer needs to buy n items, and a total of T yuan is spent. If the number of goods in the store is greater than or equal to N and the price of the cheapest N is less than or equal to T, sell the cheapest to him. Otherwise, it will not be sold.

Idea: For offline line segment trees, we take the price as the node, and then discretize it. We haven't done it for a long time. After reading the final kuangbing question, note that we have given priority to the N smallest parts.

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <queue>#define lson(x) (x<<1)#define rson(x) ((x<<1)|1)typedef long long ll;using namespace std;const int maxn = 100010;struct Node {int l, r;ll num;ll sum;int flag;} segTree[maxn<<2];int x[maxn];void pushdown(int x) {if (segTree[x].l == segTree[x].r) return;if (segTree[x].flag != -1) {segTree[lson(x)].sum = segTree[rson(x)].sum = 0;segTree[lson(x)].num = segTree[rson(x)].num = 0;segTree[lson(x)].flag = segTree[rson(x)].flag = 0;segTree[x].flag = -1;}}void pushup(int x) {if (segTree[x].l == segTree[x].r) return;segTree[x].sum = segTree[lson(x)].sum + segTree[rson(x)].sum;segTree[x].num = segTree[lson(x)].num + segTree[rson(x)].num;}void build(int x, int l, int r) {segTree[x].r = r;segTree[x].l = l;segTree[x].sum = segTree[x].num = 0;segTree[x].flag = -1;if (l == r) return;int mid = l + r >> 1;build(lson(x), l, mid);build(rson(x), mid+1, r);}void Add(int i, int c, int n) {segTree[i].sum += (ll) c * n;segTree[i].num += n;if (x[segTree[i].l] == c && x[segTree[i].r] == c)return;pushdown(i);if (c <= x[segTree[lson(i)].r]) Add(lson(i), c, n);else Add(rson(i), c, n);}ll query(int i, int n) {if (segTree[i].l == segTree[i].r) {return (ll) n * x[segTree[i].l];}pushdown(i);if (segTree[lson(i)].num >= n) return query(lson(i), n);else return segTree[lson(i)].sum + query(rson(i), n-segTree[lson(i)].num);}void clear(int i, int n) {if (segTree[i].l == segTree[i].r) {segTree[i].num -= n;segTree[i].sum = segTree[i].num * x[segTree[i].l];return;}pushdown(i);if (segTree[lson(i)].num >= n)clear(lson(i), n);else {clear(rson(i), n-segTree[lson(i)].num);segTree[lson(i)].num = segTree[lson(i)].sum = 0;segTree[lson(i)].flag = 0;}pushup(i);}struct Query {char op[10];int n;ll c;} q[maxn];int main() {int n = 0;int tot = 0;while (scanf("%s%d%lld", q[n].op, &q[n].n, &q[n].c) != EOF) {if (q[n].op[0] == 'A')x[tot++] = q[n].c;n++;}sort(x, x+tot);tot = unique(x, x+tot) - x;build(1, 0, tot-1);for (int i = 0; i < n; i++) {if (q[i].op[0] == 'A')Add(1, q[i].c, q[i].n);else {if (segTree[1].num < q[i].n) printf("UNHAPPY\n");else {if (query(1, q[i].n) > q[i].c)printf("UNHAPPY\n");else {printf("HAPPY\n");clear(1, q[i].n);}}}}return 0;}

Sgu-311 ice-cream tycoon (line segment tree)