SGU - 311 Ice-cream Tycoon(线段树)
2014-10-20 00:05
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Description
You've recently started an ice-cream business in a local school. During a day you have many suppliers delivering the ice-cream for you, and many 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
, meaning that a supplier has delivered n pieces of ice-cream priced c each to you, or
, meaning that a student wants to buy n pieces of ice-cream, having a total of t money. The latter is processed as follows: in case n cheapest pieces of ice-cream you have cost no more than t (together), you sell those n cheapest
pieces to the student; in case they cost more, she gets nothing. You start the day with no ice-cream.
For each student, output
if she gets her ice-cream, and
if she doesn't.
Input
The input file contains between 1 and 10 5 queries (inclusive), each on a separate line. The queries are formatted as described above, either
or
, 1 ≤ n, c ≤ 10 6, 1 ≤ t ≤ 10 12.
Output
For each
-query output one line, containing either the word
or the word
(answers should be in the same order as the corresponding queries).
Sample Input
题意:一个商店,有两种操作:(1)ARRIVE n c表示进货n个,每个c元。(2)BUY n t表示一个买货的人要买n个,一共拿了t元钱。如果现在店里的货的数量大于等于n且最便宜的n个的价格小于等于t则将最便宜的卖给他。否则不卖。
思路:离线的线段树,我们以价格作为结点,然后离散化,好久没做,看了final爷kuangbing的题解,注意的是优先处理最小的n个
You've recently started an ice-cream business in a local school. During a day you have many suppliers delivering the ice-cream for you, and many 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 n pieces of ice-cream priced c each to you, or
BUY nt
, meaning that a student wants to buy n pieces of ice-cream, having a total of t money. The latter is processed as follows: in case n cheapest pieces of ice-cream you have cost no more than t (together), you sell those n cheapest
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 (inclusive), 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 should be in the same order as the corresponding queries).
Sample Input
sample input | sample output |
ARRIVE 1 1 ARRIVE 10 200 BUY 5 900 BUY 5 900 BUY 5 1000 | HAPPY UNHAPPY HAPPY |
思路:离线的线段树,我们以价格作为结点,然后离散化,好久没做,看了final爷kuangbing的题解,注意的是优先处理最小的n个
#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 .op, &q .n, &q .c) != EOF) { if (q .op[0] == 'A') x[tot++] = q .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; }
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