SGU - 311 Ice-cream Tycoon(线段树)
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
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
ARRIVE nc
or
BUY nt
, 1 ≤ n, c ≤ 10 6, 1 ≤ t ≤ 10 12.
Output
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 |
HAPPY |
题意:一个商店,有两种操作:(1)ARRIVE n c表示进货n个,每一个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。假设如今店里的货的数量大于等于n且最廉价的n个的价格小于等于t则将最廉价的卖给他。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
#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;
}
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