标签：des style color io os ar for sp div
Description
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.
HAPPYif she gets her ice-cream, and
UNHAPPYif she doesn‘t.
Input
ARRIVE ncor
BUY nt, 1 ≤ n, c ≤ 10 6, 1 ≤ t ≤ 10 12.
Output
BUY-query output one line, containing either the word
HAPPYor 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 |
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
#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(线段树)
标签：des style color io os ar for sp div
原文地址：http://blog.csdn.net/u011345136/article/details/40278017
