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SGU 311. Ice-cream Tycoon（线段树）

2014-05-02 12:44 344 查看

311. Ice-cream Tycoon

[align=center]Time limit per test: 0.5 second(s)
Memory limit: 65536 kilobytes[/align]
[align=center]input: standard
output: standard[/align]

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 n c

, meaning that a supplier has delivered n pieces of ice-cream priced c each to you, or

BUY n t

, 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.

[align=left]Input[/align]

The input file contains between 1 and 105 queries (inclusive), each on a separate line. The queries are formatted as described above, either

ARRIVE n c

BUY n t

, 1 ≤ n, c ≤ 106, 1 ≤ t ≤ 1012.

[align=left]Output[/align]

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).

[align=left]Example(s)[/align]

sample input	sample output
ARRIVE 1 1 ARRIVE 10 200 BUY 5 900 BUY 5 900 BUY 5 1000	HAPPY UNHAPPY HAPPY

基础线段树。

线段树维护每个点有的个数和总和。

需要给某个点加上一个个数，清空一个区间的值，查询一个区间的总价值和。

需要离线下，离散化处理、

/* ***********************************************
Author        :kuangbin
Created Time  :2014/5/2 11:48:25
File Name     :E:\2014ACM\专题学习\数据结构\线段树\SGU311.cpp
************************************************ */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
const int MAXN = 100010;
struct Node
{
int l,r;
long long num;
long long sum;
int c;//清空标记
}segTree[MAXN*3];
int x[MAXN];
void push_down(int i)
{
if(segTree[i].l == segTree[i].r)return;
if(segTree[i].c != -1)
{
segTree[i<<1].sum = segTree[(i<<1)|1].sum = 0;
segTree[i<<1].num = segTree[(i<<1)|1].num = 0;
segTree[i<<1].c = segTree[(i<<1)|1].c = 0;
segTree[i].c = -1;
}
}
void push_up(int i)
{
if(segTree[i].l == segTree[i].r)return;
segTree[i].sum = segTree[i<<1].sum + segTree[(i<<1)|1].sum;
segTree[i].num = segTree[i<<1].num + segTree[(i<<1)|1].num;
}
void build(int i,int l,int r)
{
segTree[i].l = l;
segTree[i].r = r;
segTree[i].sum = segTree[i].num = 0;
segTree[i].c = -1;
if(l == r)return;
int mid = (l+r)/2;
build(i<<1,l,mid);
build((i<<1)|1,mid+1,r);
}
void Add(int i,int c,int n)
{
segTree[i].sum += (long long)c*n;
segTree[i].num += n;
if(x[segTree[i].l] == c && x[segTree[i].r] == c)
return;
push_down(i);
if(c <= x[segTree[i<<1].r])Add(i<<1,c,n);
else Add((i<<1)|1,c,n);
}
//查询买前n个需要的钱
long long query(int i,int n)
{
if(segTree[i].l ==segTree[i].r)
{
return (long long)n*x[segTree[i].l];
}
push_down(i);
if(segTree[i<<1].num >= n)return query(i<<1,n);
else return segTree[i<<1].sum + query((i<<1)|1,n-segTree[i<<1].num);
}
//清除前n个
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;
}
push_down(i);
if(segTree[i<<1].num >= n)clear(i<<1,n);
else
{
clear((i<<1)|1,n-segTree[i<<1].num);
segTree[i<<1].num = segTree[i<<1].sum = 0;
segTree[i<<1].c = 0;
}
push_up(i);
}
struct Query
{
char op[10];
int n;
long long c;
}q[MAXN];

int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int n = 0;
int tot = 0;
while(scanf("%s%d%I64d",&q
.op,&q
.n,&q
.c) == 3)
{
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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