SGU 311. Ice-cream Tycoon（线段树）

311. Ice-cream Tycoon

Time limit per test: 0.5 second(s)
Memory limit: 65536 kilobytes

input: standard
output: standard

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.

Input

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

or

BUY n t

, 1 ≤ n, c ≤ 10⁶, 1 ≤ t ≤ 10¹².

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).

Example(s)

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 = ;
 struct Node
 {
     int l,r;
     long long num;
     long long sum;
     int c;//清空标记
 }segTree[MAXN*];
 int x[MAXN];
 void push_down(int i)
 {
     if(segTree[i].l == segTree[i].r)return;
     if(segTree[i].c != -)
     {
         segTree[i<<].sum = segTree[(i<<)|].sum = ;
         segTree[i<<].num = segTree[(i<<)|].num = ;
         segTree[i<<].c = segTree[(i<<)|].c = ;
         segTree[i].c = -;
     }
 }
 void push_up(int i)
 {
     if(segTree[i].l == segTree[i].r)return;
     segTree[i].sum = segTree[i<<].sum + segTree[(i<<)|].sum;
     segTree[i].num = segTree[i<<].num + segTree[(i<<)|].num;
 }
 void build(int i,int l,int r)
 {
     segTree[i].l = l;
     segTree[i].r = r;
     segTree[i].sum = segTree[i].num = ;
     segTree[i].c = -;
     if(l == r)return;
     int mid = (l+r)/;
     build(i<<,l,mid);
     build((i<<)|,mid+,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<<].r])Add(i<<,c,n);
     else Add((i<<)|,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<<].num >= n)return query(i<<,n);
     else return segTree[i<<].sum + query((i<<)|,n-segTree[i<<].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<<].num >= n)clear(i<<,n);
     else
     {
         clear((i<<)|,n-segTree[i<<].num);
         segTree[i<<].num = segTree[i<<].sum = ;
         segTree[i<<].c = ;
     }
     push_up(i);
 }
 struct Query
 {
     char op[];
     int n;
     long long c;
 }q[MAXN];

 int main()
 {
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     int n = ;
     int tot = ;
     while(scanf("%s%d%I64d",&q[n].op,&q[n].n,&q[n].c) == )
     {
         if(q[n].op[] == 'A')
             x[tot++] = q[n].c;
         n++;
     }
     sort(x,x+tot);
     tot = unique(x,x+tot) - x;
     build(,,tot-);
     for(int i = ;i < n;i++)
     {
         if(q[i].op[] == 'A')
             Add(,q[i].c,q[i].n);
         else
         {
             if(segTree[].num < q[i].n)printf("UNHAPPY\n");
             else
             {
                 if(query(,q[i].n) > q[i].c)printf("UNHAPPY\n");
                 else
                 {
                     printf("HAPPY\n");
                     clear(,q[i].n);
                 }
             }
         }
     }
     return ;
 }