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  • 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、

      1 /* ***********************************************
  2 Author        :kuangbin
  3 Created Time  :2014/5/2 11:48:25
  4 File Name     :E:2014ACM专题学习数据结构线段树SGU311.cpp
  5 ************************************************ */
  6 
  7 #include <stdio.h>
  8 #include <string.h>
  9 #include <iostream>
 10 #include <algorithm>
 11 #include <vector>
 12 #include <queue>
 13 #include <set>
 14 #include <map>
 15 #include <string>
 16 #include <math.h>
 17 #include <stdlib.h>
 18 #include <time.h>
 19 using namespace std;
 20 const int MAXN = 100010;
 21 struct Node
 22 {
 23     int l,r;
 24     long long num;
 25     long long sum;
 26     int c;//清空标记
 27 }segTree[MAXN*3];
 28 int x[MAXN];
 29 void push_down(int i)
 30 {
 31     if(segTree[i].l == segTree[i].r)return;
 32     if(segTree[i].c != -1)
 33     {
 34         segTree[i<<1].sum = segTree[(i<<1)|1].sum = 0;
 35         segTree[i<<1].num = segTree[(i<<1)|1].num = 0;
 36         segTree[i<<1].c = segTree[(i<<1)|1].c = 0;
 37         segTree[i].c = -1;
 38     }
 39 }
 40 void push_up(int i)
 41 {
 42     if(segTree[i].l == segTree[i].r)return;
 43     segTree[i].sum = segTree[i<<1].sum + segTree[(i<<1)|1].sum;
 44     segTree[i].num = segTree[i<<1].num + segTree[(i<<1)|1].num;
 45 }
 46 void build(int i,int l,int r)
 47 {
 48     segTree[i].l = l;
 49     segTree[i].r = r;
 50     segTree[i].sum = segTree[i].num = 0;
 51     segTree[i].c = -1;
 52     if(l == r)return;
 53     int mid = (l+r)/2;
 54     build(i<<1,l,mid);
 55     build((i<<1)|1,mid+1,r);
 56 }
 57 void Add(int i,int c,int n)
 58 {
 59     segTree[i].sum += (long long)c*n;
 60     segTree[i].num += n;
 61     if(x[segTree[i].l] == c && x[segTree[i].r] == c)
 62         return;
 63     push_down(i);
 64     if(c <= x[segTree[i<<1].r])Add(i<<1,c,n);
 65     else Add((i<<1)|1,c,n);
 66 }
 67 //查询买前n个需要的钱
 68 long long query(int i,int n)
 69 {
 70     if(segTree[i].l ==segTree[i].r)
 71     {
 72         return (long long)n*x[segTree[i].l];
 73     }
 74     push_down(i);
 75     if(segTree[i<<1].num >= n)return query(i<<1,n);
 76     else return segTree[i<<1].sum + query((i<<1)|1,n-segTree[i<<1].num);
 77 }
 78 //清除前n个
 79 void clear(int i,int n)
 80 {
 81     if(segTree[i].l == segTree[i].r)
 82     {
 83         segTree[i].num -= n;
 84         segTree[i].sum = segTree[i].num * x[segTree[i].l];
 85         return;
 86     }
 87     push_down(i);
 88     if(segTree[i<<1].num >= n)clear(i<<1,n);
 89     else
 90     {
 91         clear((i<<1)|1,n-segTree[i<<1].num);
 92         segTree[i<<1].num = segTree[i<<1].sum = 0;
 93         segTree[i<<1].c = 0;
 94     }
 95     push_up(i);
 96 }
 97 struct Query
 98 {
 99     char op[10];
100     int n;
101     long long c;
102 }q[MAXN];
103 
104 int main()
105 {
106     //freopen("in.txt","r",stdin);
107     //freopen("out.txt","w",stdout);
108     int n = 0;
109     int tot = 0;
110     while(scanf("%s%d%I64d",&q[n].op,&q[n].n,&q[n].c) == 3)
111     {
112         if(q[n].op[0] == 'A')
113             x[tot++] = q[n].c;
114         n++;
115     }
116     sort(x,x+tot);
117     tot = unique(x,x+tot) - x;
118     build(1,0,tot-1);
119     for(int i = 0;i < n;i++)
120     {
121         if(q[i].op[0] == 'A')
122             Add(1,q[i].c,q[i].n);
123         else
124         {
125             if(segTree[1].num < q[i].n)printf("UNHAPPY
");
126             else
127             {
128                 if(query(1,q[i].n) > q[i].c)printf("UNHAPPY
");
129                 else
130                 {
131                     printf("HAPPY
");
132                     clear(1,q[i].n);
133                 }
134             }
135         }
136     }
137     return 0;
138 }
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  • 原文地址：https://www.cnblogs.com/kuangbin/p/3703956.html
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