CodeForces - 816B Karen and Coffee (线段树的区间插入+单点查询)

In the first test case, Karen knows 3 recipes.

1. The first one recommends brewing the coffee between 91 and 94 degrees, inclusive.
2. The second one recommends brewing the coffee between 92 and 97 degrees, inclusive.
3. The third one recommends brewing the coffee between 97 and 99 degrees, inclusive.

A temperature is admissible if at least 2 recipes recommend it.

She asks 4 questions.

In her first question, she wants to know the number of admissible integer temperatures between 92 and 94 degrees, inclusive. There are 3: 92, 93 and 94 degrees are all admissible.

In her second question, she wants to know the number of admissible integer temperatures between 93 and 97 degrees, inclusive. There are 3: 93, 94 and 97 degrees are all admissible.

In her third question, she wants to know the number of admissible integer temperatures between 95 and 96 degrees, inclusive. There are none.

In her final question, she wants to know the number of admissible integer temperatures between 90 and 100 degrees, inclusive. There are 4: 92, 93, 94 and 97 degrees are all admissible.

In the second test case, Karen knows 2 recipes.

1. The first one, "wikiHow to make Cold Brew Coffee", recommends brewing the coffee at exactly 1 degree.
2. The second one, "What good is coffee that isn‘t brewed at at least 36.3306 times the temperature of the surface of the sun?", recommends brewing the coffee at exactly 200000 degrees.

A temperature is admissible if at least 1 recipe recommends it.

In her first and only question, she wants to know the number of admissible integer temperatures that are actually reasonable. There are none.

题目大意：有n本时尚杂志，每本杂志都给出了咖啡的最优的冲泡温度区间 l , r 。如果有k本杂志提到了这个温度，那么这个温度就是好的。问q个询问中有多少个温度是好的？

解题思路：把 l ~ maxx的区间值赋值为1，把 r ~ maxx的区间值赋值为-1。然后询问所有温度是否大于等于 k，求出这个些温度的前缀和。每个询问就是对应前缀和的差。

AC代码：

 1 #include <iostream>
 2 #include<bits/stdc++.h>
 3 //if(~i)//当i不是-1时满足条件
 4 using namespace std;
 5 const int maxn=1e5*2+5;
 6 int n,k,q;
 7 int lazy[maxn<<2],tmp[maxn];
 8 int build(int l,int r,int t)
 9 {
10     lazy[t]=0;
11     if(l==r)
12         return 0;
13     int mid=(l+r)/2;
14     build(l,mid,t*2);
15     build(mid+1,r,t*2+1);
16     return 0;
17 }
18 int push_up(int t)
19 {
20     if(lazy[t])
21     {
22         //printf("%d==========%d\n",l,r);
23         lazy[t*2]+=lazy[t];
24         lazy[t*2+1]+=lazy[t];
25         lazy[t]=0;
26     }
27 }
28 int ins(int L,int R,int l,int r,int t,int val)
29 {
30     if(L<=l&&r<=R)
31     {
32         lazy[t]+=val;
33         return 0;
34     }
35     push_up(t);
36     int mid=(l+r)/2;
37     if(L<=mid) ins(L,R,l,mid,t*2,val);
38     if(R>mid) ins(L,R,mid+1,r,t*2+1,val);
39     return 0;
40 }
41 int query(int d,int l,int r,int t)
42 {
43     if(l==r&&l==d)
44     {
45         return lazy[t];
46     }
47     push_up(t);
48     int mid=(l+r)/2;
49     if(d<=mid) return query(d,l,mid,t*2);
50     else if(mid<d) return query(d,mid+1,r,t*2+1);
51 }
52 int main()
53 {
54 
55     while(~scanf("%d%d%d",&n,&k,&q))
56     {
57         build(1,200005,1);
58         int l,r;
59         for(int i=0; i<n; i++)
60         {
61             scanf("%d%d",&l,&r);
62             ins(l,200002,1,200005,1,1);
63             ins(r+1,200002,1,200005,1,-1);
64         }
65         tmp[0]=0;
66         for(int i=1;i<=200002;i++)
67         {
68             if(query(i,1,200005,1)>=k) tmp[i]=tmp[i-1]+1;
69             else tmp[i]=tmp[i-1];
70         }
71         for(int i=0; i<q; i++)
72         {
73             scanf("%d%d",&l,&r);
74             printf("%d\n",tmp[r]-tmp[l-1]);
75         }
76     }
77     return 0;
78 }

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