BZOJ 2501: [usaco2010 Oct]Soda Machine 离散+差分

 [usaco2010 Oct]Soda Machine

Time Limit: 3 Sec  Memory Limit: 128 MB
Submit: 266  Solved: 182
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Description

To meet the ever-growing demands of his N (1 <= N <= 50,000) cows, 
Farmer John has bought them a new soda machine. He wants to figure 
out the perfect place to install the machine. 

The field in which the cows graze can be represented as a one-dimensional 
number line. Cow i grazes in the range A_i..B_i (1 <= A_i <= B_i; 
A_i <= B_i <= 1,000,000,000) (a range that includes its endpoints), 
and FJ can place the soda machine at any integer point in the range 
1..1,000,000,000. Since cows are extremely lazy and try to move 
as little as possible, each cow would like to have the soda machine 
installed within her grazing range. 

Sadly, it is not always possible to satisfy every cow's desires. 
Thus FJ would like to know the largest number of cows that can be 
satisfied. 

To demonstrate the issue, consider four cows with grazing ranges 
3..5, 4..8, 1..2, and 5..10; below is a schematic of their grazing 
ranges: 


1 2 3 4 5 6 7 8 9 10 11 12 13
|---|---|---|---|---|---|---|---|---|---|---|---|-...
aaaaaaaaa
bbbbbbbbbbbbbbbbb
ccccc ddddddddddddddddddddd

Sample Output


As can be seen, the first, second and fourth cows share the point 5, 
but the third cow's grazing range is disjoint. Thus, a maximum of 
3 cows can have the soda machine within their grazing range. 
有N个人要去膜拜JZ,他们不知道JZ会出现在哪里,因此每个人有一个活动范围,只要JZ出现在这个范围内就能被膜拜, 
伟大的JZ当然希望膜拜他的人越多越好,但是JZ不能分身,因此只能选择一个位置出现,他最多可以被多少人膜拜呢, 
这个简单的问题JZ当然交给你了 


3

OUTPUT DETAILS:

If the soda machine is placed at location 5, cows 1, 2, and 4 can be
satisfied. It is impossible to satisfy all four cows.

Input

* Line 1: A single integer: N 

* Lines 2..N+1: Line i+1 contains two space-separated integers: A_i 
and B_i 

Output

* Line 1: A single integer representing the largest number of cows 
whose grazing intervals can all contain the soda machine. 

Sample Input

4
3 5
4 8
1 2
5 10


Sample Output

3

OUTPUT DETAILS:

If the soda machine is placed at location 5, cows 1, 2, and 4 can be
satisfied. It is impossible to satisfy all four cows.


HINT

 

Source

 
题解:
  离散化后差分,l这里+1,r+1那里-1
  即可。
 1 #include<cstring>
 2 #include<cmath>
 3 #include<iostream>
 4 #include<algorithm>
 5 #include<cstdio>
 6 #include<cstdlib>
 7 
 8 #define N 50007
 9 #define ll long long
10 using namespace std;
11 inline int read()
12 {
13     int x=0,f=1;char ch=getchar();
14     while(ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
15     while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
16     return x*f;
17 }
18 
19 int n;
20 int l[N],r[N],sum[N*2];
21 int tot,a[N*2];
22 
23 int main()
24 {
25     n=read();
26     for (int i=1;i<=n;i++)
27     {
28         l[i]=read(),r[i]=read()+1;
29         a[++tot]=l[i],a[++tot]=r[i];
30     }
31     sort(a+1,a+tot+1);
32     for (int i=1;i<=n;i++)
33     {
34         l[i]=lower_bound(a+1,a+tot+1,l[i])-a;
35         r[i]=lower_bound(a+1,a+tot+1,r[i])-a;
36         sum[l[i]]++;
37         sum[r[i]]--;
38     }
39     int ans=0;
40     for (int i=1;i<=tot;i++)
41     {
42         sum[i]=sum[i]+sum[i-1];
43         ans=max(ans,sum[i]);
44     }
45     printf("%d
",ans);
46 }
原文地址:https://www.cnblogs.com/fengzhiyuan/p/8204056.html