BZOJ3433: [Usaco2014 Jan]Recording the Moolympics

3433: [Usaco2014 Jan]Recording the Moolympics

Time Limit: 10 Sec Memory Limit: 128 MB
Submit: 55 Solved: 34
[Submit][Status]

Description

Being a fan of all cold-weather sports (especially those involving cows), Farmer John wants to record as much of the upcoming winter Moolympics as possible. The television schedule for the Moolympics consists of N different programs (1 <= N <= 150), each with a designated starting time and ending time. FJ has a dual-tuner recorder that can record two programs simultaneously. Please help him determine the maximum number of programs he can record in total.

给出n个区间[a,b).有2个记录器.每个记录器中存放的区间不能重叠.

求2个记录器中最多可放多少个区间.

Input

* Line 1: The integer N.

* Lines 2..1+N: Each line contains the start and end time of a single program (integers in the range 0..1,000,000,000).

Output

* Line 1: The maximum number of programs FJ can record.

Sample Input

6
0 3
6 7
3 10
1 5
2 8
1 9

INPUT DETAILS: The Moolympics broadcast consists of 6 programs. The first runs from time 0 to time 3, and so on.

Sample Output

4

OUTPUT DETAILS: FJ can record at most 4 programs. For example, he can record programs 1 and 3 back-to-back on the first tuner, and programs 2 and 4 on the second tuner.

HINT

Source

Silver 鸣谢Alegria_提供译文

题解：

随便打了个贪心居然就过了23333，是数据弱还是怎么？

将区间按右端点排序，维护now[0],now[1]分别表示两个记录器最后的位置

新来的线段先往now大的那个记录器放，放不下再到小的那个里，否则会出现大材小用。。。（提示：过不了样例。。。）

不知道对不对？？？挖坑，以后来想。

代码：

 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<iostream>
 7 #include<vector>
 8 #include<map>
 9 #include<set>
10 #include<queue>
11 #include<string>
12 #define inf 1000000000
13 #define maxn 1000
14 #define maxm 500+100
15 #define eps 1e-10
16 #define ll long long
17 #define pa pair<int,int>
18 #define for0(i,n) for(int i=0;i<=(n);i++)
19 #define for1(i,n) for(int i=1;i<=(n);i++)
20 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
21 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
22 #define mod 1000000007
23 using namespace std;
24 inline int read()
25 {
26     int x=0,f=1;char ch=getchar();
27     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
28     while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();}
29     return x*f;
30 }
31 int now[2],n,ans;
32 struct rec{int x,y;}a[maxn];
33 inline bool cmp(rec a,rec b)
34 {
35     return a.y<b.y;
36 }
37 int main()
38 {
39     freopen("input.txt","r",stdin);
40     freopen("output.txt","w",stdout);
41     n=read();
42     for1(i,n)a[i].x=read(),a[i].y=read();
43     sort(a+1,a+n+1,cmp);
44     now[0]=now[1]=0;
45     for1(i,n)
46      {
47          if(a[i].x>=now[1])ans++,now[1]=a[i].y;
48          else if(a[i].x>=now[0])ans++,now[0]=a[i].y;
49          if(now[0]>now[1])swap(now[0],now[1]);
50      }
51     printf("%d
",ans);
52     return 0;
53 }

View Code