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：

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<string>
#define inf 1000000000
#define maxn 1000
#define maxm 500+100
#define eps 1e-10
#define ll long long
#define pa pair<int,int>
#define for0(i,n) for(int i=0;i<=(n);i++)
#define for1(i,n) for(int i=1;i<=(n);i++)
#define for2(i,x,y) for(int i=(x);i<=(y);i++)
#define for3(i,x,y) for(int i=(x);i>=(y);i--)
#define mod 1000000007
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();}
    return x*f;
}
int now[2],n,ans;
struct rec{int x,y;}a[maxn];
inline bool cmp(rec a,rec b)
{
    return a.y<b.y;
}
int main()
{
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    n=read();
    for1(i,n)a[i].x=read(),a[i].y=read();
    sort(a+1,a+n+1,cmp);
    now[0]=now[1]=0;
    for1(i,n)
     {
         if(a[i].x>=now[1])ans++,now[1]=a[i].y;
         else if(a[i].x>=now[0])ans++,now[0]=a[i].y;
         if(now[0]>now[1])swap(now[0],now[1]);
     }
    printf("%d\n",ans);
    return 0;
}