【BZOJ】3433: [Usaco2014 Jan]Recording the Moolympics （贪心）

http://www.lydsy.com/JudgeOnline/problem.php?id=3433

想了好久啊。。。。。。。

想不出dp啊。。。。。。sad

后来看到一英文题解。。。。。。。。。。

sad。

末端点排序。。。然后对于两个录像机有有两种情况

RECODER1（当前录制节目的区间）:-----

RECODER1（当前录制节目的区间）:-----------

因右端点排序，所以下一个区间大概是:-------------------

可以看出，此时显然不能转移

当这个区间的左端点大于等于recoder1的右端点时并且小于recoder2的右端点时，显然可以转移，且一定是recoder1去录制，而非recoder2（这里是显然的）

当这个区间的左端点大于recoder2的右端点时，显然也可以转移，但是此时一定是recoder2去录制

所以我们维护当前两个recoder的右端点即可

#include <cstdio>

#include <cstring>

#include <cmath>

#include <string>

#include <iostream>

#include <algorithm>

#include <queue>

using namespace std;

#define rep(i, n) for(int i=0; i<(n); ++i)

#define for1(i,a,n) for(int i=(a);i<=(n);++i)

#define for2(i,a,n) for(int i=(a);i<(n);++i)

#define for3(i,a,n) for(int i=(a);i>=(n);--i)

#define for4(i,a,n) for(int i=(a);i>(n);--i)

#define CC(i,a) memset(i,a,sizeof(i))

#define read(a) a=getint()

#define print(a) printf("%d", a)

#define dbg(x) cout << #x << " = " << x << endl

#define printarr2(a, b, c) for1(i, 1, b) { for1(j, 1, c) cout << a[i][j]; cout << endl; }

#define printarr1(a, b) for1(i, 1, b) cout << a[i]; cout << endl

inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }

inline const int max(const int &a, const int &b) { return a>b?a:b; }

inline const int min(const int &a, const int &b) { return a<b?a:b; }



const int N=200;

int n, ans;

struct dat { int x, y; }a[N];

bool cmp(const dat &a, const dat &b) { return a.y<b.y; }



int main() {

	read(n);

	for1(i, 1, n) read(a[i].x), read(a[i].y);

	sort(a+1, a+1+n, cmp);

	int ed1=0, ed2=0;

	for1(i, 1, n) {

		int x=a[i].x;

		if(x<ed1) continue;

		if(x<ed2) ed1=ed2, ed2=a[i].y, ++ans;

		else ed2=a[i].y, ++ans;

	}

	print(ans);

	return 0;

}

 

 

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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