POJ 2376 Cleaning Shifts （贪心）

Cleaning Shifts

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 12079 Accepted: 3145

Description

Farmer John is assigning some of his N (1 <= N <= 25,000) cows to do some cleaning chores around the barn. He always wants to have one cow working on cleaning things up and has divided the day into T shifts (1 <= T <= 1,000,000), the first being shift 1 and the last being shift T. 

Each cow is only available at some interval of times during the day for work on cleaning. Any cow that is selected for cleaning duty will work for the entirety of her interval. 

Your job is to help Farmer John assign some cows to shifts so that (i) every shift has at least one cow assigned to it, and (ii) as few cows as possible are involved in cleaning. If it is not possible to assign a cow to each shift, print -1.

Input

* Line 1: Two space-separated integers: N and T 

* Lines 2..N+1: Each line contains the start and end times of the interval during which a cow can work. A cow starts work at the start time and finishes after the end time.

Output

* Line 1: The minimum number of cows Farmer John needs to hire or -1 if it is not possible to assign a cow to each shift.

Sample Input

3 101 73 66 10

Sample Output

2

Hint

This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed. 

INPUT DETAILS: 

There are 3 cows and 10 shifts. Cow #1 can work shifts 1..7, cow #2 can work shifts 3..6, and cow #3 can work shifts 6..10. 

OUTPUT DETAILS: 

By selecting cows #1 and #3, all shifts are covered. There is no way to cover all the shifts using fewer than 2 cows.

题目所求最小的cow数 覆盖1-N的区间  那么贪心思路就很明显了 按照区间起始点排序

不断的进行 选择一个后一区间与前一区间相交 且结束点最大的cow

本题需要注意的是 cow工作的是闭区间 且完全覆盖1-N区间

AC代码如下：

////  POJ 2376 Cleaning Shifts////  Created by TaoSama on 2015-02-22//  Copyright (c) 2015 TaoSama. All rights reserved.//#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>#define CLR(x,y) memset(x, y, sizeof(x))using namespace std;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;const int N = 1e5 + 10;int n, t;struct ACM {int st, ed;bool operator<(const ACM& r) const {if(st != r.st) return st < r.st;return ed > r.ed;}} a[25005];int main() {#ifdef LOCALfreopen("in.txt", "r", stdin);//freopen("out.txt","w",stdout);#endif//ios_base::sync_with_stdio(0);while(cin >> n >> t) {for(int i = 1; i <= n; ++i)cin >> a[i].st >> a[i].ed;sort(a + 1, a + 1 + n);int l = 0, r = 0, ok, idx = 1;int ans = 0;while(r < t && idx <= n) {if(a[idx].st > l + 1) break;while(a[idx].st <= l + 1 && idx <= n) {if(a[idx].ed > r) {r = a[idx].ed;ok = idx;}++idx;}l = a[ok].ed, ++ans;}if(r < t) ans = -1;cout << ans << endl;}return 0;}