标签：
POJ 2376 Cleaning Shifts（轮班打扫）
Time Limit: 1000MS Memory Limit: 65536K
|
【Description】 |
【题目描述】 |
|
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. |
农夫约翰派他的N(1 <= N <= 25,000)头牛去打扫畜棚。他希望总是有一头牛在打扫，并且把一天划分为T班倒(1 <= T <= 1,000,000)，从第1班到第N班。
每头奶牛每天只在一定的班次内干活。每头被选取打扫的奶牛会在它们的班次内一直工作。
你要做的就是帮助农夫约翰安排一些奶牛轮班，使得(i)每个班次至少有一头牛被安排，并且(ii)使尽可能少的奶牛参与打扫。如果无法为每个班次安排奶牛，则输出-1。 |
|
【Input】 |
【输入】 |
|
* 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. |
* 第1行：两个用空格分隔的整数N与T
* 第2..N+1行：每行包含一头可工作奶牛的上班与下班时间。奶牛在上班时间开始工作，并且在下班时间达到后结束工作。 |
|
【Output】 |
【输出】 |
|
* 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. |
* 第一行：约翰最少需要指派的奶牛数量，如果不能为每个班次都分配一头牛则输出-1。 |
|
【Sample Input - 输入样例】 |
【Sample Output - 输出样例】 |
|
3 10 1 7 3 6 6 10 |
2 |
|
【Hint】 |
【提示】 |
|
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. |
这个问题的输入数据比较大，使用scanf()读取比cin的速度更稳妥。
输入详解：
有3头牛与10班倒。奶牛#1可以工作在班次1..7，奶牛#2可以工作在班次3..6，并且奶牛#3可以工作在班次6..10。
输出详解：
通过安排奶牛#1和奶牛#3，所有班次可以被覆盖。无法使用2头一下的奶牛覆盖所有的班次。 |
【题解】
主要思想是贪心，不过有两种实现的方法
第一种，sort + priority_queue
按照开始班次从小到大排序，每次在当前可覆盖的班次中选取可在区间里工作的奶牛。
找出可到达的最大班次，加入到当前班次中。
重复执行即可得出答案。
第二种，非sort
开辟长度为最大班次的数组shift[T]，读取每个数据时以shift[st] = ed的形式记录。
剩下的步骤与第一种如出一辙。
选择最大班次的时候需要注意一点，每次读取的st与ed组成的区间为闭区间，即[st, ed]。
也就是说[1, 2] + [3, 4] = [1, 4]
【代码 C++】
1 #include<cstdio> 2 #include <algorithm> 3 #include <queue> 4 struct cow{ 5 int st, ed; 6 }data[25005]; 7 bool cmp(cow A, cow B){ 8 return A.st < B.st; 9 } 10 std::priority_queue<int, std::vector<int> > edShift; 11 int main(){ 12 int n, t, i, nowShift, opt; 13 scanf("%d%d", &n, &t); 14 for (i = 0; i < n; ++i) scanf("%d%d", &data[i].st, &data[i].ed); 15 std::sort(data, data + n, cmp); 16 for (i = opt = nowShift = 0; nowShift < t; edShift.pop()){ 17 while (data[i].st <= nowShift + 1 && i < n) edShift.push(data[i++].ed); 18 if (edShift.empty()) break; 19 nowShift = edShift.top(), ++opt; 20 } 21 if (nowShift < t) puts("-1"); 22 else printf("%d", opt); 23 return 0; 24 }
1 #include<cstdio> 2 #include <algorithm> 3 int shift[1000005]; 4 int main(){ 5 int n, t, i, st, ed, nowShift, nextShift, opt; 6 scanf("%d%d", &n, &t); 7 for (i = opt = nextShift = nowShift = 0; i < n; ++i){ 8 scanf("%d%d", &st, &ed); 9 shift[st] = std::max(shift[st], ed); 10 } 11 for (i = 0; i <= nowShift + 1 && nowShift < t; ++opt, nowShift = nextShift){ 12 while (i <= nowShift + 1) nextShift = std::max(nextShift, shift[i++]); 13 } 14 if (nowShift < t) puts("-1"); 15 else printf("%d", opt); 16 return 0; 17 }
POJ 2376 Cleaning Shifts（轮班打扫）
标签：
原文地址：http://www.cnblogs.com/Simon-X/p/5370258.html
