Cleaning Shifts
来源:互联网 发布:淘宝免费刷流量软件 编辑:程序博客网 时间:2024/06/05 19:21
Cleaning Shifts
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 22362 Accepted: 5584
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.
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.
* 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.
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.
#include <iostream>#include <math.h>#include <stdio.h>#include <stdlib.h>#include <algorithm>#include <set>#include <queue>#include <string.h>#define INF 0x3fffffff#define MAXN 25005typedef long long LL;using namespace std;struct Node { int b, e;};int N, T;Node node[MAXN];bool cmp(const Node &a, const Node &b) { if (a.b == b.b) { return a.e < b.e; } else { return a.b < b.b; }}int greedy() { int cur = 0; int right = 0; int left = 0; int num = 0; while (right < T) { left = right + 1; for (int i = cur; i < N; i++) { if (node[i].b <= left && left <= node[i].e) { right = max(right, node[i].e); } else if (node[i].b > left) { cur = i; break; } } if (left > right) break; else num++; } if (right == T) return num; else return -1;}int main() { //freopen("in.txt", "r", stdin); cin >> N >> T; int B = 0; int E = 0; for (int i = 0; i < N; i++) { cin >> node[i].b >> node[i].e; if (node[i].b == 1) B = 1; if (node[i].e == T) E = 1; } if (B == 0 || E == 0) { printf("-1\n"); return 0; } sort(node, node+N, cmp); int ans = greedy(); cout << ans << endl; return 0;}
阅读全文
0 0
- Cleaning Shifts
- Cleaning Shifts
- Cleaning Shifts
- - Cleaning Shifts
- Cleaning Shifts
- Cleaning Shifts
- PKU 2376 Cleaning Shifts
- POJ 2376 Cleaning Shifts
- Cleaning Shifts(p2376)
- POJ3171 Cleaning Shifts 数据结构
- POJ 2376 Cleaning Shifts
- Pku2376 Cleaning Shifts
- poj2376 Cleaning Shifts
- poj 2376 Cleaning Shifts
- POJ-2376-Cleaning Shifts
- poj 2376 Cleaning Shifts
- POJ 2376 Cleaning Shifts
- POJ2376 Cleaning Shifts 贪心
- ZOJ1025-Wooden Sticks(dp)
- 欢迎使用CSDN-markdown编辑器
- [UVA]11478 二分答案+差分约束
- windows批处理脚本循环生成文件学习心得
- SICP习题-Church计数
- Cleaning Shifts
- Android Apk 应用信息获取之 PackageManager
- 内存对齐(Memory Alignment)
- PAT甲级 1024
- ssh反向代理
- IOS 适配&开发策略
- DNS的作用
- acm 字符串
- HttpSessionListener 、HttpSessionAttributeListener以及HttpSessionBindingListener 的区别