poj 2376 Cleaning Shifts



Cleaning Shifts

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 16370 Accepted: 4176

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.

Source

USACO 2004 December Silver

题目描述：

       可简化为给出n个区间以及一个区间[1,t],要求用最少的区间覆盖[1,t],若不能覆盖输出-1，否则输出最少的区间个数。（提示：若一头牛的工作时间为（4,6），则它后面的牛最迟开始工作的时间为7）。

#include<stdio.h>#include<algorithm>using namespace std;struct A{    int s,e;} a[25005];int vis[25005];bool cmp(A a,A b){    return a.s < b.s;}int main(){    int n,T;    scanf("%d%d",&n,&T);    for(int i = 1; i <= n; i++)        scanf("%d%d",&a[i].s,&a[i].e);    sort(a + 1,a + n + 1,cmp);//for(int i = 1;i <= n;i++)//printf("%d    ^^     %d\n",a[i].s,a[i].e);int t = 0,temp = 0,ans = 0;//ans记录最少区间的个数，temp当前符合条件的区间最靠右的端点值 a[n + 1].s = 0x3f3f3f3f;int flag = 0;    for(int i = 1; i <= n; i++)    {        if(a[i].s <= t + 1)        {        if(a[i].e > temp)    //在左区间符合要求的情况下，要找区间最靠右的那个             {               temp = a[i].e;flag = 1;            }                           if(a[i + 1].s > t + 1 && flag)  //因为这里有判断第i+1项，为满足条件所以前面要先将a[n + 1].s赋值为无穷大             {                t = temp;ans++;flag = 0;                        }                        }    }    if(t < T)       //若到最后t的值小于T,那么表示不能完成         printf("-1\n");    else    printf("%d\n",ans);    return 0;}