POJ2376贪心Cleaning Shifts

Cleaning Shifts

Time Limit: 1000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

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 10
1 7
3 6
6 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.

题意：给定一个大的区间长度T，给你n个小区间，看是否可以覆盖住原来那个最大的区间。

分析：贪心题目，如果长度为10，对于这组(1,5),(6,10)

#include<stdio.h>
#include<algorithm>
#include <iostream>
using namespace std;
struct p
{
    int x,y;
} a[25001];
bool comp(p x,p y)
{
    if(x.x==y.x)
        return x.y<y.y;
    return x.x<y.x;
}
int main()
{
    int n,m,t,i;
    scanf("%d%d",&t,&n);
    for(i=0; i<t; i++)
        scanf("%d%d",&a[i].x,&a[i].y);
    sort(a,a+t,comp);
    ///完全覆盖了
    if(a[0].x==1&&a[0].y>=n)
        printf("1\n");
    ///表示开头不能覆盖
    else if(a[0].x!=1)
        printf("-1\n");
    else
    {
        ///对区间中间进行判断
        m = a[0].y;
        int sum = 1;
        ///i<t&&m<n小于测试数据组数，而且没有超过要判断区间的长度
        for(int i = 1; i < t && m < n;)
        {
            sum++;
            int mm = m;
            if (a[i].x > m + 1)///此时表明一定不能够完全覆盖
            {
                printf("-1\n");
                return 0;
            }
            ///只要满足这两个条件，则找此种情况下的最大值
            while (i < t && a[i].x <= m + 1)
            {
                mm = max(mm, a[i].y);
                i++;
            }
            m = mm;
            if(m >= n)
                break;
        }
        if (m < n)
            printf("-1\n");
        else
            printf("%d\n", sum);
    }
    return 0;
}