POJ:2367-Cleaning Shifts

传送门：http://poj.org/problem?id=2376

Cleaning Shifts

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 27272 Accepted: 6724

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.

---

解题心得：

1. 题意就是让你选择的最少的线段将整个区间完全覆盖，如果不能完全覆盖输出-1。
2. 可以使用贪心的思想，首先按照起始时间排序，如果起始时间相同，让结束时间越长的排在越前面，每次找一个线段，然后找起点在线段内，但是终点要更远的线段作为新的终点，维护远的终点，如果起点超出了当前选择的线段答案就+1。
3. 然后就是判断是否出现无线段覆盖的地方，这个就需要判断答案+1的地方和开始选择的第一个线段是否是从1开始的，主要是注意一下边界问题。

---

#include <stdio.h>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn = 25000;
int n,t;

struct The_cow{
    int s,e;
    bool operator < (const The_cow &a) const {
        if(a.s  == s)
            return a.e < e;
        return a.s > s;
    }
}cow[maxn];

int main() {
    while(scanf("%d%d",&n,&t) != EOF) {
        memset(cow,0,sizeof(cow));
        for(int i=0;i<n;i++){
            scanf("%d%d",&cow[i].s,&cow[i].e);
            if(cow[i].s > cow[i].e)
                swap(cow[i].s,cow[i].e);
        }

        sort(cow,cow+n);
        if(cow[0].s != 1) {
            printf("-1
");
            continue;
        }
        int ans = 1,pos_end = 0,temp_end = 0;//temp_end为当前记录的终点，pos_end是延伸出去的最长终点
        temp_end = cow[0].e;
        for(int i=1;i<n;i++){
            if(cow[i].s > temp_end + 1){//为啥不加等号，因为在temp_end+1的位置的起点也可以用来更新pos_end
                temp_end = pos_end;
                ans++;
            }
            if(cow[i].s <= temp_end + 1) {
                if(cow[i].e > pos_end)
                    pos_end  = cow[i].e;
                if(cow[i].e == t) {
                    ans ++;
                    temp_end = t;
                    break;
                }
            }
        }
        if(temp_end == t)
            printf("%d
",ans);
        else
            printf("-1
");
    }
    return 0;
}