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NC51192. Cleaning Shifts

描述

Farmer John is assigning some of his N 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 , 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.

输入描述

* 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.

输出描述

* 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.

示例1

输入:

3 10
1 7
3 6
6 10

输出:

2

原站题解

上次编辑到这里,代码来自缓存 点击恢复默认模板

C++14(g++5.4) 解法, 执行用时: 20ms, 内存消耗: 564K, 提交时间: 2020-09-08 15:04:47

#include<bits/stdc++.h>
using namespace std;
struct node{
    int be,ed;
};

bool cmp(node &a,node &b)
{
    if(a.be==b.be)
        return a.ed<b.ed;
    return
        a.be<b.be;
}

int main()
{
    int N,T;
    cin>>N>>T;
    node a[25500];
    for(int i=1;i<=N;i++)
    {
        cin>>a[i].be>>a[i].ed;
    }
    sort(a+1,a+1+N,cmp);
    /*for(int i=1;i<=N;i++)
    {
        cout<<a[i].be<<" "<<a[i].ed<<endl;
    }*/
    if(a[1].be!=1)
        cout<<-1<<endl;
    else
    {
        int ans=0;
        int left=1;
        int right=0;
        int cnt=1;
        while(right<T)
        {
            left=right+1;
            for(int i=cnt;i<=N;i++)
            {
                if(a[i].be<=left&&a[i].ed>=left)
                {
                    right=max(right,a[i].ed);
                }
                if(a[i].be>left)
                {
                    cnt=i;
                    break;
                }
            }
            if(left>right)
                break;
            else
                ans++;
        }
        if(right<T)
        cout<<-1<<endl;
        else
            cout<<ans<<endl;

    }
    return 0;
}

C++(g++ 7.5.0) 解法, 执行用时: 11ms, 内存消耗: 652K, 提交时间: 2022-09-21 16:09:46

#include <bits/stdc++.h>

using namespace std;
using ll = long long;
using PII = pair<int, int>;

int main() {
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int n, m;
    cin >> n >> m;
    vector<PII> a(n);
    for (auto &[x, y] : a) {
        cin >> x >> y;
    }

    sort(a.begin(), a.end(), [&](PII x, PII y){
        return x.first < y.first;
    });

    int ans = 0;
    for (int l = 1, j = 0; l <= m; l++) {
        int r = 0;
        while (j < n and a[j].first <= l) {
            r = max(r, a[j].second);
            j++;
        }
        if (r < l) {
            cout << -1 << '\n';
            return 0;
        }
        l = r;
        ans++;
    }
    cout << ans << '\n';
    return 0;
}

C++ 解法, 执行用时: 18ms, 内存消耗: 644K, 提交时间: 2022-01-30 10:08:53

#include<bits/stdc++.h>
using namespace std;
int n,m,nw,bd,ans;
struct node{int st,ed;}t[25001];
bool cmp(node x,node y){return x.st<y.st;}
int main(){
	cin>>n>>m;
	for(int i=1;i<=n;i++)cin>>t[i].st>>t[i].ed;
	sort(t+1,t+n+1,cmp);
	while(nw<=n){
		int rs=0;
		while(nw<=n&&t[nw].st<=bd+1)rs=max(rs,t[nw++].ed);
		if(!rs)break;
		++ans;
		bd=max(bd,rs);
		if(bd>=m)break;		
	}
	if(bd<m)cout<<-1;
	else cout<<ans;
	return 0;
}

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