USACO SEC.1.2 No.1 Milking Cows

题意：给出N个村民，每个村民每天早上起来在闭区间[a, b]挤牛奶，求至少一个村民挤牛奶的最长时间， 和没有村民挤牛奶的最长时间

解法：经典的区间合并问题， 将区间的端点作为事件，从左往右用扫描线进行扫描，

        需要注意的是，如果有结束事件和开始事件相同，先处理开始事件

/*
ID: lsswxr1
PROG: milk2
LANG: C++
*/
#include <iostream>
#include <vector>
#include <map>
#include <list>
#include <set>
#include <deque>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <cctype>
#include <cstdio>
#include <iomanip>
#include <cmath>
#include <cstdio>
#include <string>
#include <fstream>
using namespace std;

///宏定义
const int  INF = 1000000000;
const int MAXN = 10100;
const int maxn = MAXN;
///全局变量 和 函数

#define USACO
#ifdef USACO
#define cin fin
#define cout fout
#endif
//////////////////////////////////////////////////////////////////////////

int N;
struct events
{
    int t;
    bool flag;
    bool operator < (const events& other) const
    {
        if (t != other.t)
        {
            return t < other.t;
        }
        else
            return flag > other.flag;
    }
};

events E[maxn];

int main()
{

    
#ifdef USACO    
    ofstream fout ("milk2.out");
    ifstream fin ("milk2.in");
#endif    
    ///变量定义
    while (cin >> N)
    {
        //读入事件
        int j = 0;
        for (int i = 0; i < N; i++)
        {
            int t1, t2;
            cin >> t1 >> t2;
            E[j].t = t1;
            E[j].flag = true;
            j++;

            E[j].t = t2;
            E[j].flag = false;
            j++;
        }
        sort(E, E + j);
        
        //顺序扫描事件
        int ansNoboby = 0, ansSomebody = 0, peopleCnts = 0;
        int startTime = E[0].t;
        for (int i = 0; i < j; i++)
        {
            if (!E[i].flag)
            {
                peopleCnts--;
                if (peopleCnts == 0)
                {
                    ansSomebody = max(ansSomebody, E[i].t - startTime);
                    startTime = E[i].t;
                }
            }
            else
            {
                peopleCnts++;
                if (peopleCnts == 1)
                {                
                    ansNoboby = max(ansNoboby, E[i].t - startTime);
                    startTime = E[i].t;
                }

            }
        }
        cout << ansSomebody << " " << ansNoboby << endl;
    }

    ///结束
    return 0;
}