USACO 1.2 Milking Cows

这么水的代码也能过，USACO的数据弱爆了

/*
ID: aznfy1
PROG: milk2
LANG: C++
*/

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>

using namespace std;

struct node
{
    int start;
    int over;
}farmer[6000];

bool compare(const node &a,const node &b)
{
    if(a.start==b.start)
    return a.over>b.over;
    else
    return a.start<b.start;
}

int main()
{
    freopen("milk2.in","r",stdin);
    freopen("milk2.out","w",stdout);
    int n;
    int nowstart;
    int nowover;
    int ans1;
    int ans2;
    while(cin>>n)
    {
        ans1=0;
        ans2=0;
        for(int i=0;i<n;i++)
        cin>>farmer[i].start>>farmer[i].over;
        sort(farmer,farmer+n,compare);
        //计算最长的时间，至少有一个人在工作
        for(int i=0;i<n;i++)
        {
            nowstart=farmer[i].start;
            nowover=farmer[i].over;
            int tmp=i;
            do
            {
                if(nowover<=farmer[tmp].over&&nowover>=farmer[tmp].start)
                {
                    nowover=farmer[tmp].over;
                    tmp++;
                }
                else
                tmp++;
            }
            while(farmer[tmp].start<=nowover&&tmp<n);
            if(nowover-nowstart>ans1)
            ans1=nowover-nowstart;
        }
        //以下计算最长的时间没有一个人在工作
        int tmp=0;
        do
        {

            nowstart=farmer[tmp].start;
            nowover=farmer[tmp].over;
            do
            {
                if(nowover<=farmer[tmp].over&&nowover>=farmer[tmp].start)
                {
                    nowover=farmer[tmp].over;
                    tmp++;
                }
                else
                tmp++;
            }
            while(farmer[tmp].start<=nowover&&tmp<n);
            if(farmer[tmp].start-nowover>ans2)
            ans2=farmer[tmp].start-nowover;
        }
        while(tmp<n);
        cout<<ans1<<" "<<ans2<<endl;
    }
    return 0;
}