USACO 1.2.1 Milking Cows

题意:

        给定N个挤奶工取奶的起始时间和结束时间，求其中至少一个在取奶的最长时间和都没在取奶的最长时间。起止时间为5:00后以秒为单位的时间点，一个挤奶工对应一头奶牛。

        又一道让BUG给挂掉的题，这次还更无语。

input

4
100 200
201 301
302 402
403 503

output

100 1

        我运行的确实是"100 1"啊，可那测评系统非说我运行的是"100 0"，无奈...

  > Run 2: Execution error: Your program did not produce an answer
        that was judged as correct. The program stopped at 0.000 seconds;
        it used 3064 KB of memory. 

        Here are the respective outputs:
        ----- our output ---------
        100_1
        ---- your output ---------
        100_0
        --------------------------

        ------ Data for Run 2 ------
        4 
        100 200 
        201 301 
        302 402 
        403 503 
        ----------------------------

/*
ID: sdaujsj1
LANG: C++
TASK: milk2
*/
#include <iostream>
#include<stdlib.h>
#include<fstream>
using namespace std ;
int i, j, n ;
struct row{
    int begin, end ;
}a[5010] ;
int cmp(const void *a, const void *b){
    return (*(row *)a).begin > (*(row *)a).begin ? -1 : 1 ;
}
int find(int x){
    int k = -1, max = a[x].end ;
    for(i=x+1; i<n; i++){
        if(a[i].begin<=a[x].end){
            if(a[i].end>max){
                max = a[i].end ;
                k = i ;
            }
        }
    }
    return k ;
}
int find1(int x){
    int k ;
    for(k=x+1; k<n; k++)
        if(a[k].begin>a[x].end)    break ;
    return k ;
}
int main(){
    ifstream fin ("milk2.in");
    fin >> n ;
    for(i=0; i<n; i++)
        fin >> a[i].begin >> a[i].end ;
    qsort(a, n, sizeof(a[0]), cmp) ;
    fin.close();
    j = 0 ;
    int sum = a[0].end - a[0].begin ;
    int bsum = 0 ;
    int mbegin = a[0].begin ;
    while(true){
        int temp = j ;
        j = find(j) ;
        if(j!=-1){
            if(a[j].end - mbegin>sum)  sum = a[j].end - mbegin ;
        }
        else{
            j = find1(temp) ;
            if(j!=n){
                mbegin = a[j].begin ;
                if(a[j].begin - a[temp].end>bsum)  bsum = a[j].begin - a[temp].end ;
            }
            else    break ;
        }
    }
    ofstream fout ("milk2.out") ;
    fout << sum << " " << bsum << endl ;
    fout.close() ;
}