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题目的意思就是有一些人， 来者有先后， 给出每个人的来的时间， 和去的时间，求解第一个来的人和最后一个走的人

我的思路是直接qsort 对时间进行两次排序

附代码：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<stdlib.h>
#define Max_len 20
#define Max_num 1000

int cmp(const void *a, const void *b);
int cmp1(const void *a, const void *b);

typedef struct come{
	char str[Max_len];
	int hour, min, sec;
}come;

typedef struct back{
	char str[Max_len];
	int hour, min, sec;
}back;


int main(){
	
	come co[Max_num];
	back ba[Max_num];
	char temp[Max_len];
	int m, i, hourc, minc, secc, hourb, minb, secb;

	scanf("%d", &m);
	for(i = 0; i < m; i++){
		scanf("%s %d:%d:%d%d:%d:%d", temp, &hourc, &minc, &secc, &hourb, &minb, &secb);
		strcpy(co[i].str, temp);  co[i].hour = hourc; co[i].min = minc; co[i].sec = secc;
		strcpy(ba[i].str, temp);  ba[i].hour = hourb; ba[i].min = minb; ba[i].sec = secb;
	}
	qsort(co, m, sizeof(co[0]), cmp);
	qsort(ba, m, sizeof(ba[0]), cmp1);

	printf("%s %s\n", co[0].str, ba[0].str);
	return 0;
}

int cmp(const void *a, const void *b){
	come *c = (come *)a;
	come *d = (come *)b;
	if(c -> hour != d -> hour){
		return c -> hour - d -> hour;
	}
	else if(c -> min != d -> min){
		return c -> min - d -> min;
	}
	else{
		return c -> sec - d -> sec;
	}
}

int cmp1(const void *a, const void *b){
	back *c = (back *)a;
	back *d = (back *)b;
	if(c -> hour != d -> hour){
		return d -> hour - c -> hour;
	}
	else if(c -> min != d -> min){
		return d -> min - c -> min;
	}
	else{
		return d -> sec - c -> sec;
	}
}

pat 1006

原文：http://blog.csdn.net/huntinggo/article/details/19082129