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开门人和关门人

3
1
ME3021112225321 00:00:00 23:59:59
2
EE301218 08:05:35 20:56:35
MA301134 12:35:45 21:40:42
3
CS301111 15:30:28 17:00:10
SC3021234 08:00:00 11:25:25
CS301133 21:45:00 21:58:40

ME3021112225321 ME3021112225321
EE301218 MA301134
SC3021234 CS301133

水题，不解释了，就是简单的字符串应用。

AC代码：

#include <iostream>
#include <cstdio>
#include <string>
#include <algorithm>
using namespace std;

struct man 
{
	string id, st, et;         //人的编号，签到时间，签退时间
};
man m[10005];

bool cmp1(man a, man b){
	int x = (a.st[0]-'0')*10 + (a.st[1]-'0');
	int xx = (a.st[3]-'0')*10 + (a.st[4]-'0');
	int xxx = (a.st[6]-'0')*10 + (a.st[7]-'0');
	int y = (b.st[0]-'0')*10 + (b.st[1]-'0');
	int yy = (b.st[3]-'0')*10 + (b.st[4]-'0');
	int yyy = (b.st[6]-'0')*10 + (b.st[7]-'0');
	if(x==y && xx==yy)  return xxx < yyy;
	else if(x==y)  return xx < yy;
	return x < y;
}

bool cmp2(man a, man b){
	int x = (a.et[0]-'0')*10 + (a.et[1]-'0');
	int xx = (a.et[3]-'0')*10 + (a.et[4]-'0');
	int xxx = (a.et[6]-'0')*10 + (a.et[7]-'0');
	int y = (b.et[0]-'0')*10 + (b.et[1]-'0');
	int yy = (b.et[3]-'0')*10 + (b.et[4]-'0');
	int yyy = (b.et[6]-'0')*10 + (b.et[7]-'0');
	if(x==y && xx==yy)  return xxx > yyy;
	else if(x==y)  return xx > yy;
	return x > y;
}

int main(){
//	freopen("in.txt", "r", stdin);
	int t, n;
	scanf("%d", &t);
	while(t--){
		scanf("%d", &n);
		for(int i=0; i<n; i++)
			cin >> m[i].id >> m[i].st >> m[i].et;
		sort(m, m+n, cmp1);
		cout << m[0].id<<" ";
		sort(m, m+n, cmp2);
		cout << m[0].id << endl;
	}
	return 0;
}

HDU 1234 （浙大计算机研究生复试上机考试-2005年） 开门人和关门人 （水）

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原文地址：http://blog.csdn.net/u013446688/article/details/41122559