PAT 1036 Boys vs Girls (25)

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本文介绍了一个程序设计问题，即如何找出所有男性学生中的最低分数与所有女性学生中的最高分数，并计算两者之间的差距。该程序使用了结构体来存储学生的姓名、性别、ID和分数，并通过不同的排序方式分别找出目标学生。

This time you are asked to tell the difference between the lowest grade of all the male students and the highest grade of all the female students.

Input Specification:

Each input file contains one test case. Each case contains a positive integer N, followed by N lines of student information. Each line contains a student's name, gender, ID and grade, separated by a space, where name and ID are strings of no more than 10 characters with no space, gender is either F (female) or M (male), and grade is an integer between 0 and 100. It is guaranteed that all the grades are distinct.

Output Specification:

For each test case, output in 3 lines. The first line gives the name and ID of the female student with the highest grade, and the second line gives that of the male student with the lowest grade. The third line gives the difference grade_F-grade_M. If one such kind of student is missing, output "Absent" in the corresponding line, and output "NA" in the third line instead.

Sample Input 1:

3
Joe M Math990112 89
Mike M CS991301 100
Mary F EE990830 95

Sample Output 1:

Mary EE990830
Joe Math990112
6

Sample Input 2:

1
Jean M AA980920 60

Sample Output 2:

Absent
Jean AA980920
NA
/* 考察结构体排序 QvQ..  
 * 午睡操场传来蝉的声音 多少年后也还是很好听~~
*/
#include "iostream"
#include "vector"
#include "cstring"
#include "string"
#include "stack"
#include "algorithm"
using namespace std;
struct Node {
	char name[11];
	char gender;
	char id[11];
	int grade;
};
bool cmp(const Node &n1, const Node &n2) {
	return n1.grade > n2.grade;
}
bool cmp1(const Node &n1, const Node &n2) {
	return n1.grade < n2.grade;
}
int main() {
	int n;
	vector<Node>v1, v2;
	cin >> n;
	for (int i = 0; i < n; i++) {
		Node node;
		cin >> node.name >> node.gender >> node.id >> node.grade;
		if (node.gender == 'M') {
			v1.push_back(node);
		}
		if (node.gender == 'F') {
			v2.push_back(node);
		}
	}
	sort(v1.begin(),v1.end(),cmp1);
	sort(v2.begin(), v2.end(), cmp);
	if (v2.size()) {
		cout << v2[0].name << " " << v2[0].id << endl;
	}
	else {
		cout << "Absent" << endl;
	}
	if (v1.size()) {
		cout << v1[0].name << " " << v1[0].id << endl;
	}
	else {
		cout << "Absent" << endl;
	}
	if(v1.size()&&v2.size()){
		cout << v2[0].grade - v1[0].grade << endl;
	}
	else {
		cout << "NA" << endl;
	}
	return 0;
}