PATbasic1004

#include "stdafx.h"
#include<iostream>
#include<string>
using namespace std;

typedef struct node {
	string name;
	string number;
	int result;
	struct node *next;
}NODE;

NODE *build(int n) {
	NODE *head, *p, *nextnode;
	head = new NODE;
	head->next = NULL;
	p = head;
	for (int i = 0; i < n; i++) {
		nextnode = new NODE;
		cin >> nextnode->name;
		cin >> nextnode->number;
		cin >> nextnode->result;
		p->next = nextnode;
		p = nextnode;
	}
	p->next = NULL;
	return head;
}

void sort(NODE *head) {
	NODE *min = head->next, *max = head->next;
	NODE *p;
	for (p = head->next->next; p != NULL; p = p->next) {
		if (p->result > max->result) {
			max = p;
		}
	}
	for (p = head->next->next; p != NULL; p = p->next) {
		if (p->result < min->result) {
			min = p;
		}
	}
	cout << max->name << " " << max->number << endl;
	cout << min->name << " " << min->number << endl;
}

int main()
{
	int n;
	cin >> n;
	NODE *head = build(n);
	sort(head);
    return 0;
}

这题我做麻烦了。。。其实完全用不着链表的，用一个NODE array[]存所有的节点然后排个序OK。万恶的惯性思维。。。