http://poj.org/problem?id=1250
题意大意
住宿床位有限，按顺序入住，用ABC等代表单个人，第1次出现代表入住，第2次出现代表离开
输入：
1 ABCBCA
代表有1个床位，
A入住，
B入住，入住失败
C入住，入住失败
B离开，共1人离开（未住店）
C离开，共2人离开（未住店）
A离开
计算有几个人来了没床位离开了
思路
用2个链表存储床位上的人，等待队列的人
人如果在上述2个链表中就删除，不在就插入队列
代码
#include <iostream>
using namespace std;
struct node
{char data;node* next;node():data('\0'),next(NULL){}node(char ch):data(ch),next(NULL){}~node(){ }
};
struct list
{node* p_head;size_t listLength;list():p_head(NULL),listLength(0){}~list(){ eraseAll();}void eraseAll(){if(p_head){node *delnode = p_head, *tempnode = p_head;while(tempnode->next != NULL){tempnode = delnode->next;delete delnode;delnode = tempnode;}listLength = 0;}}void push_front(char &ch){node *tempNode = new node(ch);listLength++;tempNode->next = p_head;p_head = tempNode;}node* find(char& ch){node* temp = p_head;if(p_head){while(temp != NULL && temp->data != ch){temp = temp->next;}} else{temp = NULL;}return temp;}void delNode(char &ch){node* tempnode = p_head, *delnode;delnode = find(ch);if(delnode && delnode != p_head){while(tempnode->next != delnode){tempnode = tempnode->next;}tempnode->next = delnode->next;delete delnode;listLength--;}else{if(delnode == p_head && delnode){p_head = delnode->next;delete delnode;listLength--;}}}
};int main()
{list beds,waitlist; //床位队列，等待队列node *tempnode = NULL;size_t numsofbed, walkedaway = 0;char ch;while(cin >> numsofbed && numsofbed) //输入床位数，且不为0{cin.get(); //拿掉空格while(cin.get(ch) && ch != '\n') //输入每个人{tempnode = beds.find(ch); //去床位队列查找人if(tempnode) //找到了这个人，床位上这个人离店{beds.delNode(ch);}else //没有在床位上找到该人，则该人需要住店{if(beds.listLength < numsofbed) //床位有空余{beds.push_front(ch); //这个人住下}else //床位满了{if(waitlist.find(ch)) //这个人已经在等待队列里{waitlist.delNode(ch); //等不了了，离开等待队列walkedaway++;}else //这个人不在等待队列里，可以等待{waitlist.push_front(ch);}}}}if(walkedaway == 0){cout << "All customers tanned successfully." << endl;}else{cout << walkedaway << " customer(s) walked away." << endl;}beds.eraseAll();waitlist.eraseAll();walkedaway = 0;numsofbed = 0;}return 0;
}
