渡口管理问题.doc

渡口管理问题题目：某汽车轮渡口，过江渡船每次只能载10辆车过江。过江车辆分为客车类和货车类，上船有如下规定：同类车先到先上船，客车先于货车上渡船，且每上4辆客车，才允许上一辆货车；若等待客不足4辆，则以货车代替，若无货车等待则允许客车都上船。根据描述设计一个算法模拟渡口管理。提示： 初始化，上渡船汽车数量x，上船客车数y，上船货车数z；讨论在上渡船汽车数量小于10的情形：1 若x4，客车队列又非空，将客车队列的队头汽车出队上渡船。X 和Y增1，否则转22、若Y4，或客车队列为空队且货车队列非空，将货车队列的队头汽车出队上渡船。Y0，X和Z增1， 否则转33、若货车队列为空队且客车队非空，将客车队列的队头汽车出队上渡船。X和Y增1，Z0；否则转44、提示相应错误信息并退出程序运行。#include#include#includeusing namespace std;int main() queue q_keche, q_huoche; string input_data, huoche=huoche, keche=keche; int Max=10; int k=0,h=0; while (cininput_data) stringstream out; if (input_data = huoche) cout huoche size before adding: q_huoche.size() ; out(+h); q_huoche.push(huoche+out.str(); cout q_huoche.back() in the queue, total huoche: q_huoche.size() endl; else if (input_data = keche) cout Keche size before adding: q_keche.size() ; out(+k); q_keche.push(keche+out.str(); cout q_keche.back() in the queue, total keche: q_keche.size() endl; else break; cout 2 queues info:nTotal Keche Size: q_keche.size()endl Toatl Huoche Size: q_huoche.size()endl; if ( q_keche.empty() & (!q_huoche.empty() while (!q_huoche.empty() couthuoche Size: q_huoche.size() huoche will load on the ship: q_huoche.front()endl; q_huoche.pop(); -Max; if (Max=0) coutThe ship is full. Please wait for next ship!n.n.n.n The ship is coming, start loadingn; Max=10; else while (!q_keche.empty() | !q_huoche.empty() for (int i=0; !q_keche.empty()& i4; i+) coutKeche Size: q_keche.size() q_keche.front() will load on the ship: q_keche.front()endl; q_keche.pop(); coutKeche Sizes after loading: q_keche.size()endl; -Max; if (!q_huoche.empty() couthuoche Size: q_huoche.size() q_huoche.front() huoche will load: q_huoche.front()endl; q_huoche.pop(); cout huoche Sizes after loading: q_huoche.size()endl; -Max; if (Max=0) coutThe