B:魔兽世界之一:备战

描述

魔兽世界的西面是红魔军的司令部,东面是蓝魔军的司令部。两个司令部之间是依次排列的若干城市。 
红司令部,City 1,City 2,……,City n,蓝司令部

两军的司令部都会制造武士。武士一共有 dragon 、ninja、iceman、lion、wolf 五种。每种武士都有编号、生命值、攻击力这三种属性。 

双方的武士编号都是从1开始计算。红方制造出来的第n个武士,编号就是n。同样,蓝方制造出来的第n个武士,编号也是n。 

武士在刚降生的时候有一个生命值。 

在每个整点,双方的司令部中各有一个武士降生。 

红方司令部按照iceman、lion、wolf、ninja、dragon的顺序循环制造武士。 

蓝方司令部按照lion、dragon、ninja、iceman、wolf的顺序循环制造武士。 

制造武士需要生命元。 

制造一个初始生命值为m的武士,司令部中的生命元就要减少m个。 

如果司令部中的生命元不足以制造某个按顺序应该制造的武士,那么司令部就试图制造下一个。如果所有武士都不能制造了,则司令部停止制造武士。

给定一个时间,和双方司令部的初始生命元数目,要求你将从0点0分开始到双方司令部停止制造武士为止的所有事件按顺序输出。
一共有两种事件,其对应的输出样例如下: 

1) 武士降生 
输出样例: 004 blue lion 5 born with strength 5,2 lion in red headquarter
表示在4点整,编号为5的蓝魔lion武士降生,它降生时生命值为5,降生后蓝魔司令部里共有2个lion武士。(为简单起见,不考虑单词的复数形式)注意,每制造出一个新的武士,都要输出此时司令部里共有多少个该种武士。

2) 司令部停止制造武士
输出样例: 010 red headquarter stops making warriors
表示在10点整,红方司令部停止制造武士

输出事件时: 

首先按时间顺序输出; 

同一时间发生的事件,先输出红司令部的,再输出蓝司令部的。

输入第一行是一个整数,代表测试数据组数。

每组测试数据共两行。 

第一行:一个整数M。其含义为, 每个司令部一开始都有M个生命元( 1 <= M <= 10000)。

第二行:五个整数,依次是 dragon 、ninja、iceman、lion、wolf 的初始生命值。它们都大于0小于等于10000。输出对每组测试数据,要求输出从0时0分开始,到双方司令部都停止制造武士为止的所有事件。
对每组测试数据,首先输出"Case:n" n是测试数据的编号,从1开始 。
接下来按恰当的顺序和格式输出所有事件。每个事件都以事件发生的时间开头,时间以小时为单位,有三位。样例输入

1
20
3 4 5 6 7

样例输出

Case:1
000 red iceman 1 born with strength 5,1 iceman in red headquarter
000 blue lion 1 born with strength 6,1 lion in blue headquarter
001 red lion 2 born with strength 6,1 lion in red headquarter
001 blue dragon 2 born with strength 3,1 dragon in blue headquarter
002 red wolf 3 born with strength 7,1 wolf in red headquarter
002 blue ninja 3 born with strength 4,1 ninja in blue headquarter
003 red headquarter stops making warriors
003 blue iceman 4 born with strength 5,1 iceman in blue headquarter
004 blue headquarter stops making warriors

 

对面向对象真的是没有一点思路呀,看着答案自己把代码写了一下。

Code:

#include <iostream>
#include <cstring>
#include <string>
using namespace std;
const int WARRIOR_NUM = 5;  //士兵数量

class Headquarter;
class Warrior {
    private:
        Headquarter * pHeadquarter;
        int kindNo;     //武士种类编号
        int no;

    public:
        static string names[WARRIOR_NUM];
        static int initialLifeValue[WARRIOR_NUM];
        Warrior(Headquarter * p, int no, int kindNo_);
        void PrintResult(int nTime);
};

class Headquarter {
    private:
        int totalLifeValue;
        bool stopped;
        int totalWarriorNum;
        int color;
        int curMakingSeqIdx;
        int warriorNum[WARRIOR_NUM];
        Warrior * pWarriors[1000];
    public:
        friend class Warrior;
        static int makingSeq[2][WARRIOR_NUM];
        void Init(int color_, int lv);
        ~Headquarter();
        int Produce(int nTime);
        string GetColor();
};

Warrior::Warrior(Headquarter * p, int no_, int kindNo_) {   //Warrior类的构造函数
    no = no_;
    kindNo = kindNo_;
    pHeadquarter = p;
}

void Warrior::PrintResult(int nTime) {
    string color = pHeadquarter->GetColor();
    printf("%03d %s %s %d born with strength %d,%d %s in %s headquarter\n",
            nTime, color.c_str(), names[kindNo].c_str(), no, initialLifeValue[kindNo],
            pHeadquarter->warriorNum[kindNo], names[kindNo].c_str(), color.c_str());
}

void Headquarter::Init(int color_, int lv) {
    color = color_;
    totalLifeValue = lv;
    totalWarriorNum = 0;
    stopped = false;
    curMakingSeqIdx = 0;
    for (int i = 0; i < WARRIOR_NUM; ++i) {
        warriorNum[i] = 0;
    }
}

Headquarter::~Headquarter() {
    for (int i = 0; i < totalWarriorNum; ++i) {
        delete pWarriors[i];
    }
}

int Headquarter::Produce(int nTime) {
    if (stopped) return 0;
    int searchingTimes = 0;
    while (Warrior::initialLifeValue[makingSeq[color][curMakingSeqIdx]] > totalLifeValue &&
           searchingTimes < WARRIOR_NUM) {
        curMakingSeqIdx = (curMakingSeqIdx + 1) % WARRIOR_NUM;
        searchingTimes++;
    }
    int kindNo = makingSeq[color][curMakingSeqIdx];
    if (Warrior::initialLifeValue[kindNo] > totalLifeValue) {
        stopped = true;
        if (color == 0)
            printf("%03d red headquarter stops making warriors\n", nTime);
        else
            printf("%03d blue headquarter stops making warriors\n", nTime);
        return 0;
    }

    totalLifeValue -= Warrior::initialLifeValue[kindNo];
    curMakingSeqIdx = (curMakingSeqIdx + 1) % WARRIOR_NUM;
    pWarriors[totalWarriorNum] = new Warrior(this, totalWarriorNum+1, kindNo);
    warriorNum[kindNo]++;
    pWarriors[totalWarriorNum]->PrintResult(nTime);
    totalWarriorNum++;
    return 1;
}

string Headquarter::GetColor() {
    if (color == 0) return "red";
    else return "blue";
}

string Warrior::names[WARRIOR_NUM] = {"dragon", "ninja", "iceman", "lion", "wolf"};
int Warrior::initialLifeValue[WARRIOR_NUM];
int Headquarter::makingSeq[2][WARRIOR_NUM] = {{2, 3, 4, 1, 0}, {3, 0, 1, 2, 4}};

int main()
{
    int t;
    int m;
    Headquarter RedHead, BlueHead;
    scanf("%d", &t);
    int nCaseNo = 1;
    while (t--) {
        printf("Case:%d\n", nCaseNo++);
        scanf("%d", &m);
        for (int i = 0; i < WARRIOR_NUM; ++i) {
            scanf("%d", &Warrior::initialLifeValue[i]);
        }
        RedHead.Init(0, m);
        BlueHead.Init(1, m);
        int nTime = 0;
        while (true) {
            int tmp1 = RedHead.Produce(nTime);
            int tmp2 = BlueHead.Produce(nTime);
            if (tmp1 == 0 && tmp2 == 0) break;
            nTime++;
        }
    }
    return 0;
}

  

 

posted @ 2018-12-19 16:11  Veritas_des_Liberty  阅读(729)  评论(0编辑  收藏  举报