Problem C: The Dragon of Loowater

Once upon a time, in the Kingdom of Loowater, a minor nuisance turned into a major problem.

The shores of Rellau Creek in central Loowater had always been a prime breeding ground for geese. Due to the lack of predators, the geese population was out of control. The people of Loowater mostly kept clear of the geese. Occasionally, a goose would attack
one of the people, and perhaps bite off a finger or two, but in general, the people tolerated the geese as a minor nuisance.

One day, a freak mutation occurred, and one of the geese spawned a multi-headed fire-breathing dragon. When the dragon grew up, he threatened to burn the Kingdom of Loowater to a crisp. Loowater had a major problem. The king was alarmed, and called on his knights
to slay the dragon and save the kingdom.

The knights explained: "To slay the dragon, we must chop off all its heads. Each knight can chop off one of the dragon's heads. The heads of the dragon are of different sizes. In order to chop off a head, a knight must be at least as tall as the diameter of
the head. The knights' union demands that for chopping off a head, a knight must be paid a wage equal to one gold coin for each centimetre of the knight's height."

Would there be enough knights to defeat the dragon? The king called on his advisors to help him decide how many and which knights to hire. After having lost a lot of money building Mir Park, the king wanted to minimize the expense of slaying the dragon. As
one of the advisors, your job was to help the king. You took it very seriously: if you failed, you and the whole kingdom would be burnt to a crisp!

Input Specification:

The input contains several test cases. The first line of each test case contains two integers between 1 and 20000 inclusive, indicating the number n of heads that the dragon has, and the number m of knights in the kingdom. The next n lines
each contain an integer, and give the diameters of the dragon's heads, in centimetres. The following m lines each contain an integer, and specify the heights of the knights of Loowater, also in centimetres.

The last test case is followed by a line containing:

0 0

Output Specification:

For each test case, output a line containing the minimum number of gold coins that the king needs to pay to slay the dragon. If it is not possible for the knights of Loowater to slay the dragon, output the line:

Loowater is doomed!

Sample Input:

2 3
5
4
7
8
4
2 1
5
5
10
0 0

Output for Sample Input:

11
Loowater is doomed!


  1. /*********************************
  2. *   日期:2013-4-19
  3. *   作者:SJF0115
  4. *   题号: 题目11292 - Dragon of Loowater
  5. *   来源:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=2267
  6. *   结果:AC
  7. *   来源:UVA
  8. *   总结:
  9. **********************************/
  10. #include<stdio.h>
  11. #include<stdlib.h>
  12. int Dragon[20001],Knights[20001];
  13. //排序函数
  14. int cmp(const void *a,const void *b){
  15. return *(int *)a - *(int *)b;
  16. }
  17. int main ()
  18. {
  19. int i,j,N,M,cost,index;
  20. //freopen("C:\\Users\\XIAOSI\\Desktop\\acm.txt","r",stdin);
  21. while(scanf("%d %d",&N,&M) != EOF){
  22. if(N == 0 && M == 0){
  23. break;
  24. }
  25. //金币
  26. cost = 0;
  27. index = 0;
  28. //头
  29. for(i = 0;i < N;i++){
  30. scanf("%d",&Dragon[i]);
  31. }
  32. //勇士
  33. for(i = 0;i < M;i++){
  34. scanf("%d",&Knights[i]);
  35. }
  36. //排序
  37. qsort(Dragon,N,sizeof(int),cmp);
  38. qsort(Knights,M,sizeof(int),cmp);
  39. //贪心
  40. for(i = 0;i < M;i++){
  41. if(index >= N){
  42. break;
  43. }
  44. if(Knights[i] >= Dragon[index]){
  45. //统计所花金币
  46. cost += Knights[i];
  47. index++;
  48. }
  49. }
  50. //输出
  51. if(index >= N){
  52. printf("%d\n",cost);
  53. }
  54. else{
  55. printf("Loowater is doomed!\n");
  56. }
  57. }
  58. return 0;
  59. }

版权声明:本文博客原创文章,博客,未经同意,不得转载。

UVA它11292 - Dragon of Loowater的更多相关文章

  1. UVA 11292 Dragon of Loowater(简单贪心)

    Problem C: The Dragon of Loowater Once upon a time, in the Kingdom of Loowater, a minor nuisance tur ...

  2. uva 11292 Dragon of Loowater (勇者斗恶龙)

    Problem C: The Dragon of Loowater Once upon a time, in the Kingdom of Loowater, a minor nuisance tur ...

  3. [ACM_水题] UVA 11292 Dragon of Loowater [勇士斗恶龙 双数组排序 贪心]

    Once upon a time, in the Kingdom of Loowater, a minor nuisance turned into a major problem. The shor ...

  4. UVa 11292 - Dragon of Loowater(排序贪心)

    Once upon a time, in the Kingdom of Loowater, a minor nuisance turned into a major problem.The shore ...

  5. UVa 11292 Dragon of Loowater

    简单贪心 龙头的直径和人的佣金排序,价值小的人和直径小的配 #include<iostream> #include<cstdio> #include<cmath> ...

  6. UVA - 11292 Dragon of Loowater 贪心

    贪心策略:一个直径为X的头颅,应该让雇佣费用满足大于等于X且最小的骑士来砍掉,这样才能使得花费最少. AC代码 #include <cstdio> #include <cmath&g ...

  7. UVa 11292 Dragon of Loowater (水题,排序)

    题意:有n个条龙,在雇佣勇士去杀,每个勇士能力值为x,只能杀死头的直径y小于或等于自己能力值的龙,只能被雇佣一次,并且你要给x赏金,求最少的赏金. 析:很简单么,很明显,能力值高的杀直径大的,低的杀直 ...

  8. 贪心/思维题 UVA 11292 The Dragon of Loowater

    题目传送门 /* 题意:n个头,m个士兵,问能否砍掉n个头 贪心/思维题:两个数组升序排序,用最弱的士兵砍掉当前的头 */ #include <cstdio> #include <c ...

  9. The Dragon of Loowater

      The Dragon of Loowater Once upon a time, in the Kingdom of Loowater, a minor nuisance turned into ...

随机推荐

  1. ThinkPHP - 前置操作+后置操作

    前置操作和后置操作   系统会检测当前操作(不仅仅是index操作,其他操作一样可以使用)是否具有前置和后置操作,如果存在就会按照顺序执行,前置和后置操作的方法名是在要执行的方法前面加 _before ...

  2. SQL Server 基础 05 多链表查询和子查询

     连接查询 值得注意的是:字段前必须加表名,以便混淆 -- 多表连接查询和子查询 select * from dbo.stu_info ,dbo.sname2 -- 加连接规则的查询 where se ...

  3. instanceof 变量是否属于某一类 class 的实例

    <?phpclass MyClass{} class NotMyClass{}$a = new MyClass;$b = new NotMyClass;var_dump($a instanceo ...

  4. 【HTTP 2】HTTP/2 协议概述(HTTP/2 Protocol Overview)

    前情提要 在上一篇文章<[HTTP 2.0] 简介(Introduction)>中,我们简单介绍了 HTTP 2. 在本篇文章中,我们将会了解到 HTTP 2 协议概述部分的内容. HTT ...

  5. perl 5.22手动安装Mysql DBI和DBD

    mysql 手动安装DBI 和DBD: DBI版本: [root@dr-mysql01 DBD-mysql-4.033]# perl -MDBI -le 'print $DBI::VERSION;' ...

  6. JAVA平台上的网络爬虫脚本语言 CrawlScript

    JAVA平台上的网络爬虫脚本语言 CrawlScript 网络爬虫即自动获取网页信息的一种程序,有很多JAVA.C++的网络爬虫类库,但是在这些类库的基础上开发十分繁琐,需要大量的代码才可以完成一 个 ...

  7. 分享:json2.js源代码解读笔记

    1. 怎样理解"json" 首先应该意识到,json是一种数据转换格式,既然是个"格式",就是个抽象的东西.它不是js对象,也不是字符串,它仅仅是一种格式,一种 ...

  8. C#后台代码编写图片地址Properties.Resources._1;

    if (i == 0)            {                pictureBox1.Image = Properties.Resources._1;                ...

  9. android平板Home键的监听

    关于android4.2平板Home键的监听 1.android4.0以上的平板Home键是不能屏蔽的,至少在应用层是这样.但是可以监听Home键的触发,主要是通过广播的监听来接受Hone键的触发广播 ...

  10. Smarty - 安装+入门小例子

    环境: smarty 1.在http://www.smarty.net/download下载最新smarty包,window选择zips,linux下选择tar.gz.以windows为例,下载后解压 ...