uva11292贪心基础题目
Crawling in process... Crawling failed Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Description
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
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
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
Sample Output
11
Loowater is doomed!
题目大意:有一只邪恶的龙,勇士们要杀掉他必须要干掉他的所有头,每一个勇士可以砍掉一个直径小于他身高的头,
然后可以得到他身高的金子数量,问怎样选择才能让国王付出最少的代价杀死这条龙。
思路分析:分别对身高和头的直径排序,每一个头都让比它高一点的人杀死,这样可以保证付出最少的代价。
代码:
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
using namespace std;
const int maxn=20000+10;
int a[maxn],b[maxn];
int main()
{
int n,m,i,j;
while(scanf("%d%d",&n,&m)&&(m||n))
{
int k=0;
for(i=0;i<n;i++)
scanf("%d",&a[i]);
for(i=0;i<m;i++)
scanf("%d",&b[i]);
sort(a,a+n);
sort(b,b+m);
int t=0;
for(i=0;i<n;i++)
{
for(j=k;j<m;j++)
{
if(a[i]<=b[j])
{
t+=b[j];
k=j+1;
break;
}
}
if(j==m) break;
}
if(i<n) cout<<"Loowater is doomed!"<<endl;
else cout<<t<<endl;
}
return 0;
}
uva11292贪心基础题目的更多相关文章
- poj2709 贪心基础
D - 贪心 基础 Crawling in process... Crawling failed Time Limit:1000MS Memory Limit:65536KB 64bi ...
- hdu 1009 贪心基础题
B - 贪心 基础 Crawling in process... Crawling failed Time Limit:1000MS Memory Limit:32768KB 64bi ...
- java基础题目总结
有些基础题目由于工作中用的比较少但却又是不可少的,这样回答起来就会反应慢,不确定,不准确,特此开了文章记录遇到的不确定或者回答比较拗口的问题. 1.servlet是单例的吗,是安全的吗,是多线程吗 s ...
- iOS 面试基础题目
转载: iOS 面试基础题目 题目来自博客:面试百度的记录,有些问题我能回答一下,不能回答的或有更好的回答我放个相关链接供参考. 1面 Objective C runtime library:Obje ...
- L贪心基础
<span style="color:#330099;">/* L - 贪心 基础 Time Limit:1000MS Memory Limit:65536KB 64b ...
- Problem C: 动态规划基础题目之数字三角形
Problem C: 动态规划基础题目之数字三角形 Time Limit: 1 Sec Memory Limit: 64 MBSubmit: 208 Solved: 139[Submit][Sta ...
- 贪心法基础题目 HDU
贪心算法的基本步骤: 1.从问题的某个初始解出发.2.采用循环语句,当可以向求解目标前进一步时,就根据局部最优策略,得到一个部分解,缩小问题的范围或规模.3.将所有部分解综合起来,得到问题的最终解. ...
- sql基础题目测试及正确答案
在网上做了一套基本的sql题目,以下是我的写的答案,适合基础人员练练 --创建测试数据 use test create table Student(S# varchar(10),Sname nvarc ...
- 二叉树的基础题目学习(EPI)
1.判断是个二叉树是不是平衡二叉树. 二叉树的定义都是利用递归的方法,所以二叉树有着天然的递归属性.所以一般情况下,递归解决二叉树问题中,递归解法比较简洁.平衡二叉树的定义是左子树和右子树均是平衡二叉 ...
随机推荐
- jquery1.9学习笔记 之层级选择器(二)
子孙选择器(“祖先 子孙”) 描述:选择所有给出祖先选择器的子孙选择器. 例子: 用蓝色虚线边框标记所有表单子孙元素的输入.表单里的输入框用黄色背景. <!doctype html>< ...
- iOS学习之自定义UItableViewCell
在项目开发中,大部分情况下我们都需要自定义UITableViewCell, 今天就重点整理一下目前自己已经学过的自定义Cell的一些注意事项; 分步骤来写吧: 1.将自定义的Cell定义为属性; 2. ...
- Android ActionBar 返回为NULL
AndroidBar返回为NULL.是因为values-v11和values-v14下面的style的主题的设置的原因. 在SDK 11的设备上,会使用第二处,SDK 14会使用第三处. 只需要将三个 ...
- UIApplication详解再解-备
每个app有且只有一个UIApplication对象,当程序启动的时候通过调用UIApplicationMain方法得到的.可以通过sharedApplication方法得到. UIApplicati ...
- Lintcode--008(编辑距离)
http://www.lintcode.com/en/problem/edit-distance/ 2016-08-29 给出两个单词word1和word2,计算出将word1 转换为word2的最少 ...
- 安装notepad++之后怎样在鼠标右键上加上Edit with notepad++
在鼠标右键上加入使用notepad++编辑 我们在安装完notepad++文本编辑器之后,在一个文本文件上右键有时候并没有出现"使用notepad++编辑的选项",我们可以通过简单 ...
- android开发
从某种意义上讲,垃圾收集机制把程序员从“内存管理噩梦”中解放出来,而 Android 的进程生命周期管理机制把用户从“任务管理噩梦”中解放出来.我见过一些 Nokia S60 用户和 Windows ...
- VS2010安装与测试编译问题(fatal error LNK1123: failure during conversion to COFF: file invalid or corrupt)
由于第三方库的各种原因,与编译冲突问题,公司又决定把整个项目都统一改用VS2010来编译.所以我把我开发机上的VS2008卸载了,又重新安装了VS2010.无奈出现了COFF格式转换问题.搜索了下.完 ...
- BZOJ2750: [HAOI2012]Road
2750: [HAOI2012]Road Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 261 Solved: 113[Submit][Status ...
- 关于Set Nocount ON的性能 |c#调用存储过程的返回值总是-1
原文地址:http://www.tuicool.com/articles/qe6BZbR 也许因为它太过于简单,自己一直没能好好关注这个语句,只记得"只是"提高点性能而已.有时会在存储过程中写上几句, ...