UVA - 11292 Dragon of Loowater

Dragon of Loowater

Time Limit: 1000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

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Description

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!

---

Ondřej Lhoták

Source

Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: Problem Solving Paradigms :: Greedy ::  Involving Sorting (Or The Input Is Already Sorted)
Root :: AOAPC I: Beginning Algorithm Contests -- Training Guide (Rujia Liu) :: Chapter 1. Algorithm Design :: General Problem Solving Techniques ::  Examples
Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Problem Solving Paradigms ::  Greedy - Standard
Root :: Competitive Programming: Increasing the Lower Bound of Programming Contests (Steven & Felix Halim) :: Chapter 1. Introduction ::  Preview Contest
Root :: Prominent Problemsetters ::  Ondřej Lhoták

Root :: Competitive Programming: Increasing the Lower Bound of Programming Contests (Steven & Felix Halim) :: Chapter 3. Problem Solving Paradigms ::  Greedy

题意：

n条恶龙，m个勇士，用勇士来杀恶龙。一个勇士只能杀一个恶龙。而且勇士只能杀直径不超过自己能力值的恶龙。每个勇士需要支付能力值一样的金币。

问杀掉所有恶龙需要的最少金币。

从小到大排序+每次贪心，很简单的模拟题。代码：

#include <iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=20000+5;
int a[maxn],b[maxn];
int main()
{
    int t,n,m,i,j;
    while(scanf("%d%d",&n,&m)!=EOF&&n&&m)
    {
        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 c=0;
        for(i=0,j=0;i<n;i++,j++)
        {
            while(b[j]<a[i])
            {
                j++;
            }
            if(j>=m) break;
            c+=b[j];
        }
        if(i<n) printf("Loowater is doomed!\n");
        else printf("%d\n",c);
    }
    return 0;
}





题目本身到此结束，主要是学到了一种uva提交题目的方法。链接：http://uhunt.felix-halim.net/id/426831（建议用Google浏览器打开，360好像不友好）

在这里可以轻松提交代码，且很快返回结果（通过你在uva的邮件）。妈妈再也不用担心我在uva等几分钟出结果了。下面是这个题的邮件内容：

Hi,

This is an automated response from UVa Online Judge.

Your submission with number  14079111 for the problem  11292 - Dragon of Loowater has succeeded with verdict  Accepted.

Congratulations! Now it is time to try a new problem.

Best regards,

The UVa Online Judge team