首页 > ACM题库 > UVA > UVA-11292-Dragon of Loowater[贪心]
2014
02-24

UVA-11292-Dragon of Loowater[贪心]

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!

白皮书的第一道题,用来练练手。

题目大意:王国里有n个恶龙。每个骑士能力值为x可以干掉头的直径不超过x的恶龙,需要支付x金币,每个骑士只能使用一次。求最小的花费。

输入:

第一行为m,n.  接下来n行为每个龙头的直径,接下来m行为骑士的能力值。

//============================================================================
// Name        : uva-11292.cpp
// Author      : coder
// Version     :
// Copyright   : www.acmerblog.com
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <algorithm>
using namespace std;

int main() {
	int n,m;
	int dragons[20010], knights[20010];
	while(cin >> n >> m, n && m){
		for(int i=0; i<n; i++) cin >> dragons[i];
		for(int i=0; i<m; i++) cin >> knights[i];
		sort(dragons,dragons+n);
		sort(knights, knights+m);
		int ans = 0, i,k=0;
		for(i=0; i<n && k<m; i++){
			//找到最小可以用的骑士
			while(k<m && knights[k++] < dragons[i]);
			if(k <= m)
				ans += knights[k-1];
		}
		//是否杀死所有的恶龙
		if(i == n) cout << ans << endl;
		else cout << "Loowater is doomed!" << endl;
	}
	return 0;
}

 


  1. 你的理解应该是:即使主持人拿走一个箱子对结果没有影响。这样想,主持人拿走的箱子只是没有影响到你初始选择的那个箱子中有奖品的概率,但是改变了其余两个箱子的概率分布。由 1/3,1/3 变成了 0, 2/3

  2. 可以参考算法导论中的时间戳。就是结束访问时间,最后结束的顶点肯定是入度为0的顶点,因为DFS要回溯

  3. 第二个方法挺不错。NewHead代表新的头节点,通过递归找到最后一个节点之后,就把这个节点赋给NewHead,然后一直返回返回,中途这个值是没有变化的,一边返回一边把相应的指针方向颠倒,最后结束时返回新的头节点到主函数。