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!
题目大意：
给定N只Dargon，杀死他们分别要Ni的费用
给定M个Knight， 他们的能力分别为Mi，每个Knight只能雇佣一次。
雇佣的cost 与 Mi 相同
问能杀死所有的Dragon的最少费用。
若不能杀死，输出Loowater is doomed!
1 /*************************************************
2 Copyright: Call_Me_BIGBALLON
3 Author: BIGBALLON
4 Date: 2014-05-02
5 Description: UVa 11292(贪心)
6 **************************************************/
7
8 #include<cstdio>
9 #include<cstring>
10 #include<algorithm>
11
12 using namespace std;
13 #define CLR(a,b) memset(a,b,sizeof(a))
14
15 int M,N;
16 int Dragon[20002],Knight[20002];
17 int sum1,sum2,res;
18
19 int main()
20 {
21 while(scanf("%d %d",&N,&M)!=EOF)
22 {
23 if(N == 0 && M == 0)
24 break;
25 sum1 = sum2 = 0;
26 for(int i=0;i<N;i++)
27 {
28 scanf("%d",&Dragon[i]);
29 sum1 += Dragon[i];
30 }
31 for(int i=0;i<M;i++)
32 {
33 scanf("%d",&Knight[i]);
34 sum2 += Knight[i];
35 }
36 if(M<N)
37 {
38 printf("Loowater is doomed!\n");
39 continue;
40 }
41 else if(sum1 > sum2)
42 {
43 printf("Loowater is doomed!\n");
44 continue;
45 }
46 else
47 {
48 sort(Dragon,Dragon+N);
49 sort(Knight,Knight+M);
50 int cur = 0;
51 int cost = 0;
52 for(int i=0;i<M;i++)
53 {
54 if(Dragon[cur] <= Knight[i])
55 {
56 cur++;
57 cost += Knight[i];
58 }
59 if(cur == N)
60 break;
61 }
62 if(cur < N) printf("Loowater is doomed!\n");
63 else printf("%d\n",cost);
64 }
65
66 }
67
68 return 0;
69 }
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