[UVA][greedy] 11292 - Dragon of Loowater＠Morris' Blog｜PChome Online 人新台

Morris' Blog 我在摸索我存在的意、生命的意、涵及值。有真正神手般的害，套用模式是大部分的情。那害的我存在的意何，托出神的存在？有候不能怪人不我，我已始省思，什我有人的原因，而是全要靠人。有候我我不同，那是因我做不到很多人能到的事情。我不到、玩不起、得不明、理解得不多，我缺少的文能力，就是一切一切能的判定。此，我不得能做些人有助的事情。我知道我自己是不服的性格，但在我不得不。我有能力，因此我必。我到底世界作什？充不定性，而我是法出拔萃，想一般人似乎也回不了了。越大的挑需要越力的支持，而我的支持是什？最近人，但都是失的局，人能接受，不需要言，不需要通，一切所想都表不出。
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2012-12-17 22:08:25| 人759| 回0 | 上一篇 | 下一篇

[UVA][greedy] 11292 - Dragon of Loowater

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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!

Ondej Lhoták

路上搜到的翻如下。

你的王国里有一条n个头的恶龙，你希望雇一些骑士把它杀死（即砍掉所有头）。村里有m个骑士可以雇佣，一个能力值为x的骑士可以砍掉恶龙一个直径不超过x的头，且需要支付x个金币。如何雇佣骑士才能砍掉恶龙的所有头，且需要支付的金币最少？注意，一个骑士只能砍一个头（且不能被雇佣两次）。

#include <stdio.h>
#include <algorithm>
using namespace std;

int main() {
    int n, m, i;
    int A[20000], B[20000];
    while(scanf("%d %d", &n, &m) == 2) {
        if(!n && !m)    break;
        if(n > m) {
            puts("Loowater is doomed!");
            continue;
        }
        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 idx = 0, sum = 0;
        for(i = 0; i < n; i++) {
            while(idx < m) {
                if(B[idx] >= A[i])
                    break;
                idx++;
            }
            if(idx == m)    break;
            sum += B[idx];
            idx++;
        }
        if(i == n)
            printf("%d\n", sum);
        else
            puts("Loowater is doomed!");
    }
    return 0;
}