UVa11292 The Dragon of Loowater(sort排序直接比较)

UVa11292

The Dragon of Loowater(sort排序)

Problem 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!
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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!’.
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Sample Input
2 3
5
4
7
8
4
2 1
5
5
10
0 0

Sample Output
11
Loowater is doomed!

题意：已知m个骑士kni[i]【是他的攻击和血量】、n个恐龙dra[i]，求骑士可以杀死所有恐龙同时保证每个骑士的血量大于恐龙攻击的最小血量之和。其中，每个骑士只能用一次。sort升序排序直接比较。

#include <algorithm>#include <iostream>#include <cstring>#include <cstdio>#include <ctime>#include <set>#include <map>#include <cmath>using namespace std;#define BUG puts("here!!!")#define W(a) while(a--)#define FRead() freopen("in", "r", stdin)#define FWrite() freopen("out", "w", stdout)#define Cls(a) memset((a), 0, sizeof(a))#define Clr(a, x) memset((a), (x), sizeof(a))#define pi 3.14159265359typedef long long LL;const int maxn = 20100;int n, h, k;int dra[maxn],kni[maxn];int main(){    int n,m;    while(~scanf("%d %d",&n,&m) && (n||m)){        memset(kni,0,sizeof(kni));        memset(dra,0,sizeof(dra));        for(int i=1; i<=n; i++){            scanf("%d",&dra[i]);        }        for(int i=1; i<=m; i++){            scanf("%d",&kni[i]);        }        sort(dra+1,dra+1+n);        sort(kni+1,kni+1+m);        int sum = 0;        int i,j;        for(i=1,j=1; i<=m; i++){            if(kni[i]>=dra[j] && j<=n){                sum += kni[i];                dra[j++] = 0;            }        }        if(dra[n]==0)            printf("%d\n",sum);        else             printf("Loowater is doomed!\n");    }    return 0;}