【贪心】 hdu3979 Monster

Monster

http://acm.hdu.edu.cn/showproblem.php?pid=3979

Problem Description

One day, v11 encounters a group of monsters in a foreast. In order to defend the homeland, V11 picks up his weapon and fights!

All the monsters attack v11 at the same time. Every enemy has its HP, and attack value ATK. In this problem, v11 has his ATK and infinite HP. The damage (also means reduction for HP) is exactly the ATK the attacker has. For example, if v11's ATK is 13 and the monster's HP is 27, then after v11's attack, the monster's HP become 27 - 13 = 14 and vice versa.

v11 and the monsters attack each other at the same time and they could only attack one time per second. When the monster's HP is less or equal to 0 , we think this monster was killed, and obviously it would not attack any more. For example, v11's ATK is 10 and a monster's HP is 5, v11 attacks and then the monster is killed! However, a monster whose HP is 15 will be killed after v11 attack for two times. v11 will never stop until all the monsters are killed ! He wants to minimum the HP reduction for the fight! Please note that if in some second, some monster will soon be killed , the monster's attack will works too.

 

Input

The first line is one integer T indicates the number of the test cases. (T <=100)

Then for each case, The first line have two integers n (0<n<=10000), m (0<m<=100), indicates the number of the monsters and v11's ATK . The next n lines, each line has two integers hp (0<hp<=20), g(0<g<=1000) ,indicates the monster's HP and ATK.

 

Output

Output one line.

First output “Case #idx: ”, here idx is the case number count from 1. Then output the minimum HP reduction for v11 if he arrange his attack order optimal .

 

Sample Input

   
   
   
   

    
    
    
    2 
3 1 
1 10 
1 20 
1 40 
1 10 
7 3
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case #1: 110 
Case #2: 3
   
   
   
   

贪心，首先要计算出kill每个怪兽所需要的时间t，然后按怪兽的伤害和被杀所要的比率排序，当然优先斩杀比率高的怪兽了。

注意要用long long。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define LL long long
struct node
{
    LL hp;
    LL g;
    LL t;
}monster[10005];
bool cmp(const struct node &a,const struct node &b)
{
    return (double)a.g/a.t>(double)b.g/b.t;
}
int main()
{
    int T,n,m;
    scanf("%d",&T);
    for(int tt=1;tt<=T;++tt)
    {
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;++i)
        {
            scanf("%I64d%I64d",&monster[i].hp,&monster[i].g);
            monster[i].t=(monster[i].hp/m+((monster[i].hp%m)?1:0));
        }
        sort(monster,monster+n,cmp);
        LL reduction=0,t=0;
        for(int i=0;i<n;++i)
        {
            t+=monster[i].t;
            reduction+=(t*monster[i].g);
        }
        printf("Case #%d: %I64d\n",tt,reduction);
    }
    return 0;
}

来源： http://blog.csdn.net/acm_ted/article/details/7773137