HDOJ----------1009

题目：

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 52127    Accepted Submission(s):
17505

Problem Description

FatMouse prepared M pounds of cat food, ready to trade
with the cats guarding the warehouse containing his favorite food,
JavaBean.
The warehouse has N rooms. The i-th room contains J[i] pounds of
JavaBeans and requires F[i] pounds of cat food. FatMouse does not have to trade
for all the JavaBeans in the room, instead, he may get J[i]* a% pounds of
JavaBeans if he pays F[i]* a% pounds of cat food. Here a is a real number. Now
he is assigning this homework to you: tell him the maximum amount of JavaBeans
he can obtain.

Input

The input consists of multiple test cases. Each test
case begins with a line containing two non-negative integers M and N. Then N
lines follow, each contains two non-negative integers J[i] and F[i]
respectively. The last test case is followed by two -1's. All integers are not
greater than 1000.

Output

For each test case, print in a single line a real
number accurate up to 3 decimal places, which is the maximum amount of JavaBeans
that FatMouse can obtain.

Sample Input

5 3

7 2

4 3

5 2

20 3

25 18

24 15

15 10

-1 -1

Sample Output

13.333

31.500

分析：本题用到的是贪心算法，猫粮换成Java豆的比例越大应越先被兑换。在此我通过每个结构体保存每个屋子中的J[i]与F[i]及其兑换比例scale,然后利用sort将所有结构体中的scale按从大到小进行排序。之所以这样做的原因是，根据scale排序后，不会打乱原先每个屋子J[i]与F[i]及其兑换比例scale的对应关系,即排序的过程中结构体的结构不发生变换，只不过是根据结构体中的scale变量给所有结构体排一下序而已。最后的换Java豆的工作也简单多了。

代码如下：

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

 const int maxN =  + ;

 struct warehouse{
     int J;
     int F;
     double scale;
 }House[maxN];

 bool cmp(const struct warehouse a, const struct warehouse b) {
     return a.scale > b.scale;
 }

 int main() {
     int M, N;
     double ans;
     while(scanf("%d %d", &M, &N) == ){
         if(M == - && N == -) break;
         //输入
         for(int i = ; i < N; i++) {
             scanf("%d %d", &House[i].J, &House[i].F);
             House[i].scale = (double)House[i].J/House[i].F;
         }
         //将所有屋子中的猫粮与Java豆兑换的比例排序
         sort(House, House + N, cmp);
 //        for(int i = 0; i < N; i++)
 //            printf("%.3lf\t", House[i].scale);
         //按比例从大到小分配猫粮
         ans = 0.0;
         int pos = ;
         while(M >  && N > ){//猫粮换完，或者Java豆已经没有时应该终止循环
             if(M > House[pos].F)
                 ans += House[pos].J;        //若猫粮充足，直接将屋子的Java豆兑换下来
             else
                 ans += (double)House[pos].J * M / House[pos].F; //能兑换的猫粮不足，这时应该按比例来兑换Java豆
             M -= House[pos].F;
             N--;
             pos++;//到下一家
         }
         //输出
         printf("%.3lf\n", ans);
     }
     return ;
 }

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