HDOJ----------1009，hdoj1003

HDOJ----------1009，hdoj1003

題目：

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：本題用到的是貪心算法，貓糧換成Java豆的比例越大應越先被兌換。在此我通過每個結構體保存每個屋子中的J[i]與F[i]及其兌換比例scale,然後利用sort將所有結構體中的scale按從大到小進行排序。之所以這樣做的原因是，根據scale排序後，不會打亂原先每個屋子J[i]與F[i]及其兌換比例scale的對應關系,即排序的過程中結構體的結構不發生變換，只不過是根據結構體這種的scale變量給所有結構體排一下序而已。最後的換Java豆的工作也簡單多了。   代碼如下：

 1 #include<cstdio>
 2 #include<algorithm>
 3 using namespace std;
 4 
 5 const int maxN = 1000 + 5;
 6 
 7 struct warehouse{
 8     int J;
 9     int F;
10     double scale;
11 }House[maxN];
12 
13 bool cmp(const struct warehouse a, const struct warehouse b) {
14     return a.scale > b.scale;
15 }
16 
17 int main() {
18     int M, N;
19     double ans;
20     while(scanf("%d %d", &M, &N) == 2){
21         if(M == -1 && N == -1) break;
22         //輸入
23         for(int i = 0; i < N; i++) {
24             scanf("%d %d", &House[i].J, &House[i].F);
25             House[i].scale = (double)House[i].J/House[i].F;
26         }
27         //將所有屋子中的貓糧與Java豆兌換的比例排序
28         sort(House, House + N, cmp);
29 //        for(int i = 0; i < N; i++)
30 //            printf("%.3lf\t", House[i].scale);
31         //按比例從大到小分配貓糧
32         ans = 0.0;
33         int pos = 0;
34         while(M > 0 && N > 0){//貓糧換完，或者Java豆已經沒有時應該終止循環
35             if(M > House[pos].F)
36                 ans += House[pos].J;        //若貓糧充足，直接將屋子的Java豆兌換下來
37             else
38                 ans += (double)House[pos].J * M / House[pos].F; //能兌換的貓糧不足，這時應該按比例來兌換Java豆
39             M -= House[pos].F;
40             N--;
41             pos++;//到下一家
42         }
43         //輸出
44         printf("%.3lf\n", ans);
45     }
46     return 0;
47 }

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