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.
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:
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 //输入 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 //将所有屋子中的猫粮与Java豆兑换的比例排序 28 sort(House, House + N, cmp); 29 // for(int i = 0; i < N; i++) 30 // printf("%.3lf\t", House[i].scale); 31 //按比例从大到小分配猫粮 32 ans = 0.0; 33 int pos = 0; 34 while(M > 0 && N > 0){//猫粮换完,或者Java豆已经没有时应该终止循环 35 if(M > House[pos].F) 36 ans += House[pos].J; //若猫粮充足,直接将屋子的Java豆兑换下来 37 else 38 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 39 M -= House[pos].F; 40 N--; 41 pos++;//到下一家 42 } 43 //输出 44 printf("%.3lf\n", ans); 45 } 46 return 0; 47 }
2015-07-02文