• HDOJ 1009. Fat Mouse' Trade 贪心 结构体排序


    FatMouse' Trade

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 56784    Accepted Submission(s): 19009
    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
     
    Author
    CHEN, Yue
     
    Source
     

     

            简单水题,相同的题目有:HDOJ 1009、ZOJ 2109。
            题目大意:胖老鼠拿M磅猫粮去贿赂守卫仓库的猫,仓库有N个房间,第i个房间有J[i]磅JavaBean且需要贿赂F[i]磅猫粮。可以选择只要J[i]*a%磅JavaBean,这样只需要贿赂F[i]*a%磅猫粮。请计算可以换取JavaBean的最大数量。
            我们可以用P[i]来表示每磅猫粮能换多少JavaBean,即J[i]/F[i],可以看做性价比。对P排个序,优先选择性价比最高的进行交易,算是贪心算法的思想。需要用到结构体排序。
     
     1 #include <stdio.h>
     2 #include <algorithm>
     3 struct room{int J,F;double P;}a[1010];
     4 bool cmp(const room&a, const room&b){return a.P>b.P;}
     5 
     6 int main()
     7 {
     8     int M, N;
     9     while(scanf("%d%d",&M,&N)&&~M) {
    10         for(int i=0; i<N; i++) {
    11             scanf("%d%d", &a[i].J, &a[i].F);
    12             a[i].P=(double)a[i].J/a[i].F;
    13         }
    14         std::sort(a,a+N,cmp);
    15         double maxf=.0;
    16         for(int i=0; i<N; i++)
    17             if(M>a[i].F) {
    18                 M-=a[i].F;
    19                 maxf+=a[i].J;
    20             }else{
    21                 maxf+=(double)a[i].J * M /a[i].F;
    22                 break;
    23             }
    24         printf("%.3f
    ",maxf);
    25     }
    26     return 0;
    27 }

     

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  • 原文地址:https://www.cnblogs.com/BlackStorm/p/4940846.html
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