HDU1009 貪心思想

標籤：貪心思想

FatMouse‘ TradeTime Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64uSubmit Status Practice HDU 1009

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 37 24 35 220 325 1824 1515 10-1 -1 

 

Sample Output

 13.33331.500 

 

這個題的大意是：它有M貓食，N個房間，每個房間有f[i]貓食，j[i]量的 javabean，按一定的比例拿貓食來保護javabean。問一共有M貓食最多能保護多少javabean。

簡單的貪心，只需按每個房間的javabean和貓食的比例從高到底排序，然後貪心即可。

代碼中我處理了輸入時f[i]等於0的情況，但交上去WA，我刪去之後就AC了，看來沒必要處理啊，，

#include <stdio.h>#include <string.h>#include <math.h>#include <algorithm>using namespace std;struct food{int f,j;double ave;}s[1010];int cmp(const food &a,const food &b){return a.ave>b.ave;}int main(){int i,k,n,m;double ans;while(scanf("%d%d",&m,&n)){if(m==-1 && n==-1)break;for(i=0;i<n;i++){scanf("%d%d",&s[i].j,&s[i].f);s[i].ave=(double)s[i].j/s[i].f;}sort(s,s+n,cmp);ans=0;for(i=0;i<n;i++){if(s[i].f<=m){ans+=(double)s[i].j;m-=s[i].f;}else{ans+=s[i].ave*m;break;}}printf("%.3f\n",ans);}return 0;}

HDU1009 貪心思想