贪心初步-FatMouse' Trade
这道题是真的很简单,只需要用优先队列以单位价值作为排序规则就可以解决。
FatMouse' Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 46804 Accepted Submission(s): 15704
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
Author
CHEN, Yue
Source
Recommend
JGShining
#include <cstdio> #include<queue> #define maxn 1005; using namespace std; struct Food{ int hav,ned; double va; }; bool operator < (struct Food a,struct Food b){ if(a.va < b.va) return true; else return false; } int main(){ int m,n; struct Food ha[1005]; while(scanf("%d %d",&m,&n) && m!=-1&&n!=-1){ priority_queue<struct Food> lis; for(int i = 0;i < n;i++){ scanf("%d %d",&ha[i].hav,&ha[i].ned); } for(int i = 0;i < n;i++){ ha[i].va = ha[i].hav*1.0/ha[i].ned; lis.push(ha[i]); } double cnt = 0.0; while(!lis.empty() && m){ struct Food t; t = lis.top(); if(m >= t.ned){ m -= t.ned; cnt += t.hav; lis.pop(); }else{ cnt += t.va * m; m = 0; lis.pop(); } } printf("%.3f\n",cnt); } return 0; }