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;
}
