HDOJ 1009（貪心）

題意描述

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.

思路

一道貪心的題目，我們可以按他的收益率來排序，每次選收益率最高的那一個。通過看其他人的做法，學會了自定義排序的方法。

bool cmp(Node x,Node y){
	return x.rate>y.rate;
}

通過重寫cmp函數，可以實現對node結構體按收益率降序排序。

AC代碼

#include<vector>
#include<iostream>
#include<cstring> 
#include<queue>
#include<set>
#include<algorithm>
#define IOS ios::sync_with_stdio(false); cin.tie(0); 
using namespace std;
typedef pair<int,int> PII;
typedef long long ll;
const int N=1005;
const int INF=0x3f3f3f3f; 
int m,n;
int f[N],j[N];
typedef struct Node{
	int j;
	int f;
	double rate;
};
Node node[N];
bool cmp(Node x,Node y){
	return x.rate>y.rate;
}
int main() {
	IOS;
	while(cin>>m>>n&&m!=-1&&n!=-1){
		for(int i=0;i<n;i++){
			int a,b;
			cin>>a>>b;
			node[i].j=a,node[i].f=b;
			node[i].rate=(double)a/b;
		}
		sort(node,node+n,cmp);
		double sum=0;
		for(int i=0;i<n;i++){
			if(m>node[i].f){
				sum+=node[i].j;
				m-=node[i].f;
			}else{
				sum+=node[i].rate*m;
				m=0;
			}
		}
		printf("%.3f\n",sum)
	}
    return 0;
}