ZOJ _3607_Lazier Salesgirl(贪心)
2013-05-29 23:38
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题型:贪心
题目:
Description
Kochiya Sanae is a lazy girl who makes and sells bread. She is an expert at bread making and selling. She can sell the i-th customer a piece of bread for price pi. But she is so lazy that she will fall asleep if no customer
comes to buy bread for more than w minutes. When she is sleeping, the customer coming to buy bread will leave immediately. It's known that she starts to sell bread now and the i-th customer come after ti minutes.
What is the minimum possible value of w that maximizes the average value of the bread sold?
Input
There are multiple test cases. The first line of input is an integer T ≈ 200 indicating the number of test cases.
The first line of each test case contains an integer 1 ≤ n ≤ 1000 indicating the number of customers. The second line contains n integers 1 ≤ pi ≤ 10000. The third line contains n integers 1 ≤ ti ≤
100000. The customers are given in the non-decreasing order of ti.
Output
For each test cases, output w and the corresponding average value of sold bread, with six decimal digits.
Sample Input
Sample Output
题意:
一个小姑娘非常懒但是卖面包非常厉害。她能在以Pi的价格将面包卖给第i个人。但是她一旦在w的时间段内木有顾客,她就会睡着,一觉不起。。。
现在给出买个每个顾客的价钱和相关时间点,现在想要求出在最少的w里卖出最大的面包价钱的平均值,问这一平均值和w是多少。
分析:
贪心问题,只需关心最符合题意的情况。但是有一些注意点:
1、当前对象的w是之前所有时间间隔中最大的,不然就卖不到这了;
2、当前对象的w一定要比卖到下一个顾客的时间间隔要小,否则就会买个下一个人,就平均值不一定是最小了。
方法,用一个结构体分别存入卖到当前顾客的平均值、w、和到下一个人的时间间隔,然后由大到小排序,最后从头找到符合w<下一个人的时间间隔的即可。
代码:
题目:
Description
Kochiya Sanae is a lazy girl who makes and sells bread. She is an expert at bread making and selling. She can sell the i-th customer a piece of bread for price pi. But she is so lazy that she will fall asleep if no customer
comes to buy bread for more than w minutes. When she is sleeping, the customer coming to buy bread will leave immediately. It's known that she starts to sell bread now and the i-th customer come after ti minutes.
What is the minimum possible value of w that maximizes the average value of the bread sold?
Input
There are multiple test cases. The first line of input is an integer T ≈ 200 indicating the number of test cases.
The first line of each test case contains an integer 1 ≤ n ≤ 1000 indicating the number of customers. The second line contains n integers 1 ≤ pi ≤ 10000. The third line contains n integers 1 ≤ ti ≤
100000. The customers are given in the non-decreasing order of ti.
Output
For each test cases, output w and the corresponding average value of sold bread, with six decimal digits.
Sample Input
2 4 1 2 3 4 1 3 6 10 4 4 3 2 1 1 3 6 10
Sample Output
4.000000 2.500000 1.000000 4.000000
题意:
一个小姑娘非常懒但是卖面包非常厉害。她能在以Pi的价格将面包卖给第i个人。但是她一旦在w的时间段内木有顾客,她就会睡着,一觉不起。。。
现在给出买个每个顾客的价钱和相关时间点,现在想要求出在最少的w里卖出最大的面包价钱的平均值,问这一平均值和w是多少。
分析:
贪心问题,只需关心最符合题意的情况。但是有一些注意点:
1、当前对象的w是之前所有时间间隔中最大的,不然就卖不到这了;
2、当前对象的w一定要比卖到下一个顾客的时间间隔要小,否则就会买个下一个人,就平均值不一定是最小了。
方法,用一个结构体分别存入卖到当前顾客的平均值、w、和到下一个人的时间间隔,然后由大到小排序,最后从头找到符合w<下一个人的时间间隔的即可。
代码:
#include <vector> #include <list> #include <map> #include <set> #include <deque> #include <stack> #include <queue> #include <bitset> #include <algorithm> #include <functional> #include <numeric> #include <utility> #include <iostream> #include <iomanip> #include <cstdio> #include <cmath> #include <cstdlib> #include <ctime> #include <string> #include <cstring> #include <cctype> #include <assert.h> using namespace std; struct node{ double ave;//平均值 double w; double nextT;//到下一个顾客的时间间隔 }a[1234]; double cmp(node a,node b){//按比较的优先级写比较函数 if(a.ave==b.ave){ return a.w<b.w;//当ave相同时,w小的优先 } else return a.ave>b.ave; } int main(){ int t,n; double price[1234],time[1234]; while(~scanf("%d",&t)){ while(t--){ scanf("%d",&n); for(int i=0;i<n;i++){ scanf("%lf",&price[i]); } for(int i=0;i<n;i++){ scanf("%lf",&time[i]); } double sum=0; for(int i=0;i<n;i++){ sum+=price[i]; a[i].ave=sum/(i+1); if(i==0){ a[i].w=time[i]; } else a[i].w=time[i]-time[i-1]; if(i==n-1){ a[i].nextT=999999;//最后一个顾客后的时间定为无限大 } else a[i].nextT=time[i+1]-time[i]; } int max=a[0].w; for(int i=0;i<n;i++){ if(max>a[i].w){ a[i].w=max; } else{ max=a[i].w; } } sort(a,a+n,cmp); /*for(int i=0;i<n;i++){ printf("***%.3lf %d %d\n",a[i].ave,a[i].w,a[i].nextT); }*/ for(int i=0;i<n;i++){ if(a[i].w<a[i].nextT){ printf("%.6lf %.6lf\n",a[i].w,a[i].ave); break; } } } } return 0; }
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