hdu 1070 milk
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Milk
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 38 Accepted Submission(s) : 21
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Problem Description
Here are some rules:
1. Ignatius will never drink the milk which is produced 6 days ago or earlier. That means if the milk is produced 2005-1-1, Ignatius will never drink this bottle after 2005-1-6(inclusive).
2. Ignatius drinks 200mL milk everyday.
3. If the milk left in the bottle is less than 200mL, Ignatius will throw it away.
4. All the milk in the supermarket is just produced today.
Note that Ignatius only wants to buy one bottle of milk, so if the volumn of a bottle is smaller than 200mL, you should ignore it.
Given some information of milk, your task is to tell Ignatius which milk is the cheapest.
Input
Each test case starts with a single integer N(1<=N<=100) which is the number of kinds of milk. Then N lines follow, each line contains a string S(the length will at most 100 characters) which indicate the brand of milk, then two integers for the brand: P(Yuan) which is the price of a bottle, V(mL) which is the volume of a bottle.
Output
Sample Input
22Yili 10 500Mengniu 20 10004Yili 10 500Mengniu 20 1000Guangming 1 199Yanpai 40 10000
Sample Output
MengniuMengniu
这道题也是感觉和背包没关系的,用结构排序就可以做了,根据价格排序,若价格相同则根据体积排序,注意若体积小于200,则应该人为将单价改成无限大。
#include<iostream>#include<cmath>#include<algorithm>#include<iomanip>#include<cstring>#include<cstdio>#include<queue>using namespace std;struct milk{ char na[111]; int price; int vol; double pp;};bool cmp1(milk x,milk y){ if(x.pp==y.pp)return x.vol>y.vol; return x.pp<y.pp;}int main(){ int n; cin>>n; while(n--) { int m; cin>>m; int i,j,k; milk st[111]; for(i=1;i<=m;i++) { cin>>st[i].na>>st[i].price>>st[i].vol; if(st[i].vol<200) st[i].pp=99999999; else if(st[i].vol<=1200) st[i].pp=st[i].price*1.0/(st[i].vol*1.0/200); else st[i].pp=st[i].price*1.0/(1200*1.0/200); } sort(st+1,st+1+m,cmp1); cout<<st[1].na<<endl; } return 0;}
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