Problem A: Auction
Recently the auction house has introduced a new type of auction, the lowest price auction. In this new system, people compete for the lowest bid price, as opposed to what they did in the past. What an amazing thing! Now you could buy cool stuff with one penny. Your task is to write the software to automate this auction system.
First the auctioneer puts an upper limit on bid price for each item. Only positive price less than or equal to this price limit is a valid bid. For example, if the price limit is 100, then 1 to 100, inclusive, are all valid bid prices. Bidder can not put more than one bid for the same price on a same item. However they can put many bids on a same item, as long as the prices are different. After all bids are set, the auctioneer chooses the winner according to the following rules:
1. If any valid price comes from only one bidder, the price is a "unique bid". If there are unique bids, then the unique bid with the lowest price wins. This price is the winning price and the only bidder is the winning bidder.
2. If there are no unique bids, then the price with fewest bids is the winning bid. If there are more than one price which has the same lowest bid count, choose the lowest one. This price is the winning price. The bidder who puts this bid first is the winning bidder.
Given the price limit and all the bids that happen in order, you will determine the winning bidder and the winning price.
Input Description
Standard input will contain multiple test cases. The first line of the input is a single integer T (1 <= T <= 10) which is the number of test cases. T test cases follow, each preceded by a single blank line.
The first line of each test case contains two integers: U (1 <= U <= 10,000), the price upper limit and M (1 <= M <= 100,000), the total number of bids. M lines follow, each of which presents a single bid. The bid contains the bidder's name (consecutive non-whitespace characters) and the price P (1 <= P <= U), separated with a single space. All bids in the input are guaranteed to be valid ones.
Output Description
For each test case, print the sentence "The winner is W." on the first line, and "The price is P." on the second. Replace W and P with the winning bidder’s name and the winning price.
Sample Input
2
3 3
Alice 1
Bob 2
Carl 3g
3 6
Alice 1
Alice 2
Alice 3
Bob 1
Bob 3
Carl 3
Sample Output
Case 1:
The winner is Alice.
The price is 1.
Case 2:
The winner is Alice.
The price is 2.
题目如上，想了一下，写了这样的一个算法：
#include <string>
#include <vector>
#include <algorithm>
#include <list>
using namespace std;
struct bid_info
{
string name;
int value;
int order;
bool operator ==(const struct bid_info& bi) const {
if(value == bi.value) return true;
return false;
}
};
typedef vector
<struct bid_info> bid_list; bool comp_bid(const struct bid_info &b1 , const struct bid_info & b2 ){
if(b1.value == b2.value)
return b1.order == b2.order;
return b1.value < b2.value;
}
void output(int i , bid_list::iterator it)
{
cout <<"Case " << i <<":" <<endl;
cout <<"The winner is " << it->name <<"." << endl;
cout <<"The price is " << it->value <<"." <<endl;
}
void find_winner(int i, bid_list & list)
{
bid_list::iterator first = list.begin();
while(first!=list.end())
{
if(count(list.begin(),list.end(),*first) == 1) {
output(i,first);
return ;
} else first++;
}
output(i,list.begin());
} void print(struct bid_info & bidinfo)
{
cout <<bidinfo.name <<" " << bidinfo.value <<endl;
}
int main()
{
int i,n ,j;
int up, m ;
bid_list list;
struct bid_info bidinfo;
cin
>> n ;for(i = 0 ; i < n ; i ++ )
{
cin >> up >> m ;
list.erase(list.begin(),list.end());
for(j = 0 ; j < m ; j ++)
{
cin >> bidinfo.name;
cin >> bidinfo.value;
if(bidinfo.value <= up && bidinfo.value > 0 )
list.push_back(bidinfo);
}
stable_sort(list.begin(),list.end(), comp_bid);
//for_each(list.begin(),list.end(),print);
cout <<endl;
find_winner(i+1,list);
}
return 0;
}
这个算法耗时的地方主要在sort 和 find_winner 中对于每个元素进行遍历。 find_winner 的时间复杂度应该是O(n^2) 。 根据评测获胜者的规则以及输入的特殊性，可以改进如下：
#include <string> using namespace std;
#define NMAX 10000
struct bid_info
{
string name;
int count;
};
void output(int i , int value, string name)
{
cout <<endl;
cout <<"Case " << i << ":" <<endl;
cout <<"The winner is " << name <<"." <<endl ;
cout <<"The price is " << value <<"." <<endl;
}
int main()
{
struct bid_info bids[NMAX];
int i,j;
int up, n,m ;
int value;
int index;
string name;
bool bfirst ;
cin
>> n ;for(i = 0 ; i < n ; i ++ )
{
for(j = 0 ; j < NMAX ; j ++)
bids[j].count = 0 ;
cin >> up >> m ;
for(j = 0 ; j < m ; j ++)
{
cin >> name;
cin >> value;
if(value <= up && value > 0 )
{
bids[value].count ++;
if(bids[value].count == 1)
bids[value].name = name;
}
}
bfirst = false;
for(j = 0 ; j < NMAX ; j ++)
{
if(bids[j].count == 0) continue;
else if(bids[j].count == 1) {
bfirst = false;
output( i + 1 , j , bids[j].name);
break;
} else if(bfirst == false) {
bfirst = true;
index = j;
}
}
if(bfirst == true)
{
output(i +1, index, bids[index].name);
}
}
return 0;
}
对于每种价格只是保持第一个bid 的信息。通过scan 整个数组可以找出获胜者！