1022. Poor contestant Prob
Constraints
Time Limit: 1 secs, Memory Limit: 32 MB
Description
As everybody known, “BG meeting” is very very popular in the ACM training team of ZSU.
After each online contest, they will go out for “smoking”. Who will be the poor ones that have to BG the others? Of course, the half who solve less problems.
The rule runs well when the number of the contestants is even. But if the number is odd, it is impossible to divide them into two equal parts. It gives a dilemma to the BG meeting committee. After a careful discussion with Mr. Guo, a new rule emerged: if the number of the contestant is odd, the committee will first sort the contestants according to the number of problems they solved, and then they will pick out the middle one. This poor boy or girl will have no chance to attend the BG meeting.
Strange rule, isn`t it?
As the number of the contestants is becoming more and more large, the committee need to write a program which will pick out the poor one efficiently.
Note that: Every contestant solves different number of problems. The total number of the contestants will not exceed 10^5.
Input
There are several cases in the input. The first line of the input will be an integer M, the number of the cases.
Each case is consisted of a list of commands. There are 3 types of commands.
1. Add xxx n : add a record to the data base, where xxx is the name of the contestant, which is only consisted of at most 10 letters or digits, n is the number of problems he/she solved. (Each name will appear in Add commands only once).
2.Query :
3.End :End of the case.
Output
1.For the Query command: If the current number of contestants is odd, the program should output the poor contestant’s name currently even if there is only one contestants, otherwise, just out put “No one!” (without quotes).
2.For the End command:
If the total number of contestants in the data base is even, you should out put “Happy BG meeting!!”(without quotes),otherwise, you should out put the “xxx is so poor. ”(without quotes) where xxx is the name of the poor one.
3.Each case should be separated by a blank line.
Sample Input
2 Add Magicpig 100 Add Radium 600 Add Kingfkong 300 Add Dynamic 700 Query Add Axing 400 Query Add Inkfish 1000 Add Carp 800 End Add Radium 100 Add Magicpig 200 End
Sample Output
No one! Axing Radium is so poor. Happy BG meeting!!
Problem Source
ZSUACM Team Member
这个题目思路并不复杂，就是建两个优先队列，一个最大堆，一个最小堆，在插入过程不断维护，使得两个堆size尽可能相同。
问题就是，做完之后会超时，百度了一下，发现是那个scanf和printf能更省时（蛋疼，我真不爱用这个）。然后还得把所有的string都变成char[]
终于给过了。。。。。。
#include <iostream>
#include <queue>
#include <cstring>
#include <vector>
#include <cstdio>
#include <string>
using namespace std;
struct contestant
{
string name;
int num;
};
//建立一个最大堆和最小堆(优先队列实现)
struct Greater
{
bool operator()(const contestant &a,const contestant &b)
{
return a.num < b.num;
}
};
struct Less
{
bool operator()(const contestant &a,const contestant &b)
{
return a.num > b.num;
}
};
int main()
{
int M;
scanf("%d",&M);
while(M--)
{
priority_queue<contestant,vector<contestant>,Less> MIN;
priority_queue<contestant,vector<contestant>,Greater> MAX;
//string command;
char command[20];
while(scanf("%s",command))
{
if(strcmp(command,"Add")==0)
{
char name1[20];
scanf("%s",name1);
string name(name1);
int number;
scanf("%d",&number);
const contestant tmp={name,number};
if(MAX.size()==MIN.size())
{
if(MAX.empty() || tmp.num < MIN.top().num)
MAX.push(tmp);
else if(tmp.num > MIN.top().num)
{
MAX.push(MIN.top());
MIN.pop();
MIN.push(tmp);
}
}
else if(MAX.size() > MIN.size())
{
if(tmp.num > MAX.top().num)
MIN.push(tmp);
else if(tmp.num < MAX.top().num)
{
MIN.push(MAX.top());
MAX.pop();
MAX.push(tmp);
}
}
}
else if(strcmp(command,"Query")==0)
{
if(MAX.size() == MIN.size())
printf("No one!
");
else
{
printf("%s
",MAX.top().name.c_str());
}
}
else if(strcmp(command,"End")==0)
{
if(MAX.size() == MIN.size())
printf("Happy BG meeting!!
");
else printf("%s is so poor.
",MAX.top().name.c_str());
break;
}
}
if(M != 0)
printf("
");
}
return 0;
}