A Contesting Decision POJ 1581 ZOJ 1764 UVA 2832

本文介绍了一个用于自动评估编程竞赛成绩的软件系统,该系统旨在解决传统评分过程中的复杂性和不可靠性问题。通过模块化设计,系统能准确计算参赛队伍解决问题的数量及罚时,最终确定胜者。

Judging a programming contest is hard work, with demanding contestants, tedious decisions,and monotonous work. Not to mention the nutritional problems of spending 12 hours with only donuts, pizza, and soda for food. Still, it can be a lot of fun.
Software that automates the judging process is a great help, but the notorious unreliability of some contest software makes people wish that something better were available. You are part of a group trying to develop better, open source, contest management software, based on the principle of modular design. 
Your component is to be used for calculating the scores of programming contest teams and determining a winner. You will be given the results from several teams and must determine the winner. 
Scoring 
There are two components to a team's score. The first is the number of problems solved. The second is penalty points, which reflects the amount of time and incorrect submissions made before the problem is solved. For each problem solved correctly, penalty points are charged equal to the time at which the problem was solved plus 20 minutes for each incorrect submission. No penalty points are added for problems that are never solved. 
So if a team solved problem one on their second submission at twenty minutes, they are charged 40 penalty points. If they submit problem 2 three times, but do not solve it, they are charged no penalty points. If they submit problem 3 once and solve it at 120 minutes, they are charged 120 penalty points. Their total score is two problems solved with 160 penalty points. 
The winner is the team that solves the most problems. If teams tie for solving the most problems,then the winner is the team with the fewest penalty points.

Input

For the programming contest your program is judging, there are four problems. You are guaranteed that the input will not result in a tie between teams after counting penalty points. 
Line 1 < nTeams > 
Line 2 - n+1 < Name > < p1Sub > < p1Time > < p2Sub > < p2Time > ... < p4Time > 

The first element on the line is the team name, which contains no whitespace.Following that, for each of the four problems, is the number of times the team submitted a run for that problem and the time at which it was solved correctly (both integers). If a team did not solve a problem, the time will be zero. The number of submissions will be at least one if the problem was solved.

Output

The output consists of a single line listing the name of the team that won, the number of problems they solved, and their penalty points.

Sample Input

4
Stars 2 20 5 0 4 190 3 220
Rockets 5 180 1 0 2 0 3 100
Penguins 1 15 3 120 1 300 4 0
Marsupials 9 0 3 100 2 220 3 80

Sample Output

Penguins 3 475
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>

using namespace std;

struct Team {
	char name[100];
	int solveNum;
	int totalTime;

	 friend bool operator<(const Team& m, const Team& n){

        if(m.solveNum != n.solveNum)
        {
			return m.solveNum < n.solveNum;
		}

		if (m.solveNum == n.solveNum)
		{
			return m.totalTime > n.totalTime;
		}

		return false;
	}
} teams;

int main() 
{
	int n;
	scanf("%d", &n);

    priority_queue<Team> q;

	int i, j;
	for (i = 0; i < n; ++i) {
		scanf("%s", teams.name);
		teams.solveNum = 0;
		teams.totalTime = 0;

		for (j = 0; j < 4; ++j) {
			int a, b;
			scanf("%d %d", &a, &b);

			if (b) {
				teams.solveNum += 1;
				teams.totalTime += ((a - 1) * 20 + b);
			}
            q.push(teams);
		}

	}
    Team t = q.top();
	printf("%s %d %d\n", t.name, t.solveNum, t.totalTime);

	return 0;
}

 

Task 3: Simulate Election (4 marks) To select the winner of the election, instant-runoff voting (IRV) is used. This method selects a winner over the course of a number of rounds. The following procedure is repeated until there is only one candidate left: Assign each ballot to the candidate who is ranked highest on that ballot and who has not yet been eliminated. Eliminate the candidate(s) with the lowest number of assigned ballots in that round, so they are not counted in future rounds. For this task, you must run a simulated IRV for an election given a list of valid positional ballots. Write a function simulate_election(ballots, parties) You need to write a function that returns the full history of elimination rounds for the given election. Arguments: ballots: a list of valid positional ballots. parties: the ordered list of parties contesting the election. Returns: a list of elimination round history, where each round consists of a tuple with: A dictionary mapping each remaining candidate to their vote total A set of candidates to be eliminated after that round. Details: If two candidates are tied for the lowest vote total in a round, they are both eliminated simultaneously If there is no unique winner, your program should raise a TiedElectionError as defined at the top of the provided code. As a reminder, you can raise an error using the built in Python statement raise TiedElectionError. This is a custom Exception type that we have defined for you. You can use it like any other Exception , and there is no need to understand how we have defined it. Interested students can learn more about custom object and classes in this optional worksheet. Example Calls:
最新发布
10-17
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