【CS50x】Runoff 题解(下)

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前言

CS50x 是哈佛大学推出的一门知名公开课,本课程是一门计算机科学的导论课程,适合于对计算机科学感兴趣的任何人学习,不需要任何基础。通过学习本课程有助于对计算机科学的体系建立一个基本的概念,其学习内容如下:

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Runoff

上篇文章我们总结了 Runoff 投票机制,大概需要以下步骤:

  • 读取候选人信息
  • 得到未淘汰的候选人的票数,总结为偏好数组
  • 判断是否有胜者产生
  • 找到最小值
  • 判断是否为平局
  • 淘汰票数最少的候选人
#include <cs50.h>
#include <stdio.h>
#include <string.h>

// 选民和候选人最大值
#define MAX_VOTERS 100
#define MAX_CANDIDATES 9

// 偏好数组
int preferences[MAX_VOTERS][MAX_CANDIDATES];

typedef struct
{
    string name;
    int votes;
    bool eliminated;
} candidate;
candidate candidates[MAX_CANDIDATES];

// Numbers of voters and candidates
int voter_count;
int candidate_count;

// Function prototypes
bool vote(int voter, int rank, string name);
void tabulate(void);
bool print_winner(void);
int find_min(void);
bool is_tie(int min);
void eliminate(int min);
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其中着重要注意的就是偏好数组,preference[i][j] 代表第 i 个选民的第 j 个喜欢。

我们可以根据过程写出 main 函数,首先获取用户输入:

int main(int argc, string argv[])
{
    // Check for invalid usage
    if (argc < 2)
    {
        printf("Usage: runoff [candidate ...]\n");
        return 1;
    }

    // Populate array of candidates
    candidate_count = argc - 1;
    if (candidate_count > MAX_CANDIDATES)
    {
        printf("Maximum number of candidates is %i\n", MAX_CANDIDATES);
        return 2;
    }
    for (int i = 0; i < candidate_count; i++)
    {
        candidates[i].name = argv[i + 1];
        candidates[i].votes = 0;
        candidates[i].eliminated = false;
    }

    voter_count = get_int("Number of voters: ");
    if (voter_count > MAX_VOTERS)
    {
        printf("Maximum number of voters is %i\n", MAX_VOTERS);
        return 3;
    }

    // Keep querying for votes
    for (int i = 0; i < voter_count; i++)
    {

        // Query for each rank
        for (int j = 0; j < candidate_count; j++)
        {
            string name = get_string("Rank %i: ", j + 1);

            // Record vote, unless it's invalid
            if (!vote(i, j, name))
            {
                printf("Invalid vote.\n");
                return 4;
            }
        }

        printf("\n");
    }
  
  	...
}
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然后写出主要部分:

// 循环直到赢家产生
while (true)
{
  // 计算剩余候选人的票数
  tabulate();

  // 检查是否有人胜利
  bool won = print_winner();
  if (won)
  {
    break;
  }

  // 找到票数最少的候选人,同时判断是否平局
  int min = find_min();
  bool tie = is_tie(min);

  // 如果平局,剩下的所有人都赢了
  if (tie)
  {
    for (int i = 0; i < candidate_count; i++)
    {
      if (!candidates[i].eliminated)
      {
        printf("%s\n", candidates[i].name);
      }
    }
    break;
  }

  // 淘汰票数最少的候选人
  eliminate(min);

  // 重复该过程
  for (int i = 0; i < candidate_count; i++)
  {
    candidates[i].votes = 0;
  }
}

return 0;
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分别实现其中的函数部分,首先是 vote 函数,我们只需要检查候选人是否存在即可:

// Record preference if vote is valid
bool vote(int voter, int rank, string name)
{
    // TODO
    for (int i = 0; i < candidate_count; i++)
    {
        if (strcmp(candidates[i].name, name) == 0)
        {
            preferences[voter][rank] = i;
            return true;
        }
    }
    return false;
}
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然后是最重要的制表环节,我们需要根据每一个选民的偏好数组找到还未被淘汰的候选人,为其加上一票:

// Tabulate votes for non-eliminated candidates
void tabulate(void)
{
    // TODO
    for (int i = 0; i < voter_count; i++)
    {
        for (int j = 0; j < candidate_count; j++)
        {
            if (!candidates[preferences[i][j]].eliminated)
            {
                candidates[preferences[i][j]].votes++;
                break;
            }
        }
    }
    return;
}
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检查是否有赢家产生,得到超过50%的票数:

// Print the winner of the election, if there is one
bool print_winner(void)
{
    // TODO
    for (int i = 0; i < candidate_count; i++)
    {
        if (candidates[i].votes > voter_count / 2)
        {
            printf("%s\n", candidates[i].name);
            return true;
        }
    }
    return false;
}
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找到得票最少的候选人:

// Return the minimum number of votes any remaining candidate has
int find_min(void)
{
    // TODO
    int min = voter_count;
    for (int i = 0; i < candidate_count; i++)
    {
        if (candidates[i].votes < min && !candidates[i].eliminated)
        {
            min = candidates[i].votes;
        }
    }
    return min;
}
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判断是否平局:

// Return true if the election is tied between all candidates, false otherwise
bool is_tie(int min)
{
    // TODO
    for (int i = 0; i < candidate_count; i++)
    {
        if (candidates[i].votes != min && !candidates[i].eliminated)
        {
            return false;
        }
    }
    return true;
}
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淘汰得票最少的候选人:

// Eliminate the candidate (or candidates) in last place
void eliminate(int min)
{
    // TODO
    for (int i = 0; i < candidate_count; i++)
    {
        if (candidates[i].votes == min && !candidates[i].eliminated)
        {
            candidates[i].eliminated = true;
        }
    }
    return;
}
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测试样例

./runoff Alice Bob Charlie
Number of voters: 5
Rank 1: Alice
Rank 2: Charlie
Rank 3: Bob

Rank 1: Alice
Rank 2: Charlie
Rank 3: Bob

Rank 1: Bob
Rank 2: Charlie
Rank 3: Alice

Rank 1: Bob
Rank 2: Charlie
Rank 3: Alice

Rank 1: Charlie
Rank 2: Alice
Rank 3: Bob

Alice
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转载自juejin.im/post/7105757691175911432