Runoff: 課題
決選投票のプログラムを作ること。前回のPluralityと違うところは、投票数の大多数を確保できた候補者はすぐ投票が決まるが、誰も大多数を獲得できなかった場合は、Runoffを発動し、投票数が最も少なかった候補者に投票された表を見に行き、2番目の候補を集計に加える。これを繰り返して、候補者を決定する。
https://cs50.harvard.edu/x/2021/psets/3/runoff/
Runoff: 回答例
#include <cs50.h>
#include <stdio.h>
#include <string.h>
// Max voters and candidates
#define MAX_VOTERS 100
#define MAX_CANDIDATES 9
// preferences[i][j] is jth preference for voter i
int preferences[MAX_VOTERS][MAX_CANDIDATES];
// Candidates have name, vote count, eliminated status
typedef struct
{
string name;
int votes;
bool eliminated;
}
candidate;
// Array of candidates
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);
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");
}
// Keep holding runoffs until winner exists
while (true)
{
// Calculate votes given remaining candidates
tabulate();
// Check if election has been won
bool won = print_winner();
if (won)
{
break;
}
// Eliminate last-place candidates
int min = find_min();
bool tie = is_tie(min);
// If tie, everyone wins
if (tie)
{
for (int i = 0; i < candidate_count; i++)
{
if (!candidates[i].eliminated)
{
printf("%s\n", candidates[i].name);
}
}
break;
}
// Eliminate anyone with minimum number of votes
eliminate(min);
// Reset vote counts back to zero
for (int i = 0; i < candidate_count; i++)
{
candidates[i].votes = 0;
}
}
return 0;
}
// Record preference if vote is valid
bool vote(int voter, int rank, string name)
{
for (int i = 0; i < candidate_count; i++)
{
if (strcmp(name, candidates[i].name) == 0)
{
preferences[voter][rank] = i;
return true;
}
}
return false;
}
// Tabulate votes for non-eliminated candidates
void tabulate(void)
{
for (int i = 0; i < voter_count; i++)
{
for (int j = 0; j < candidate_count; j++)
{
int k = preferences[i][j];
if (candidates[k].eliminated == false)
{
candidates[k].votes++;
break;
}
}
}
return;
}
// Print the winner of the election, if there is one
bool print_winner(void)
{
// Majory (50% or more) of votes?
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;
}
// Return the minimum number of votes any remaining candidate has
int find_min(void)
{
// TODO
int vote_num = voter_count;
for (int i = 0; i < candidate_count; i++)
{
if (candidates[i].votes < vote_num && candidates[i].eliminated == false)
{
vote_num = candidates[i].votes;
}
}
return vote_num;
}
// Return true if the election is tied between all candidates, false otherwise
bool is_tie(int min)
{
// TODO
int tie_count = 0;
int on_run = 0;
for (int i = 0; i < candidate_count; i++)
{
if (candidates[i].eliminated == false && candidates[i].votes == min)
{
tie_count++;
on_run++;
}
else if (candidates[i].eliminated == false)
{
on_run++;
}
}
// Checks for tie
if (tie_count == on_run)
{
return true;
}
return false;
}
// Eliminate the candidate (or candidates) in last place
void eliminate(int min)
{
// TODO
for (int i = 0; i < candidate_count; i++)
{
if (candidates[i].eliminated == false && candidates[i].votes == min)
{
candidates[i].eliminated = true;
}
}
return;
}
未希 諒
香港在住の34歳サラリーマン。戦略家、写真家、旅人、サウナー。 2025年までに季節を選びながら世界中を飛び回る仕事をするために準備中。その道中で発見した気づきや気になる話題を中心に発信しています。2010年より旅をスタートし通算500箇所の宿に宿泊。年間の3分の1をホテルで暮らす。
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