Question

我在pset3中使用CS50的tideman遇到了一些问题。除了最后一个，check50中的所有内容都是正确的：

:( print_winner prints winner of election when some pairs are tied.
    print_winner did not print winner of election

这是我的代码：

#include <cs50.h>
#include <stdio.h>
#include <string.h>

// Max number of candidates
#define MAX 9

// preferences[i][j] is number of voters who prefer i over j
int preferences[MAX][MAX];

// locked[i][j] means i is locked in over j
bool locked[MAX][MAX];

// Each pair has a winner, loser
typedef struct
{
    int winner;
    int loser;
}
pair;

// Array of candidates
string candidates[MAX];
pair pairs[MAX * (MAX - 1) / 2];

int pair_count;
int candidate_count;

// Function prototypes
bool vote(int rank, string name, int ranks[]);
void record_preferences(int ranks[]);
void add_pairs(void);
void sort_pairs(void);
void lock_pairs(void);
void print_winner(void);

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

    // Populate array of candidates
    candidate_count = argc - 1;
    if (candidate_count > MAX)
    {
        printf("Maximum number of candidates is %i\n", MAX);
        return 2;
    }
    for (int i = 0; i < candidate_count; i++)
    {
        candidates[i] = argv[i + 1];
        //printf("Candidate %i: %s\n", i + 1, argv[i + 1]);
    }

    // Clear graph of locked in pairs
    for (int i = 0; i < candidate_count; i++)
    {
        for (int j = 0; j < candidate_count; j++)
        {
            locked[i][j] = false;
        }
    }

    pair_count = 0;
    int voter_count = get_int("Number of voters: ");

    // Query for votes
    for (int i = 0; i < voter_count; i++)
    {
        // ranks[i] is voter's ith preference
        int ranks[candidate_count];

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

            if (!vote(j, name, ranks))
            {
                printf("Invalid vote.\n");
                return 3;
            }
        }

        record_preferences(ranks);

        printf("\n");
    }

    add_pairs();
    sort_pairs();
    lock_pairs();
    print_winner();
    return 0;
}

// Update ranks given a new vote
bool vote(int rank, string name, int ranks[])
{
    // TODO
    for (int i = 0; i < candidate_count; i++)
    {
        if (strcmp(name, candidates[i]) == 0)
        {
            ranks[rank] = i;

            return true;
        }
    }

    return false;
}

// Update preferences given one voter's ranks
void record_preferences(int ranks[])
{
    // TODO
    int counter = 0;
    for (int i = 0; i < candidate_count; i++)
    {
        int j = 1;
        for (j += counter; j < candidate_count; j++)
        {
            preferences[ranks[i]][ranks[j]] += 1;
        }
        counter++;
    }

    return;
}

// Record pairs of candidates where one is preferred over the other
void add_pairs(void)
{
    // TODO
    int k = 0;
    int l = 1;
    for (int i = 0; i < candidate_count - 1; i++)
    {
        for (int j = l; j < candidate_count; j++)
        {
            if (preferences[i][l] > preferences[l][i])
            {
                pairs[k].winner = i;
                pairs[k].loser = l;
                pair_count++;
                k++;
            }
            else if (preferences[i][l] < preferences[l][i])
            {
                pairs[k].loser = i;
                pairs[k].winner = l;
                pair_count++;
                k++;
            }
            if (l < candidate_count - 1)
            {
                l++;
            }
        }
    }

    return;
}

// Sort pairs in decreasing order by strength of victory
//int strength [MAX];
pair temp[1];
void sort_pairs(void)
{
    // TODO

    for (int j = 0; j < candidate_count - 1; j++)
    {
        //if (strength[j] < strength[j + 1])
        if (preferences[pairs[j].winner][pairs[j].loser] < preferences[pairs[j + 1].winner][pairs[j + 1].loser])
        {
            temp[0] = pairs[j + 1];
            pairs[j + 1] = pairs[j];
            pairs[j] = temp[0];
        }
    }

    return;
}

// Lock pairs into the candidate graph in order, without creating cycles
pair first[1];
void lock_pairs(void)
{
    // TODO

    first[0] = pairs[0];
    locked[pairs[0].winner][pairs[0].loser] = true;
    for (int i = 1; i < candidate_count; i++)
    {
        if (pairs[i].loser == pairs[i - 1].winner)
        {
            first[0] = pairs[i];
        }
        locked[pairs[i].winner][pairs[i].loser] = true;
    }

    for (int j = 1; j < candidate_count; j++)
    {
        if (pairs[j].loser == first[0].winner)
        {
            locked[pairs[j].winner][pairs[j].loser] = false;
        }
    }

    return;
}

// Print the winner of the election
void print_winner(void)
{
    // TODO

    printf("%s\n", candidates[first[0].winner]);

    return;
}

据我所知，如果成对绑在一起，它们具有相同的强度吗？实力是仅意味着对获胜者的总体偏好，还是获胜者与失败者之间的差异？例如，如果查理（Charlie）赢得爱丽丝（Alice）7-2，则力量是5还是7？

我不太了解是什么原因导致此错误，请帮忙吗？

Answer 1

只是碰到了这个线程，这挽救了我的理智：

https://www.reddit.com/r/cs50/comments/enm5t8/pset3_tideman_winner_when_some_pairs_are_tied/

事实证明，您毕竟需要考虑多个来源！

pset3中CS50的tideman出现问题（当某些对并列时，不是打印出选举的获胜者）

1 个答案: