我目前正在学习CS50,这是哈佛大学的一个基于在线的编程入门模块。当我为一个叫做"Tideman“的问题写代码时,我对lock_pairs函数有很多困难,因为我对编码非常陌生。问题的描述可以在这里找到:https://cs50.harvard.edu/x/2020/psets/3/tideman/。
在@TomKarze的一些帮助下,我设法稍微改进了我的代码,但由于某些原因,我现在无法满足问题的一个要求,即我的代码不能像现在应该的那样工作,但我非常肯定几天前它工作了。即使我使用我的原始代码(没有Tom的输入),它仍然不再工作,所以我真的很困惑(就像...我是不是看到了什么?)。
我有问题的函数是:
bool iscycle(int index, bool visited[])
{
if (visited[index])
{
return true;
}
visited[index] = true;
for (int i = 0; i < candidate_count; i++)
{
if (locked[index][i] && iscycle(i, visited))
{
return true;
}
}
return false;
}
// Lock pairs into the candidate graph in order, without creating cycles
void lock_pairs(void)
{
for (int i = 0; i < pair_count; i++)
{
locked[pairs[i].winner][pairs[i].loser] = true;
bool visited[candidate_count];
for (int j = 0; j < candidate_count; j++)
{
visited[j] = false;
}
if (iscycle(i, visited))
{
locked[pairs[i].winner][pairs[i].loser] = false;
}
}
return;
}
我的整个代码(包括上面的函数)如下:
#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];
}
// 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[])
{
for (int i = 0; i < candidate_count; i++)
{
if (strcmp(candidates[i], name) == 0)
{
ranks[rank] = i;
return true;
}
}
return false;
}
// Update preferences given one voter's ranks
void record_preferences(int ranks[])
{
for (int i = 0; i < candidate_count; i++)
{
for (int j = 0; j < candidate_count; j++)
{
if (i < j)
{
preferences[ranks[i]][ranks[j]]++;
}
}
}
return;
}
// Record pairs of candidates where one is preferred over the other
void add_pairs(void)
{
for (int i = 0; i < candidate_count; i++)
{
for (int j = i + 1; j < candidate_count; j++)
{
if (preferences[i][j] > preferences[j][i])
{
pairs[pair_count].winner = i;
pairs[pair_count].loser = j;
pair_count++;
}
else if (preferences[i][j] < preferences[j][i])
{
pairs[pair_count].winner = j;
pairs[pair_count].loser = i;
pair_count++;
}
}
}
return;
}
// Sort pairs in decreasing order by strength of victory
void sort_pairs(void)
{
int swapcounter = -1;
pair toswap;
do
{
swapcounter = 0;
for (int i = 0; i < pair_count - 1; i++)
if (preferences[pairs[i].winner][pairs[i].loser] < preferences[pairs[i + 1].winner][pairs[i + 1].loser])
{
toswap = pairs[i];
pairs[i] = pairs[i + 1];
pairs[i + 1] = toswap;
swapcounter++;
}
}
while (swapcounter != 0);
return;
}
bool iscycle(int index, bool visited[])
{
if (visited[index])
{
return true;
}
visited[index] = true;
for (int i = 0; i < candidate_count; i++)
{
if (locked[index][i] && iscycle(i, visited))
{
return true;
}
}
return false;
}
// Lock pairs into the candidate graph in order, without creating cycles
void lock_pairs(void)
{
for (int i = 0; i < pair_count; i++)
{
locked[pairs[i].winner][pairs[i].loser] = true;
bool visited[candidate_count];
for (int j = 0; j < candidate_count; j++)
{
visited[j] = false;
}
if (iscycle(i, visited))
{
locked[pairs[i].winner][pairs[i].loser] = false;
}
}
return;
}
// Print the winner of the election
void print_winner(void)
{
for (int i = 0; i < candidate_count; i++)
{
for (int j = 0; j < candidate_count; j++)
{
if (locked[j][i] == false)
{
printf("%s\n", candidates[i]);
}
}
}
return;
}
当我通过Check50运行它时,一个内置的检查器来查看我的代码是否满足了问题的要求,我得到的结果是:
:) tideman.c exists
:) tideman compiles
:) vote returns true when given name of candidate
:) vote returns false when given name of invalid candidate
:) vote correctly sets rank for first preference
:) vote correctly sets rank for all preferences
:) record_preferences correctly sets preferences for first voter
:) record_preferences correctly sets preferences for all voters
:) add_pairs generates correct pair count when no ties
:) add_pairs generates correct pair count when ties exist
:) add_pairs fills pairs array with winning pairs
:) add_pairs does not fill pairs array with losing pairs
:) sort_pairs sorts pairs of candidates by margin of victory
:) lock_pairs locks all pairs when no cycles
:( lock_pairs skips final pair if it creates cycle
lock_pairs did not correctly lock all non-cyclical pairs
:) lock_pairs skips middle pair if it creates a cycle
:) print_winner prints winner of election when one candidate wins over all others
:) print_winner prints winner of election when some pairs are tied
我无法理解我逻辑中的缺陷在哪里。
附注:如果任何人能花时间向我解释一下,从效率和“代码设计”的角度来看,什么时候递归比迭代更可取,反之亦然,我也会非常感激!
发布于 2020-10-19 01:40:21
我一直在努力使用这个函数。基本上,在这个递归函数中,如果其中一个获胜者等于失败者,那么所有的循环都被锁定,则必须返回TRUE。但是您正在检查已访问的数组,我也使用了“已访问”路径,并且我花了一些时间来寻找另一种方式:)
https://stackoverflow.com/questions/62283700
复制相似问题