PTA浙江大学数据结构习题——第九周

排好序后，统计每一个连续的相同数字的个数。

#include 

using namespace std;

const int N = 100010;

int n;
int a[N];

void quick_sort(int l, int r)
{
    if (l >= r) return;
    
    int x = a[l + r >> 1], i = l - 1, j = r + 1;
    while (i < j)
    {
        do i ++;    while (a[i] < x);
        do j --;    while (a[j] > x);
        if (i < j)  swap(a[i], a[j]);
    }
    
    quick_sort(l, j);
    quick_sort(j + 1, r);
}

int main()
{
    cin >> n;
    for (int i = 0; i < n; i ++)    cin >> a[i];
    
    quick_sort(0, n - 1);
    
    int cur = a[0], num = 0;
    for (int i = 0; i < n; i ++)
    {
        if (a[i] != cur && num != 0)
        {
            if (cur != -1)  printf("%d:%dn", cur, num);
            num = 0;
            cur = a[i];
        }
        num ++;
    }
    printf("%d:%dn", cur, num);
    
    return 0;
}

用结构体定义需要排序的属性，将所给的用户属性计算出来之后，直接调用 sort 函数。

#include 
#include 

using namespace std;

const int N = 10010, K = 6;

int n, k, m;
int p[K];

struct User
{
    int total_score, finish_num, id;
    int score[K];
    int valid;
    
    // valid为1的排在valid为0的前面，然后分数递减排序，完成题目递减排序，id升序排列
    bool operator< (const User &t) const
    {
        if (valid != t.valid)   return valid > t.valid;
        if (total_score != t.total_score)   return total_score > t.total_score;
        if (finish_num != t.finish_num)     return finish_num > t.finish_num;
        return id < t.id;
    }
} user[N];

int main()
{
    cin >> n >> k >> m;
    for (int i = 1; i <= k; i ++)    cin >> p[i];
    for (int i = 1; i <= n; i ++)
        for (int j = 1; j <= k; j ++)
            user[i].score[j] = -2;
    while (m --)
    {
        int u_id, p_id, p_s;
        cin >> u_id >> p_id >> p_s;
        
        if (p_s >= 0 && !user[u_id].valid)   // 是否在排行榜
            user[u_id].valid = 1;
        
        if (p_s > user[u_id].score[p_id])   // 是否更新答题分数
            user[u_id].score[p_id] = p_s;
        
        user[u_id].id = u_id;   // id
    }
    
    for (int i = 1; i <= n; i ++)
    {
        if (!user[i].valid) continue;
        for (int j = 1; j <= k; j ++)
        {
            if (user[i].score[j] >= 0)
                user[i].total_score += user[i].score[j];
            if (user[i].score[j] == p[j])
                user[i].finish_num ++;
        }
    }
    
    sort(user + 1, user + n + 1);
    
    int rank = 1, pre = user[1].total_score;
    for (int i = 1; user[i].valid == 1; i ++)
    {
        if (user[i].total_score != pre)
        {
            rank = i;
            pre = user[i].total_score;
        }
        
        printf("%d %05d %d", rank, user[i].id, user[i].total_score);
        for (int j = 1; j <= k; j ++)
        {
            int t = user[i].score[j];
            if (t == -2)    printf(" -");
            else if (t == -1)   printf(" 0");
            else if (t >= 0)    printf(" %d", t);
        }
        puts("");
    }
    
    return 0;
}

待排序列中有若干个环，分为几种情况

（1）环中仅有一个元素，则不需要交换

（2）环中有 0，且元素数目为 n(n > 1) ，交换次数为 n - 1

（3）环中没有 0，且元素数目为 n(n > 1) ，交换次数为 n + 1 （先将 0 与环中任意元素交换，然后再把这个包含 0 的新环交换好）

#include 

using namespace std;

const int N = 100010;

int n;
int a[N];
bool st[N];

int main()
{
    cin >> n;
    for (int i = 0; i < n; i ++)    cin >> a[i];
    
    int res = 0, cur = 0;
    for (int i = 0; !st[i]; i = a[i])
    {
        st[i] = true;
        cur ++;
    }
    res += cur - 1;
    
    for (int i = 0; i < n; i ++)
    {
        if (!st[i])
        {
            cur = 0;
            for (int j = i; !st[j]; j = a[j])
            {
                st[j] = true;
                cur ++;
            }
            if (cur == 1)   continue;
            res += cur + 1;
        }
    }
    cout << res << endl;
    
    return 0;
}