20180323算法设计分析HW3 分治上机三题

算法设计分析HW3 分治上机三题

冯浩然 1600013009

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1 Ultra-QuikSort2

• 实际是求逆序对个数。尽管明知道这是考分治要卡时间，但第一次还是作死地试了试 O(n2) O ( n 2 ) 的冒泡，然后还不死心的试了试带swap标记的优化版冒泡，果然都死了。
• 于是乖乖使用分治。算法实质就是mergesort，但需要对merge函数动一些手脚。在对数组 A,B A , B 进行归并时，此时已经归并了 i i 个元素。如果发现A[index1]>B[index2] A [ i n d e x 1 ] > B [ i n d e x 2 ] ，就说明 A[index1] A [ i n d e x 1 ] 与他后面的 B[0] B [ 0 ] 到 B[index2] B [ i n d e x 2 ] 的元素构成 (index2−i) ( i n d e x 2 − i ) 个逆序对那么就给 rev_cnt r e v _ c n t 变量加上 (index2−i) ( i n d e x 2 − i )
• 复杂度 O(nlogn) O ( n l o g n )
• 代码如下

#include 
#include 
using namespace std;
#define MAXN 500000 + 10

typedef long long ll;
ll n, rev_cnt;
ll c[MAXN], c_tmp[MAXN];

ll merge(int l, int r, int mid)
{
    ll cnt = 0;

    ll i, j, index1, index2;
    for (j = l; j <= r; j++)
        c_tmp[j] = c[j];

    index1 = l;
    index2 = mid + 1;
    i = l;
    while (index1 <= mid && index2 <= r){
        if (c_tmp[index1] <= c_tmp[index2])
            c[i++] = c_tmp[index1++];
        else{
            rev_cnt += index2 - i;
            c[i++] = c_tmp[index2++];
        }
    }
    while(index1 <= mid)
        c[i++] = c_tmp[index1++];
    while(index2 <= r)
        c[i++] = c_tmp[index2++];

    return cnt;
}

ll merge_sort(int l, int r)
{
    ll mid, cnt = 0;
    if (l < r){
        mid = (l + r) / 2;
        cnt += merge_sort(l, mid);
        cnt += merge_sort(mid + 1, r);
        cnt += merge(l, r, mid);
    }
    return cnt;
}

int main()
{


    while (cin >> n){
        if (n == 0)
            break;

        for (int i = 0; i < n; i++){
            scanf(" %lld", &c[i]);
        }

        rev_cnt = 0;
        merge_sort(0, n - 1);
        //rev_cnt = merge_sort(0, n - 1);
        /*
        int rev_cnt = 0;        
        bool no_swap;
        for (int i = 0; i < n - 1; i++){
            no_swap = true;
            for (int j = n - 1; j >= i; j--){
                if (c[j] < c[j - 1]){
                    int tmp = c[j];
                    c[j] = c[j - 1];
                    c[j - 1] = tmp;
                    no_swap = false;
                    rev_cnt++;
                }
            }
        }
        */


        printf("%lld\n", rev_cnt);
    }

    return 0;

}

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2 最近点对问题

• 课上老师详细分析过此题的算法，只要复现即可
• 具体操作为将点集对半划分-两边递归各算各的-算一下分别在两边的离得比较近的点。需要注意的是在找分别在两边的点对时，可以以距离mid距离不大于 δ δ 为界。
• 复杂度 O(nlogn) O ( n l o g n )
• 代码如下

#include 
#include 
#include 
#include 
using namespace std;
#define MAXN 200000 + 10
typedef long long ll;

int N;
struct point
{
    ll x, y;

    bool operator < (const point &b){
        return x < b.x;
    }


}P[MAXN];

double dis(const point &a, const point &b){
    return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}

double divide_findmin(int l, int r)
{
    if (r - l == 1){
        return dis(P[l], P[r]);
    }

    if (r - l == 2){
        double d[3];
        d[0] = dis(P[l], P[l + 1]);
        d[1] = dis(P[l], P[r]);
        d[2] = dis(P[l + 1], P[r]);

        return min(min(d[0], d[1]), d[2]);
    }


    int mid = (l + r) / 2;
    double min_dis1, min_dis2, min_dis3, min_dis;

    min_dis1 = divide_findmin(l, mid);
    min_dis2 = divide_findmin(mid + 1, r);
    min_dis = min(min_dis1, min_dis2); 

    /*find the min_dis near the line, which is \delta_3*/
    point tmp[r - l];
    int index1 = 0, index2 = 0;
    for (int i = mid; (i >= l) && ((P[mid].x - P[i].x) < min_dis); i--)
        tmp[index1++] = P[i];
    for (int i = mid + 1; (i <= r) && ((P[i].x - P[mid].x) < min_dis); i++)
        tmp[index1 + (index2++)] = P[i];
    for (int i = 0; i < index1; i++)
        for (int j = index1; j < index1 + index2; j++){
            min_dis3 = dis(tmp[i], tmp[j]);
            min_dis = min(min_dis, min_dis3);
        }

    return min_dis;
}

int main()
{
    scanf(" %d", &N);
    for (int i = 0; i < N; i++){
        scanf(" %lld%lld", &P[i].x, &P[i].y);
    }
    sort(P, P + N);

    double min_dis = divide_findmin(0, N - 1);
    printf("%.3lf\n", min_dis);

    return 0;
}

3 集合求交

• 上次的作业题了

• 先对 B B 排序，然后在其中顺次查找A A 中的每个元素。如果找到，则说明在交集里含有这个元素

• 复杂度为 O(nlogm)=O(nloglogn) O ( n l o g m ) = O ( n l o g l o g n )

• 代码如下

  
  #include 
  
  
  #include 
  
  
  #include 
  
  using namespace std;
  
  #define MAXN 100000 + 10
  
  
  int a[MAXN], b[MAXN];
  int c[MAXN];
  int n, m;
  
  bool bin_search(int k, int l, int r) {
  
  while (l < r) {
      int mid = (l + r) / 2;
      if (b[mid] < k)
          l = mid + 1;
      else
          r = mid;
  }
  
  if (b[r] == k)
      return true;
  return false;
  }
  
  int main()
  {
  
  scanf(" %d%d", &n, &m);
  for (int i = 0; i < n; i++)
      scanf(" %d", &a[i]);
  for (int i = 0; i < m; i++)
      scanf(" %d", &b[i]);
  
  sort(b, b + m);
  int c_index = 0;
  for (int i = 0; i < n; i++) {
      //printf("End of search\n");
      if (bin_search(a[i], 0, m - 1))
          c[c_index++] = a[i];
  }
  
  //sort(c, c + c_index);
  //for (int i = 0; i < c_index; i++)
  //  printf("%d ", c[i]);
  //printf("End\n");
  printf("%d\n", c_index);
  
  return 0;
  }

  ​