灯塔(LightHouse)

Description

As shown in the following figure, If another lighthouse is in gray area, they can beacon each other.

For example, in following figure, (B, R) is a pair of lighthouse which can beacon each other, while (B, G), (R, G) are NOT.

Input

1st line: N

2nd ~ (N + 1)th line: each line is X Y, means a lighthouse is on the point (X, Y).

Output

How many pairs of lighthourses can beacon each other

( For every lighthouses, X coordinates won‘t be the same , Y coordinates won‘t be the same )

Example

Input

3
2 2
4 3
5 1

Output

1

Restrictions

For 90% test cases: 1 <= n <= 3 * 10⁵

For 95% test cases: 1 <= n <= 10⁶

For all test cases: 1 <= n <= 4 * 10⁶

For every lighthouses, X coordinates won‘t be the same , Y coordinates won‘t be the same.

1 <= x, y <= 10^8

Time: 2 sec

Memory: 256 MB

Hints

The range of int is usually [-2³¹, 2³¹ - 1], it may be too small.

视频里有讲解我尽然没看到，醉了。

就是求一下逆序对就行。

 1 #include <cstdlib>
 2 #include <iostream>
 3 #include <cstdio>
 4 #include <cstring>
 5 using namespace std;
 6 #define LL long long
 7 const int max_size = 4 * 1e6;
 8 LL y_val[max_size];
 9 LL tmp_arry[max_size];
10 ///加速代码，why 百度
11 const int SZ = 1<<20;
12 struct fastio{
13     char inbuf[SZ];
14     char outbuf[SZ];
15     fastio(){
16         setvbuf(stdin,inbuf,_IOFBF,SZ);
17         setvbuf(stdout,outbuf,_IOFBF,SZ);
18     }
19 }io;
20 struct Point
21 {
22     LL x, y;
23 } p[max_size];
24
25 int cmp(const void *a, const void *b)
26 {
27     struct Point *c = (Point *)a;
28     struct Point *d = (Point *)b;
29     if(c->x != d->x)
30         return c->x - d->x;
31     else
32         return d->y - c->y;
33 }
34
35 LL Merge(LL *arr, LL beg, LL mid, LL end, LL *tmp_arr)
36 {
37     memcpy(tmp_arr+beg, arr+beg, sizeof(LL)*(end - beg + 1));
38     LL i = beg;
39     LL j = mid+1;
40     LL k = beg;
41     LL inversion = 0;
42     while(i <= mid && j <= end)
43     {
44         if(tmp_arr[i] <= tmp_arr[j])
45             arr[k++] = tmp_arr[i++];
46         else
47         {
48             arr[k++] = tmp_arr[j++];
49             inversion += mid - i + 1;
50         }
51     }
52
53     while(i <= mid)
54         arr[k++] = tmp_arr[i++];
55     while(j <= end)
56         arr[k++] = tmp_arr[j++];
57     return inversion;
58 }
59
60 LL MergeInversion(LL *arr, LL beg, LL end, LL *tmp_arr)
61 {
62     LL inversions = 0;
63     if(beg < end)
64     {
65         LL mid = (beg + end) >> 1;
66         inversions += MergeInversion(arr, beg, mid, tmp_arr);
67         inversions += MergeInversion(arr, mid+1, end, tmp_arr);
68         inversions += Merge(arr, beg, mid, end, tmp_arr);
69     }
70     return inversions;
71 }
72
73 int main()
74 {
75     LL n;
76     cin >> n;
77     for(int i = 0; i < n; i++)
78     {
79         cin >> p[i].x >> p[i].y;
80     }
81
82     qsort(p, n, sizeof(p[0]), cmp);
83     for(int i = 0; i < n; i++)
84     {
85         y_val[i] = p[i].y;
86     }
87
88     memcpy(tmp_arry, y_val, sizeof(LL)*n);
89     cout << n*(n-1) /2 - MergeInversion(y_val, 0, n-1, tmp_arry) << endl;
90     return 0;
91 }

然后再对代码进行优化吧，我的这段过不了最后一个点。因为多做了排序操作，没必要。