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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.
1st line: N
2nd ~ (N + 1)th line: each line is X Y, means a lighthouse is on the point (X, Y).
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 )
Input
3
2 2
4 3
5 1
Output
1
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
The range of int is usually [-231, 231 - 1], it may be too small.
第一眼看到这题时第一反应是用树状数组，但是发现数据太大，开不了10的6次方的二维数组，看了别人的博客才知道是用归并排序。
先理解一下题意，要求有多少对两两互相照亮的灯塔，怎么样才能是两两互相照亮的呢，两点的斜率为正，也就是按x递增排序，y也递增。
这样咱们就可以把n个点对x进行排序，找到y有多少对是顺序对就行，这就可以用到归并排序，在对左右两个集合合并时，i，j分别为左右两个集合的
指针，如果le[i]<ri[j]，正序对加上n2-j+1对，就这样去考虑。还有就是那里不能用<algorithm>，可以用stdlib.h里的qsort（）。
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 #include <cstdlib> 5 6 using namespace std; 7 8 9 10 struct w 11 { 12 int x; 13 int y; 14 }k[5100000]; 15 int yi[5100000]; 16 long long ans; 17 int le[5000005],ri[5000005]; 18 19 int cmp( const void *a ,const void *b) 20 { 21 return (*(w *)a).x > (*(w *)b).x ; 22 } 23 24 25 void merge(int l,int mi,int r) 26 { 27 int i,j,p,n1=mi-l+1,n2=r-mi; 28 const int MAX=500000005; 29 30 for (int i=1;i<=n1;i++) 31 le[i]=k[l+i-1].y; 32 for (int i=1;i<=n2;i++) 33 ri[i]=k[mi+i].y; 34 35 le[n1+1]=MAX;ri[n2+1]=MAX; 36 i=1;j=1; 37 for (int p=l;p<=r;p++) 38 { 39 if (le[i]>ri[j]) 40 k[p].y=ri[j++]; 41 else 42 { 43 k[p].y=le[i++]; 44 ans+=n2-j+1; 45 // cout <<ans<<endl; 46 } 47 } 48 } 49 void mergesort(int l,int r) 50 { 51 if (l==r) 52 return; 53 int mid=l+(r-l)/2; 54 mergesort(l,mid); 55 mergesort(mid+1,r); 56 merge(l,mid,r); 57 } 58 59 int main() 60 { 61 int n; 62 cin>>n; 63 for (int i=0;i<n;i++) 64 cin>>k[i].x>>k[i].y; 65 qsort(k,n,sizeof(k[0]),cmp); 66 ans=0; 67 mergesort(0,n-1); 68 cout <<ans<<endl; 69 return 0; 70 }
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原文地址：http://www.cnblogs.com/arno-my-boke/p/4760450.html
