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
/*@Date : 2018-02-05 15:02:45@Author : 酸饺子 (changzheng300@foxmail.com)@Link : https://github.com/SourDumplings@Version : $Id$
*//*
https://dsa.cs.tsinghua.edu.cn/oj/problem.shtml?id=1144
解答：
对于n对灯塔， 先把灯塔们按x坐标排序，然后观察y坐标中有多少逆序对*/#include
#include
#include
#include using namespace std;const int maxn = 4000001;const int SZ = 1<<20; //快速io
struct fastio{char inbuf[SZ];char outbuf[SZ];fastio(){setvbuf(stdin,inbuf,_IOFBF,SZ);setvbuf(stdout,outbuf,_IOFBF,SZ);}
}io;struct LightHouse
{int x, y;
} data[maxn];int cmpx(const void *a, const void *b)
{return ((LightHouse *)a)->x - ((LightHouse *)b)->x;
}int cmpy(const void *a, const void *b)
{return ((LightHouse *)a)->y - ((LightHouse *)b)->y;
}int binary_search(LightHouse A[], int l, int h, int value)
{while (l < h){int mi = (l + h) >> 1;value < A[mi].y ? h = mi : l = mi + 1;}return --l;
}long long calc_Inv_y(int start, int stop)
{if (stop - start < 2) return 0;int mid = (stop + start) >> 1;// printf("start = %d, stop = %d, mid = %d\n", start, stop, mid);long long cross = 0;LightHouse *temp = new LightHouse[mid-start];for (int i = start, count = 0; i != mid;)temp[count++] = data[i++];qsort(temp, mid-start, sizeof(temp[0]), cmpy);for (int i = mid; i != stop; ++i){int j = binary_search(temp, 0, mid-start, data[i].y) + 1;cross += j;}delete [] temp;long long left = calc_Inv_y(start, mid);long long right = calc_Inv_y(mid, stop);// printf("left = %lld, right = %lld, cross = %lld\n", left, right, cross);return cross + left + right;
}int main(int argc, char const *argv[])
{int n;scanf("%d", &n);int X, Y;for (int i = 0; i != n; ++i){scanf("%d %d", &X, &Y);data[i].x = X;data[i].y = Y;}qsort(data, n, sizeof(data[0]), cmpx);printf("%lld\n", calc_Inv_y(0, n));return 0;
}