51nod 1278 The Circle (sort + Modify Step)

Topic Meaning:

http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1278

there are n circles on the plane, and their center is on the x-axis, giving the center and radius of all the circles, and asking how many pairs of circles are absent. For example: 4 circles are located at 1, 2, 3, 4 positions, with radii of 1, 1, 2, 1, then {1, 2}, {1, 3} {2, 3} {2, 4} {3, 4} 5 pairs have intersections, only {1, 4} are separated. Input

Line 1th: A number N, indicating the number of circles (1 <= N <= 50000) 2-n + 1 lines: 2 numbers per line p, r Middle with a space, p for the position of the center, R for the radius of the circle (1 <= P, R <= 10^9)

Output

The output has a total number of circles to be absent.

Input Example

41 12 13) 24 1

Output Example

1

Topic Analysis:

/**
* Method One
* Divide the circle into left and right short points, represented by line segments, and now deform for line intersection problems
* Left endpoint Left=c-r, right endpoint right=c+r, sorted by left endpoint ascending, equal
Sort by the right endpoint in ascending order, or scan, for each line I, when a later bar
* The left end of line J is greater than the right end of I, then all subsequent segments are
* Time complexity of O (Nlog (n) +n^2)
* The data for this question is 50000, the normal O (n*n) time will definitely be timed out
* Now we can consider the method of changing the Step size optimization algorithm (can only reduce the coefficient, but sometimes really good),
* I optimize in the query, the other step is 100, if the current J satisfies the condition, then first satisfies
* The condition of J must be within 100 of the J before, only the first 100 of the query J can be,
* Time Complexity O (N*log (n) +n* (n/100+100)); Reduce your n^2 factor by 100 times times
*************************************************************************
* Method Two thank Dava
* Place the start and end points of all circles in an array, sorted by size (the left and right short points of the circle)
* If the starting point of a circle is equal to the end of another circle, the starting point is in front
* Define Structure body
*struct{
* A long long d;//segment node
* int tmp;//marks the beginning (0) and the end point (1),
*｝
* Record all start and end points directly and quickly,
* Make the number of num= round, meet the starting point num--, meet the end Sum+=num
*/

AC Code, Method 1

#include <iostream> #include <string> #include <algorithm>using namespace std;struct node{int left, right;};    Node P[50005];int CMP (node A,node b) {if (a.left<b.left) return 1;    else if (a.left==b.left&&a.right<b.right) return 1; return 0;}    int main () {int c,r,n;            while (Cin>>n) {for (int i=0;i<n;i++) {cin>>c>>r;            P[i].left=c-r;        P[i].right=c+r;        } sort (p,p+n,cmp);        int i,j,sum=0,k;            for (i=0;i<n;i++) {for (j=i+100;j<n;j+=100) {////step = (p[j].left>p[i].right) break;            } if (j>n) j=n; for (k=j-1;k>0&&k>j-101;k--) {//Scan 100 elements between if (p[k].left<=p[i].right) {sum +=n-(k+1);                Break    }}} cout<<sum<<endl; } return 0;}

AC code, method two

#include <iostream> #include <algorithm>using namespace std;struct node{    long long d;//line node    int flag;//Mark Start (0) and end point (1),}p[100005];int CMP (node A,node b) {    if (A.D<B.D) return 1;    else if (a.d==b.d&&a.flag==0) return 1;    return 0;} int main () {    int c,r,n;    while (cin>>n) {        int k=0;        for (int i=0;i<n;i++) {            cin>>c>>r;            P[k].d=c-r;            p[k++].flag=0;//beginning            p[k].d=c+r;            p[k++].flag=1;//End        }        //cout<<k<<endl;        Sort (p,p+k,cmp);        int sum=0,num=n;        for (int i=0;i<k;i++) {            //cout<<p[i].d<< "" <<p[i].flag<<endl;            if (p[i].flag==0) num--;            else if (p[i].flag==1) sum+=num;        }        cout<<sum<<endl;    }    return 0;}

51nod 1278 The Circle (sort + Modify Step)