Clique Problem Solving Report

Clique Topic description

There are \ (n\) points on the number axis, the coordinates of the i\ points are \ (x_i\), and the weights are \ (w_i\). Two points \ (i\),\ (j\) exist between an edge and only if \ (ABS (X_i-x_j) >=w_i+w_j\). You need to ask for the maximum number of points in this chart. (a group is a set of vertices with edges between 22)

Input data

The first line is an integer \ (n\), and the next \ (n\) line is two integers per line ( x_i,w_i\).

Output data

An integer on one line indicates the answer

Data range

For \ (20\%\) data, \ (n<=10\).
For \ (60\%\) data, \ (n<=1000\).
For \ (100\%\) data, \ (n<=200000\), \ (0<=|x_i|,w_i<=10^9\).

A nature was found during the examination

If three points are left-to-right, point 1 points 2 have edges and point 2 points 3 to the right you can get points 1 points 3 have edges

So can \ (n^2\) run topo the longest road (from left to right with a forward edge)

Want to get (set\) optimize the edge but didn't think.

The result is a small explosion of 30.

The positive solution is not thought of it.

If a point is considered as an interval of the form \ ([x-w,x+w]\) , then the two points have a connecting edge equivalent to the interval no intersection.

So the problem turns into the number of intervals that don't overlap.

Can be simple greedy, can also DP

The complexity is \ (O (NLOGN) \)

Code:

#include <cstdio>#include <algorithm>const int N=2e5+10;struct node{    int l,r;    bool friend operator <(node n1,node n2){return n1.r<n2.r;}}seg[N];int rr[N],dp[N],f[N],pos[N],n;int max(int x,int y){return x>y?x:y;}int main(){    scanf("%d",&n);    for(int x,w,i=1;i<=n;i++)    {        scanf("%d%d",&x,&w);        seg[i].l=x-w,seg[i].r=x+w;        rr[i]=seg[i].r;    }    std::sort(seg+1,seg+1+n);    std::sort(rr+1,rr+1+n);    rr[0]=seg[1].r-1;    for(int i=1;i<=n;i++)    {        int l=0,r=i-1;        while(l<r)        {            int mid=l+r+1>>1;            if(seg[i].l>=rr[mid])                l=mid;            else                r=mid-1;        }        pos[i]=l;    }    for(int i=1;i<=n;i++)        dp[i]=f[pos[i]]+1,f[i]=max(f[i-1],dp[i]);    printf("%d\n",f[n]);    return 0;}

2018.10.13

Clique Problem Solving Report