CF527D

题面

这题还挺水的，把那个式子稍微变形一下就可以的到xi-wi>=xj+wj,易知：若把每个点看做一条线段，左端点是xi-wi,右端点是xi+wi，就只要求最多的不重叠的线段数就可以了，然后就是乱搞了

#include <cstdio>
#include <algorithm>
using namespace std;
const int N=;
int n;
struct point{int x,w;}a[N];
inline bool cmp(point a,point b){return a.x<b.x;}
struct node{int l,r;}line[N];
inline bool cmp1(node a,node b){return (a.r!=b.r)?(a.r<b.r):(a.l<b.l);}
int main()
{
    int i,re=,tmp=-2e9; scanf("%d",&n); for(i=;i<=n;i++)scanf("%d%d",&a[i].x,&a[i].w); sort(a+,a+n+,cmp);
    for(i=;i<=n;i++)line[i].l=a[i].x-a[i].w,line[i].r=a[i].x+a[i].w; sort(line+,line+n+,cmp1);
    for(i=;i<=n;i++)
    {
        if(tmp<=line[i].l){tmp=line[i].r; re++;}
    }printf("%d\n",re);
}