10000.cpp

# pragma GCC optimize ("O3")
# pragma GCC optimize ("Ofast")
# pragma GCC optimize ("unroll-loops")
# pragma GCC target("sse,sse2,sse3,ssse3,sse4,avx,avx2")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>

using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;

typedef unsigned int ui;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef tuple<int, int, int> ti3;
typedef tuple<int, int, int, int> ti4;
typedef stack<int> si;
typedef queue<int> qi;
typedef priority_queue<int> pqi;
typedef pair<ll, ll> pll;
typedef vector<ll> vl;
typedef tuple<ll, ll, ll> tl3;
typedef tuple<ll, ll, ll, ll> tl4;
typedef stack<ll> sl;
typedef queue<ll> ql;
typedef priority_queue<ll> pql;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>ordered_set;

const int dx[4] = { 1,0,-1,0 };
const int dy[4] = { 0,1,0,-1 };
const int ddx[8] = { 0,0,1,1,1,-1,-1,-1 }, ddy[8] = { 1,-1,1,0,-1,1,0,-1 };
ll POW(ll a, ll b, ll MMM) { ll ret = 1; for (; b; b >>= 1, a = (a*a) % MMM)if (b & 1)ret = (ret*a) % MMM; return ret; }
ll GCD(ll a, ll b) { return b ? GCD(b, a%b) : a; }
ll LCM(ll a, ll b) { if (a == 0 || b == 0)return a + b; return a / GCD(a, b) * b; }
ll INV(ll a, ll m) {
	ll m0 = m, y = 0, x = 1;
	if (m == 1)	return 0;
	while (a > 1) {
		ll q = a / m;
		ll t = m;
		m = a % m, a = t;
		t = y;
		y = x - q * y;
		x = t;
	}
	if (x < 0) x += m0;
	return x;
}
pll EXGCD(ll a, ll b) {
	if (b == 0) return { 1,0 };
	auto t = EXGCD(b, a%b);
	return { t.second,t.first - t.second*(a / b) };
}
bool OOB(ll x, ll y, ll N, ll M) { return 0 > x || x >= N || 0 > y || y >= M; }
#define X first
#define Y second
#define rep(i,a,b) for(int i = a; i < b; i++)
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define sz(a) ((int)(a.size()))
#define sf1(a) cin >> a
#define sf2(a,b) cin >> a >> b
#define sf3(a,b,c) cin >> a >> b >> c
#define sf4(a,b,c,d) cin >> a >> b >> c >> d
#define sf5(a,b,c,d,e) cin >> a >> b >> c >> d >> e
#define sf6(a,b,c,d,e,f) cin >> a >> b >> c >> d >> e >> f
#define pf1(a) cout << (a) << ' '
#define pf2(a,b) cout << (a) << ' ' << (b) << ' '
#define pf3(a,b,c) cout << (a) << ' ' << (b) << ' '<< (c) << ' '
#define pf4(a,b,c,d) cout << (a) << ' ' << (b) << ' '<< (c) << ' '<< (d) << ' '
#define pf5(a,b,c,d,e) cout << (a) << ' ' << (b) << ' '<< (c) << ' '<< (d) << ' '<< (e) << ' '
#define pf6(a,b,c,d,e,f) cout << (a) << ' ' << (b) << ' '<< (c) << ' '<< (d) << ' '<< (e) << ' ' << (f) << ' '
#define pf0l() cout << '\n';
#define pf1l(a) cout << (a) << '\n'
#define pf2l(a,b) cout << (a) << ' ' << (b) << '\n'
#define pf3l(a,b,c) cout << (a) << ' ' << (b) << ' '<< (c) << '\n'
#define pf4l(a,b,c,d) cout << (a) << ' ' << (b) << ' '<< (c) << ' '<< (d) << '\n'
#define pf5l(a,b,c,d,e) cout << (a) << ' ' << (b) << ' '<< (c) << ' '<< (d) << ' '<< (e) << '\n'
#define pf6l(a,b,c,d,e,f) cout << (a) << ' ' << (b) << ' '<< (c) << ' '<< (d) << ' '<< (e) << ' ' << (f) << '\n'
#define pfvec(V) for(auto const &t : V) pf1(t)
#define pfvecl(V) for(auto const &t : V) pf1(t); pf0l()

pll c[300005];
int isdivide[300005]; // 해당 원이 2개로 갈라졌는가.
bool cmp(pll a, pll b){
  if(a.X-a.Y != b.X-b.Y)
    return a.X-a.Y < b.X-b.Y;
  return a.X > b.X;
}
int n;
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed;
  cout.precision(20);
///////////////////////////////////////////////
  sf1(n);
  fill(isdivide,isdivide+n+1,1);
  c[0] = {0,1e12}; // 구현의 편의를 위해 아주 큰 원 하나로 다 감쌈
  rep(i,1,n+1) sf2(c[i].X,c[i].Y);
  sort(c,c+n+1,cmp);
//  rep(i,0,n+1) pf2l(c[i].X,c[i].Y);
//  return 0;
  stack<pll> S;
  S.push({0,c[0].X-c[0].Y});
  int cnt = n+1;
  rep(i,1,n+1){
    int cur = S.top().X;
    while(c[i].X > c[cur].X+c[cur].Y){ // c[cur]의 외부에 c[i]가 있는 동안 stack에서 pop
      if(S.top().Y == c[cur].X+c[cur].Y) cnt++;
      S.pop();
      cur = S.top().X;
    }
    if(c[i].X <= c[cur].X+c[cur].Y){ // c[cur]의 내부에 c[i]가 있을 경우
      if(c[i].X-c[i].Y == S.top().Y){ // 아귀가 딱 맞는 경우
        S.top().Y = c[i].X+c[i].Y;
      }
//      else{
//        isdivide[cur] = 0;
//      }
    }
    S.push({i,c[i].X-c[i].Y});
  }
  int cur;
  while(!S.empty()){
    cur = S.top().X;
    if(S.top().Y == c[cur].X+c[cur].Y) cnt++;
    S.pop();    
  }
  pf1l(cnt);
}