题解【洛谷P6875】[COCI2013-2014#6] KRUŽNICE

题面

首先，答案最小为 (n+1)。

注意到圆与圆之间互不相交，因此只有圆的直径上被小圆全部覆盖答案才会 (+1)。

如何判断这一情况？

首先可以将坐标离散化，然后按半径大小从小到大加入圆（因为半径大的圆只会被半径比它小的圆覆盖直径），用线段树维护即可。

由于我们维护的是线段之间的关系，因此加入直径时左端点要 (+1)。

#include <bits/stdc++.h>
#define DEBUG fprintf(stderr, "Passing [%s] line %d
", __FUNCTION__, __LINE__)
#define File(x) freopen(x".in","r",stdin); freopen(x".out","w",stdout)
#define DC int T = gi <int> (); while (T--)

using namespace std;

typedef long long LL;
typedef pair <int, int> PII;
typedef pair <int, PII> PIII;

template <typename T>
inline T gi()
{
	T f = 1, x = 0; char c = getchar();
	while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
	while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return f * x;
}

const int INF = 0x3f3f3f3f, N = 600003, M = N << 1;

int n;
struct Circle
{
	int l, r, bj;
} a[N];
int b[N];

inline bool cmp(Circle x, Circle y)
{
	return x.bj < y.bj;
}

namespace SGT
{
	bool tr[N << 2];
#define ls(p) (p << 1)
#define rs(p) (p << 1 | 1)
	void pushup(int p) {tr[p] = tr[ls(p)] & tr[rs(p)];}
	void pushdown(int p) {tr[ls(p)] |= tr[p], tr[rs(p)] |= tr[p];}
	void modify(int ql, int qr, int l, int r, int p)
	{
		if (ql <= l && r <= qr) {tr[p] = 1; return;}
		pushdown(p);
		int mid = (l + r) >> 1;
		if (ql <= mid) modify(ql, qr, l, mid, ls(p));
		if (qr > mid) modify(ql, qr, mid + 1, r, rs(p));
		pushup(p);
		return ;
	}
	bool query(int ql, int qr, int l, int r, int p)
	{
		if (ql <= l && r <= qr) return tr[p];
		pushdown(p);
		int mid = (l + r) >> 1;
		bool ans = 1;
		if (ql <= mid) ans = query(ql, qr, l, mid, ls(p));
		if (qr > mid) ans &= query(ql, qr, mid + 1, r, rs(p));
		return ans;
	}
} //namespace SGT

int main()
{
//	File("");
	n = gi <int> ();
	for (int i = 1; i <= n; i+=1)
	{
		int x = gi <int> (), r = gi <int> ();
		a[i] = (Circle){x - r, x + r, r};
		b[i * 2 - 1] = x - r, b[i * 2] = x + r;
	}
	sort(b + 1, b + 1 + n * 2);
	int tot = unique(b + 1, b + 1 + n * 2) - b - 1;
	for (int i = 1; i <= n; i+=1)
		a[i].l = lower_bound(b + 1, b + 1 + tot, a[i].l) - b, a[i].r = lower_bound(b + 1, b + 1 + tot, a[i].r) - b;
	sort(a + 1, a + 1 + n, cmp);
	int ans = n + 1;
	for (int i = 1; i <= n; i+=1)
	{
		bool all1 = SGT :: query(a[i].l + 1, a[i].r, 1, n * 2, 1);
		if (all1) ++ans;
		SGT :: modify(a[i].l + 1, a[i].r, 1, n * 2, 1);
	}
	printf("%d
", ans);
	return 0;
}