bzoj3663

几何+lis

很巧妙。直接做很困难，那么我们转化一下，把每个点能看见的圆弧画出来。只有这些圆弧相交时才满足条件。

那么也就是找出圆上尽量多两两相交的区间。

所以我们先按左端点极角排序，然后固定一个必须选的区间，找出所有和它相交的区间，按右端点做lis就行了。

#include<bits/stdc++.h>
using namespace std;
const int N = ;
const double pi = acos(-);
struct data {
    double l, r;
    int id;
} x[N], a[N];
int n, ans;
double R;
int tree[N << ], dp[N];
void update(int l, int r, int x, int pos, int t)
{
    if(l == r)
    {
        tree[x] = t;
        return;
    }
    int mid = (l + r) >> ;
    if(pos <= mid)
        update(l, mid, x << , pos, t);
    else
        update(mid + , r, x <<  | , pos, t);
    tree[x] = max(tree[x << ], tree[x <<  | ]);
}
int query(int l, int r, int x, int a, int b)
{
    if(l > b || r < a)
        return ;
    if(l >= a && r <= b)
        return tree[x];
    int mid = (l + r) >> ;
    return max(query(l, mid, x << , a, b),
               query(mid + , r, x <<  | , a, b));
}
int lis(int t)
{
    if(!t) return ;
    int ans = ;
    dp[] = ;
    memset(tree, , sizeof(tree));
    update(, n, , x[].id, dp[]);
    for(int i = ; i <= t; ++i)
    {
        dp[i] = query(, n, , , x[i].id) + ;
        ans = max(ans, dp[i]);
        update(, n, , x[i].id, dp[i]);
    }
    return ans;
}
bool cp1(data x, data y)
{
    return x.r < y.r;
}
bool cp(data x, data y)
{
    return x.l < y.l;
}
int main()
{
    scanf("%d%lf", &n, &R);
    for(int i = ; i <= n; ++i)
    {
        double x, y; scanf("%lf%lf", &x, &y);
        double degx = atan2(y, x);
        double degc = acos(R / sqrt(x * x + y * y));
        a[i].l = degx - degc;
        a[i].r = degx + degc;
        while(a[i].l <= -pi)
            a[i].l +=  * pi;
        while(a[i].r >= pi)
            a[i].r -=  * pi;
        if(a[i].l > a[i].r)
            swap(a[i].l, a[i].r);
    }
    sort(a + , a + n + , cp1);
    for(int i = ; i <= n; ++i)
        a[i].id = i;
    sort(a + , a + n + , cp);
    for(int i = ; i <= n; ++i)
    {
        int t = ;
        for(int j = i + ; j <= n; ++j)
            if(a[j].l <= a[i].r && a[j].r >= a[i].r)
                x[++t] = a[j];
        ans = max(ans, lis(t) + );
    }
    printf("%d\n", ans);
    return ;
}