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 = 2010;
const double pi = acos(-1);
struct data {
    double l, r;
    int id;
} x[N], a[N];
int n, ans;
double R;
int tree[N << 2], 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) >> 1;
    if(pos <= mid)
        update(l, mid, x << 1, pos, t);
    else
        update(mid + 1, r, x << 1 | 1, pos, t);
    tree[x] = max(tree[x << 1], tree[x << 1 | 1]);
}
int query(int l, int r, int x, int a, int b)
{
    if(l > b || r < a) 
        return 0;
    if(l >= a && r <= b) 
        return tree[x];
    int mid = (l + r) >> 1;
    return max(query(l, mid, x << 1, a, b),
               query(mid + 1, r, x << 1 | 1, a, b));
}
int lis(int t)
{
    if(!t) return 0;
    int ans = 1;
    dp[1] = 1;
    memset(tree, 0, sizeof(tree));
    update(1, n, 1, x[1].id, dp[1]);
    for(int i = 2; i <= t; ++i)
    {
        dp[i] = query(1, n, 1, 1, x[i].id) + 1;
        ans = max(ans, dp[i]);
        update(1, n, 1, 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 = 1; 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 += 2 * pi;
        while(a[i].r >= pi)
            a[i].r -= 2 * pi;
        if(a[i].l > a[i].r)
            swap(a[i].l, a[i].r);
    }
    sort(a + 1, a + n + 1, cp1);
    for(int i = 1; i <= n; ++i)
        a[i].id = i;
    sort(a + 1, a + n + 1, cp);
    for(int i = 1; i <= n; ++i)
    {
        int t = 0;
        for(int j = i + 1; 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) + 1);
    }
    printf("%d\n", ans);
    return 0;
}