CF1142C U2

题目链接：洛谷 codeforces

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$y>x^2+bx+c$也就是$y-x^2>bx+c$

左边是点，右边是直线.

维护上凸包.

虽然这么简单但就是做不出来。

#include<cstdio>
#include<algorithm>
#define Rint register int
using namespace std;
typedef long long LL;
const int N = 100003;
struct Point {
    LL x, y;
    inline Point(LL _x = 0, LL _y = 0): x(_x), y(_y){}
    inline Point operator - (const Point &o) const {return Point(x - o.x, y - o.y);}
    inline LL operator * (const Point &o) const {return x * o.y - y * o.x;}
    inline bool operator < (const Point &o) const {return x < o.x || x == o.x && y > o.y;}
} p[N], q[N], stk[N];
int n, m, top;
int main(){
    scanf("%d", &n);
    for(Rint i = 1;i <= n;i ++){
        scanf("%I64d %I64d", &p[i].x, &p[i].y);
        p[i].y -= p[i].x * p[i].x;
    }
    sort(p + 1, p + n + 1);
    q[m = 1] = p[1];
    for(Rint i = 2;i <= n;i ++) if(p[i].x != p[i - 1].x) q[++ m] = p[i];
    for(Rint i = 1;i <= m;i ++){
        while(top > 1 && (stk[top] - stk[top - 1]) * (q[i] - stk[top - 1]) >= 0) -- top;
        stk[++ top] = q[i];
    }
    printf("%d
", top - 1);
} // nantf tai qiang le!