BZOJ 1007. [HNOI2008]水平可见直线

可见的直线为一下凸壳。

先按斜率和截距从小到大排序，再用单调栈判断交点的相对位置即可。

#include <bits/stdc++.h>

const int N = 5e4 + 7;
const double eps = 1e-7;

inline int dcmp(double x) {
    if (fabs(x) < eps) return 0;
    return x < 0 ? -1 : 1;
}

struct P {
    double x, y;
    int id;
    inline bool operator < (const P &rhs) const {
        if (dcmp(x - rhs.x) == 0) return y < rhs.y;
        return x < rhs.x;
    }
} p[N];

int st[N], top;

inline double crossx(const P &a, const P &b) {
    return (a.y - b.y) / (b.x - a.x);
}

void ins(int id) {
    const P &cur = p[id];
    while (top) {
        if (dcmp(p[st[top]].x - cur.x) == 0) top--;
        else if (top > 1 && dcmp(crossx(p[st[top]], cur) - crossx(p[st[top]], p[st[top - 1]])) <= 0) top--;
        else break;
    }
    st[++top] = id;
}

int ans[N];

int main() {
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%lf%lf", &p[i].x, &p[i].y);
        p[i].id = i;
    }
    std::sort(p + 1, p + 1 + n);
    for (int i = 1; i <= n; i++)
        ins(i);
    for (int i = 1; i <= top; i++)
        ans[p[st[i]].id] = 1;
    for (int i = 1; i <= n; i++)
        if (ans[i])
            printf("%d ", i);
    return 0;
}