HDU5033-Building(维护单调栈)

题目链接

题意：给出n座大楼的位置以及高度，再给出m个人的位置，查询给出的人的位置所能看到的最大的仰角是多少。

思路：维护每两座的楼之间的斜率，使之成为一个凸面，用栈来维护，听了GG小伙伴的思路，可以将人当作高度为0的大楼来带入计算。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;

const int MAXN = 200005;
const double pi = acos(-1.0);

struct Build{
    Build() {
        memset(ang, 0, sizeof(ang));
    }
    double x, h;
    double ang[2];
    bool pos;
    int id;
}b[MAXN], q[MAXN];

int t, n, m;

bool cmp1(Build a, Build b) {
    return a.x < b.x;
}

bool cmp2(Build a, Build b) {
    return a.id < b.id;
}

double deal(Build a, Build b) {
    double dx = fabs(b.x - a.x);
    double dy = b.h - a.h;
    return dy / dx;
}

int main() {
    int cas, t = 1;
    scanf("%d", &cas);
    while (cas--) {
        scanf("%d", &n);
        for (int i = 0; i < n; i++) {
            scanf("%lf%lf", &b[i].x, &b[i].h);
            b[i].pos = false;
            b[i].id = i;
        }
        scanf("%d", &m);
        for (int i = n; i < n + m; i++) {
            scanf("%lf", &b[i].x);
            b[i].h = 0;
            b[i].pos = true;
            b[i].id = i;
        }

        n += m;
        sort(b, b + n, cmp1);
        q[0] = b[0];
        int top = 0;
        for (int i = 1; i < n; i++) {
            if (b[i].pos == false) {
                while (top && deal(b[i], q[top]) < deal(q[top], q[top - 1]))
                    top--;
                q[++top] = b[i];
            }
            else {
                int tmp = top;
                while (tmp && deal(b[i], q[tmp]) < deal(b[i], q[tmp - 1]))
                    tmp--;
                b[i].ang[0] = deal(b[i], q[tmp]);
            }
        }

        q[0] = b[n - 1];
        top = 0;
        for (int i = n - 2; i >= 0; i--) {
            if (b[i].pos == false) {
                while (top && deal(b[i], q[top]) < deal(q[top], q[top - 1]))
                    top--;
                q[++top] = b[i];
            }
            else {
                int tmp = top;
                while (tmp && deal(b[i], q[tmp]) < deal(b[i], q[tmp - 1]))
                    tmp--;
                b[i].ang[1] = deal(b[i], q[tmp]);
            }
        }

        printf("Case #%d:\n", t++);
        sort(b, b + n, cmp2);
        for (int i = 0; i < n; i++) {
            if (b[i].pos == true) {
                double ans = pi - atan(b[i].ang[0]) - atan(b[i].ang[1]);
                ans = ans / pi * 180;
                printf("%.4lf\n", ans);
            }
        }
    }
    return 0;
}