HDU 5033 Building(单调栈维护凸包)

盗张图：来自http://blog.csdn.net/xuechelingxiao/article/details/39494433

题目大意：有一排建筑物坐落在一条直线上，每个建筑物都有一定的高度，给出一个X坐标，高度为0，问X位置能看到的视角是多少度。如图：

图一：  

图二：

图一为样例一，图二为样例三，红色部分为高楼，蓝色虚线为视角的最大范围。

思路：维护一个上凸的凸壳，递减的。也就是这样的。

分别找一下左边的和右边的就可以求出来答案

#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 100005;
const double PI = acos(-1);
struct Node {
    int x, h;
    bool operator < (const Node &a) const {
        return x < a.x;
    }
}node[maxn<<2], stack[maxn<<2];
int T, n, q;
double ans[maxn];

int check(Node &a, Node &b, Node c) {  
    if (c.h <= 0)  
        c.h = 0;  
    return (long long)(b.x - a.x) * (c.h - a.h) >= (long long)(c.x - a.x) * (b.h - a.h);//前面一定要加上long long ，或者用double也行，不然他的乘积有可能爆int 
} 
double getAngle(const Node &p1, const Node &p2)
{
    return atan((double)(p2.x - p1.x) / (double)p1.h);
}
void solve()
{
    int head = 0;
    for (int i = 0; i < n + q; i++)
    {
        if (node[i].h <= 0)
        {
            while (head >= 2 && check(stack[head - 2], stack[head - 1], node[i]))
                head--;
            ans[-node[i].h] += getAngle(stack[head - 1], node[i]);
        }
        else
        {
            while (head && stack[head - 1].h <= node[i].h)
                head--;
            while (head >=2 && check(stack[head - 2], stack[head - 1], node[i]))
                head--;
            stack[head++] = node[i];
        }
    }
}
int main()
{
    int kase = 0;
    scanf("%d", &T);
    while (T--)
    {
        scanf("%d", &n);
        for (int i = 0; i < n; i++)
            scanf("%d %d", &node[i].x, &node[i].h);
        scanf("%d", &q);
        for (int i = 0; i < q; i++)
        {
            scanf("%d", &node[i + n].x);
            node[i + n].h = -i;
        }
        memset(ans, 0, sizeof(ans));
        sort(node, node + n + q);
        solve();
        reverse(node, node + n + q);
        for (int i = 0; i < n + q; i++)
            node[i].x = 10000000 - node[i].x;
        solve();
        printf("Case #%d:
", ++kase);
        for (int i = 0; i < q; i++)
            printf("%.10f
", ans[i] * 180.0 / PI);
    }
    return 0;
}