HDU 5033 Building（北京网络赛B题）

HDU 5033 Building

题目链接

思路：利用单调栈维护建筑建的斜线，保持斜率单调性，然后可以把查询当成高度为0的建筑，和建筑和在一起考虑，从左往右和从右往左各扫一遍即可

代码：

#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <algorithm>
using namespace std;

const int N = 200005;
const double pi = acos(-1.0);
const double eps = 1e-8;

int t, n, q;
struct Build {
double x, h;
bool isq;
double s[2];
int id;
} b
, Q
;

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

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

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

int main() {
int cas = 0;
scanf("%d", &t);
while (t--) {
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%lf%lf", &b[i].x, &b[i].h);
b[i].isq = false;
b[i].id = i;
}
scanf("%d", &q);
for (int i = n; i < n + q; i++) {
scanf("%lf", &b[i].x);
b[i].h = 0;
b[i].isq = true;
b[i].id = i;
}
n += q;
sort(b, b + n, cmp);
Q[0] = b[0];
int top = 0;
int u = 0;
for (int i = 1; i < n; i++) {
if (b[i].isq == false) {
while (top && cal(b[i], Q[top]) < cal(Q[top], Q[top - 1]))
top--;
Q[++top] = b[i];
} else {
int tmp = top;
while (tmp && cal(b[i], Q[tmp]) < cal(b[i], Q[tmp - 1]))
tmp--;
b[i].s[u] = cal(b[i], Q[tmp]);
}
}
reverse(b, b + n);
Q[0] = b[0];
top = 0;
u = 1;
for (int i = 1; i < n; i++) {
if (b[i].isq == false) {
while (top && cal(b[i], Q[top]) < cal(Q[top], Q[top - 1]))
top--;
Q[++top] = b[i];
} else {
int tmp = top;
while (tmp && cal(b[i], Q[tmp]) < cal(b[i], Q[tmp - 1]))
tmp--;
b[i].s[u] = cal(b[i], Q[tmp]);
}
}
sort(b, b + n, cmpid);
printf("Case #%d:\n", ++cas);
for (int i = 0; i < n; i++) {
if (b[i].isq) {
double ans = pi - atan(b[i].s[0]) - atan(b[i].s[1]);
ans = ans / pi * 180;
printf("%.10lf\n", ans);
}
}
}
return 0;
}