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题目链接

  本题是2005年icpc欧洲区域赛中欧赛区的题目

题意

  某国是一个长n公里，宽m公里的矩形，边平行于坐标轴，左上角的坐标为(0,0)，右下角的坐标为(n,m)。该国领土上有一些核电厂。核电厂有大小之分，小核电厂周围0.58公里内都不能种植大麦，大核电厂周围1.31公里内都不能种植大麦。你的任务是计算出可以种植大麦的总面积。

分析

  经典题目：求圆的面积并，用离散化扫描法求解，详细思路《算法竞赛入门经典–训练指南》第433页有写，这里只说一些编程细节：
  调用数学运算函数sqrt、acos这种函数时一定要消除数值异常，推荐这样写：sqrt(max(x, 0.))，acos(max(min(x, 1.), -1.))。
  计算两圆的交点也要消除数值异常：

#define eps 1e-9
double dx = cx2-cx1, dy = cy2-cy1, d = sqr(dx*dx + dy*dy);
if (d > r1+r2-eps || d < eps) return;

  某个扫描段只有一个上/下圆弧时面积的计算要特殊考虑。

测试数据

  给一份测试数据。

AC 代码

#include <iostream>
#include <iomanip>
#include <algorithm>
#include <cmath>
using namespace std;#define rs .58
#define rl 1.31
#define eps 1e-9
#define M 40200
#define N 102
#define sqr(x) sqrt(max(x, 0.))
#define aco(x) acos(max(min(x, 1.), -1.))
int xs[N], ys[N], xl[N], yl[N], a[N], b[N], m, n, ks, kl, t, c; double x[M];struct arc {int cx, cy; bool s, t; double y;double area(double x1, double x2) const {double r = s ? rs : rl, dx1 = x1 - cx, dx2 = x2 - cx,y1 = cy - sqr(r*r - dx1*dx1), y2 = cy - sqr(r*r - dx2*dx2), dy1 = y1 - cy, dy2 = y2 - cy;double t = aco((dx1*dx2+dy1*dy2)/sqr(dx1*dx1+dy1*dy1)/sqr(dx2*dx2+dy2*dy2));return .5 * (r*r*t - abs(dx1*dy2-dx2*dy1));}double yy(double x) const {double r = s ? rs : rl, dx = x - cx;return t ? cy + sqr(r*r - dx*dx) : cy - sqr(r*r - dx*dx);}bool operator< (const arc& rhs) const {return y < rhs.y;}
} q[N<<2];bool cmps(int i, int j) {return xs[i] < xs[j];
}bool cmpl(int i, int j) {return xl[i] < xl[j];
}void intersect(int cx1, int cy1, int cx2, int cy2, double r1, double r2) {double dx = cx2-cx1, dy = cy2-cy1, d = sqr(dx*dx + dy*dy);if (d > r1+r2-eps || d < eps) return;double d1 = .5 * (d + (r1*r1 - r2*r2)/d), s = sqr(r1*r1-d1*d1), x0 = cx1 + dx*d1/d;d = dy*s/d; x[t++] = x0 - d; x[t++] = x0 + d;
}void intersect(int cx, int cy, double r) {if (cy < r) {double s = sqr(r*r-cy*cy); x[t++] = cx - s; x[t++] = cx + s;}if (cy + r > m) {double s = sqr(r*r-(m-cy)*(m-cy)); x[t++] = cx - s; x[t++] = cx + s;}
}void add_arc(int cx, int cy, bool s, double x) {double r = s ? rs : rl;if (abs(x-cx) > r-eps) return;double d = sqr(r*r-(x-cx)*(x-cx));if (cy > d) q[c++] = {cx, cy, s, false, cy-d};if (cy+d < m) q[c++] = {cx, cy, s, true, cy+d};
}double solve() {for (int i=t=0; i<ks; ++i) {cin >> xs[i] >> ys[i], x[t++] = xs[i]-rs, x[t++] = xs[i]+rs, a[i] = i, intersect(xs[i], ys[i], rs);for (int j=0; j<i; ++j) intersect(xs[i], ys[i], xs[j], ys[j], rs, rs);}for (int i=0; i<kl; ++i) {cin >> xl[i] >> yl[i], x[t++] = xl[i]-rl, x[t++] = xl[i]+rl, b[i] = i, intersect(xl[i], yl[i], rl);for (int j=0; j<i; ++j) intersect(xl[i], yl[i], xl[j], yl[j], rl, rl);for (int j=0; j<ks; ++j) intersect(xs[j], ys[j], xl[i], yl[i], rs, rl);}x[t++] = 0.; x[t++] = n; sort(a, a+ks, cmps); sort(b, b+kl, cmpl); sort(x, x+t);double s = m*n; int h1 = 0, h2 = 0;for (int i=1; i<t; ++i) {double x1 = x[i-1], x2 = x[i], xc = .5*(x1+x2); c = 0;if (x1 > x2-eps || x1 < -eps || x2 > n+eps) continue;while (h1 < ks && xs[a[h1]]+rs < x1-eps) ++h1;for (int j=h1; j<ks && xs[a[j]]-rs < x2+eps; ++j) add_arc(xs[a[j]], ys[a[j]], true, xc);while (h2 < kl && xl[b[h2]]+rl < x1-eps) ++h2;for (int j=h2; j<kl && xl[b[j]]-rl < x2+eps; ++j) add_arc(xl[b[j]], yl[b[j]], false, xc);sort(q, q+c);for (int j=0, cc=0; j<c; ++j) if (q[j].t) {if (cc) {if (--cc == 0) s -= q[j].area(x1, x2) + .5*(x2-x1)*(q[j].yy(x1)+q[j].yy(x2));} else s -= q[j].area(x1, x2) + .5*(x2-x1)*(q[j].yy(x1)+q[j].yy(x2));} else if (++cc == 1) s -= q[j].area(x1, x2) - .5*(x2-x1)*(q[j].yy(x1)+q[j].yy(x2)-(j==c-1 ? 2.*m : 0.));}return s;
}int main() {cout << fixed << setprecision(2);while (cin >> n >> m >> ks >> kl && (n || m || ks || kl)) cout << solve() << endl;return 0;
}