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求 n 个圆的面积并
算法参照 这篇博客
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
using namespace std;
const int inf = 0x3f;
const int INF = 0x3f3f3f3f;
const int maxn = 1005;
const double eps = 1e-6;
typedef pair<double, double> PDD;
int n, L, R;
int x[maxn], y[maxn], r[maxn];
int nx[maxn], ny[maxn], nr[maxn];
int xl[maxn], xr[maxn];
int s[maxn];
PDD se[maxn];
inline bool cmp(int a, int b)
{
if(x[a] - r[a] == x[b] - r[b])
return x[a] + r[a] < x[b] + r[b];
return x[a] - r[a] < x[b] - r[b];
}
inline bool cmp0(int a, int b)
{
return r[a] > r[b];
}
inline double f(double v)
{
int sz = 0, i, j;
double ret = 0.0;
for(int i = L; i < R; i++)
{
if(v <= xl[i] || v >= xr[i]) continue;
j = s[i];
double d = sqrt(r[j] - (v - x[j]) * (v - x[j]));
se[sz].first = y[j] - d;
se[sz].second = y[j] + d;
++sz;
}
sort(se, se + sz);
for(i = 0; i < sz; i++)
{
double nowl, nowr;
nowl = se[i].first;
nowr = se[i].second;
for(j = i + 1; j < sz; ++j)
{
if(se[j].first > nowr) break;
else nowr = max(nowr, se[j].second);
}
ret += nowr - nowl;
i = j - 1;
}
return ret;
}
inline double rsimp(double l, double m, double r, double sl,
double sm, double sr, double tot)
{
double m1 = (l + m) * 0.5, m2 = (m + r) * 0.5;
double s0 = f(m1), s2 = f(m2);
double gl = (sl + sm + s0 * 4.0)*(m-l),
gr = (sm + sr + s2 * 4.0)*(r-m);
if(fabs(gl + gr - tot) < eps) return gl + gr;
return rsimp( l, m1, m, sl, s0, sm, gl) + rsimp( m, m2,r, sm, s2, sr, gr);
}
void work()
{
sort(s, s+n, cmp);
double lo, hi, ans = 0.0;
int i, j;
for(i = 0; i < n; i++)
{
xl[i] = x[s[i]] - r[s[i]];
xr[i] = x[s[i]] + r[s[i]];
r[s[i]] *= r[s[i]];
}
for(i = 0; i < n; i++)
{
int ilo, ihi;
ilo = xl[i];
ihi = xr[i];
for(j = i+1; j < n; j++)
{
if(xl[j] > ihi) break;
ihi = max(ihi, xr[j]);
}
lo = ilo, hi = ihi;
L = i; R = j;
double mid = (lo + hi) * 0.5;
double sl = f(lo), sm= f(mid), sr = f(hi);
double tot = sl + sr + sm * 4.0;
ans += rsimp(lo, mid, hi, sl, sm, sr, tot);
i = j - 1;
}
printf("%.3f\n", ans / 6.0);
}
int main()
{
int i, j, k;
while(~scanf("%d", &n))
{
i = 0; j = 0; k = 0;
for(i = 0; i < n; i++)
{
scanf("%d %d %d", &x[i], &y[i], &r[i]);
s[i] = i;
}
sort(s, s+n, cmp0);
for(i = 0; i < n; i++)
{
for(k = 0; k < j; k++)
{
if((nx[k]-x[s[i]])*(nx[k]-x[s[i]])+
(ny[k]-y[s[i]])*(ny[k]-y[s[i]])
<= (nr[k]-r[s[i]])*(nr[k]-r[s[i]]))
break;
}
if(k == j)
{
nx[j] = x[s[i]];
ny[j] = y[s[i]];
nr[j] = r[s[i]];
s[j] = j;
j++;
}
}
n = j;
for(i = 0; i < n; i++)
{
x[i] = nx[i];
y[i] = ny[i];
r[i] = nr[i];
}
work();
}
return 0;
}
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原文地址:http://www.cnblogs.com/tenlee/p/4544651.html