题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6354

题目:

Everything Has Changed(HDU6354+圆交+求周长)-LMLPHP

Everything Has Changed(HDU6354+圆交+求周长)-LMLPHP

题意:用一堆圆来切割一个圆心为原点,半径为R的圆A,问切割完毕后圆A外围剩余部分的周长(图中的红线部分)。

思路:首先判定圆与圆A的关系,这题我们只需要与A内切、相交的圆。然后就是求每个圆把圆A切割掉多少周长,增加了多少周长(因为圆A被切割的部分在切割后绝对是内凹的,此时周长是增加的),内切的时候直接加上切割圆的周长(如最上面的那个小圆),相交的圆部分我采用的方法是用余弦定理(A的半径记为R,切割圆半径为r,二者的圆心距离为d,圆心的连线与 圆A和一个交点的夹角为a,则2*d*R*cosa=R*R+d*d-r*r)求出夹角a,再用弧长公式l=a*r求出弧长最后进行加减即可。

代码实现如下:

 #include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std; typedef long long ll;
typedef pair<ll, ll> pll;
typedef pair<ll, int> pli;
typedef pair<int, ll> pil;;
typedef pair<int, int> pii;
typedef unsigned long long ull; #define lson i<<1
#define rson i<<1|1
#define bug printf("*********\n");
#define FIN freopen("D://code//in.txt", "r", stdin);
#define debug(x) cout<<"["<<x<<"]" <<endl;
#define IO ios::sync_with_stdio(false),cin.tie(0); const double eps = 1e-;
const int mod = ;
const int maxn = 1e6 + ;
const double pi = acos(-);
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f; typedef struct stu {
double x,y;
} point; double Distance(point a,point b) {
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
} double Inter(point a,double R,point b,double r) { //变化的周长
double dis=Distance(a,b);
double angle1=acos((R*R+dis*dis-r*r)/(2.0*R*dis));
double angle2=acos((r*r+dis*dis-R*R)/(2.0*r*dis));
double s=r*angle2*-R*angle1*;
return s;
} int t, m, R;
double x, y, r, ans;
stu o, p; int main() {
//FIN;
scanf("%d", &t);
while(t--) {
scanf("%d%d", &m, &R);
ans = * pi * R;
o.x = , o.y = ;
for(int i = ; i <= m; i++) {
scanf("%lf%lf%lf", &x, &y, &r);
p.x = x, p.y = y;
double d = Distance(o, p);
if(d - R - r >= eps) continue; //外离
if(fabs(R - r) - d > eps) continue; //内离
if(R == r + d) { //内切
ans += * pi * r;
} else { //相交
ans += Inter(o, R, p, r);
}
}
printf("%.12f\n", ans);
}
return ;
}
05-13 04:55