poj 1418 || zoj 1696 Viva Confetti

去北京前做的这个题，一直木有A。。。因为有一些情况纠结了，见下图。

黄色的盘子是第一个放的。这样的话，其他盘盘都把它的边缘给覆盖掉了，但是它依然是可见的！

这点需要处理下。

我的做法是，求圆和圆相交的交点，然后计算交点在某个圆内，按极角排序在同一个圆周上的点（记得去重），然后计算每小段弧的中点，然后看这个中点在几个圆盘里，记录这个点和对应的id（可以记成一个点一个id）。最后扫一遍中点，未被遮住的就算到答案里。

另外需要特殊处理的就是，如果一个盘盘未被覆盖，就是它是完整的，或者它完全被覆盖，这样的话计算交点是算不出来的，需要特判。

精度开12-15都没问题，开11就WA了。

#include <set>#include <map>#include <queue>#include <stack>#include <math.h>#include <stdio.h>#include <stdlib.h>#include <iostream>#include <limits.h>#include <string.h>#include <string>#include <algorithm>#define MID(x,y) ( ( x + y ) >> 1 )#define L(x) ( x << 1 )#define R(x) ( x << 1 | 1 )#define FOR(i,s,t) for(int i=(s); i<(t); i++)#define BUG puts("here!!!")#define STOP system("pause")#define file_r(x) freopen(x, "r", stdin)#define file_w(x) freopen(x, "w", stdout)using namespace std;const int MAX = 110;const double eps = 1e-12;bool dy(double x,double y){return x > y + eps;}// x > y bool xy(double x,double y){return x < y - eps;}// x < y bool dyd(double x,double y){ return x > y - eps;}// x >= y bool xyd(double x,double y){return x < y + eps;} // x <= y bool dd(double x,double y) {return fabs( x - y ) < eps;}  // x == struct point{double x, y;point(){}point(double xx, double yy):x(xx), y(yy){}bool operator==(const point &a){return dd(a.x, x) && dd(a.y, y);}void get(){scanf("%lf%lf", &x, &y);}};double crossProduct(point a,point b,point c)//向量 ac 在 ab 的方向 顺时针是正 {return (c.x - a.x)*(b.y - a.y) - (b.x - a.x)*(c.y - a.y);}double disp2p(point a,point b) //  a b 两点之间的距离 {return sqrt( ( a.x - b.x ) * ( a.x - b.x ) + ( a.y - b.y ) * ( a.y - b.y ) );}double disp2p_sqr(point a,point b){return ( ( a.x - b.x ) * ( a.x - b.x ) + ( a.y - b.y ) * ( a.y - b.y ) );}point l2l_inst_p(point u1,point u2,point v1,point v2){point ans = u1;double t = ((u1.x - v1.x)*(v1.y - v2.y) - (u1.y - v1.y)*(v1.x - v2.x))/((u1.x - u2.x)*(v1.y - v2.y) - (u1.y - u2.y)*(v1.x - v2.x));ans.x += (u2.x - u1.x)*t;ans.y += (u2.y - u1.y)*t;return ans;}struct NODE{point p;int id;};struct circle{point c;double r;int id;void get(){c.get();scanf("%lf", &r);}};circle cir[MAX];NODE a[MAX*MAX*2];point p[MAX][MAX*MAX*2];int len[MAX];bool c2c_inst(point a,double r1,point b,double r2){if( xy(disp2p(a,b),r1+r2) && dy(disp2p(a,b),fabs(r1 - r2)) )return true;return false;}void l2c_inst_p(point c,double r,point l1,point l2,point &p1,point &p2){point p = c;double t;p.x += l1.y - l2.y;p.y += l2.x - l1.x;p = l2l_inst_p(p,c,l1,l2);t = sqrt(r*r - disp2p(p,c)*disp2p(p,c))/disp2p(l1,l2);p1.x = p.x + (l2.x - l1.x)*t;p1.y = p.y + (l2.y - l1.y)*t;p2.x = p.x - (l2.x - l1.x)*t;p2.y = p.y - (l2.y - l1.y)*t;}void c2c_inst_p(point c1,double r1,point c2,double r2,point &p1,point &p2){point u,v;double t;t = (1 + (r1*r1 - r2*r2)/disp2p(c1,c2)/disp2p(c1,c2))/2;u.x = c1.x + (c2.x - c1.x)*t;u.y = c1.y + (c2.y - c1.y)*t;v.x = u.x + c1.y - c2.y;v.y = u.y - c1.x + c2.x;l2c_inst_p(c1,r1,u,v,p1,p2);}point C;bool cmp(const point& a,const point& b)  {  double t1 = atan2(a.y - C.y, a.x - C.x);double t2 = atan2(b.y - C.y, b.x - C.x);    if( dd(t1, t2) ) return xy(fabs(a.x),fabs(b.x));    return xy(t1, t2);  }bool cmp_equal(point &a, point &b){return dd(a.x, b.x) && dd(a.y, b.y);}bool cmp_NODE(NODE a, NODE b){if( a.id == b.id ){if( dd(a.p.x, b.p.x) )return xy(a.p.y, b.p.y);return xy(a.p.x, b.p.x);}return a.id < b.id;}bool cmp_NODE_equal(NODE a, NODE b){return dd(a.p.x, b.p.x) && a.id == b.id && dd(a.p.y, b.p.y);}point foot_line(point a,point l1,point l2)//ac在l1l2的逆时针方向 {point c;    c.x = a.x - l2.y + l1.y;    c.y = a.y + l2.x - l1.x;return c;}bool inst[MAX];bool see[MAX];point getmid(point a, point b, circle &c){point mid = point((a.x+b.x)/2, (a.y+b.y)/2);if( mid == c.c )mid = foot_line(c.c, a, b);point p1, p2;l2c_inst_p(c.c, c.r, c.c, mid, p1, p2);return dy(crossProduct(b, a, p1), 0) ? p2 : p1; }bool c2c_ainb(point a,double r1,point b,double r2){return xyd(disp2p(a,b),r2 - r1);//a在b中，如果是包括内切，用xyd }bool check(NODE a, int n){FOR(i, a.id+1, n)if( xy(disp2p_sqr(a.p, cir[i].c), cir[i].r*cir[i].r) )return false;return true;}bool check_cover_all(circle c, int n){FOR(i, c.id+1, n)if( c2c_ainb(c.c, c.r, cir[i].c, cir[i].r) )return false;return true;}int solve(int n){int sum = 0;memset(inst, false, sizeof(inst));memset(len, 0, sizeof(len));FOR(i, 0, n){FOR(k, i+1, n)if( c2c_inst(cir[i].c, cir[i].r, cir[k].c, cir[k].r) ){point a, b;inst[i] = inst[k] = true;c2c_inst_p(cir[i].c, cir[i].r, cir[k].c, cir[k].r, a, b);p[i][len[i]++] = p[k][len[k]++] = a;p[i][len[i]++] = p[k][len[k]++] = b;}}int l = 0;FOR(i, 0, n){C = cir[i].c;sort(p[i], p[i]+len[i], cmp);len[i] = unique(p[i], p[i]+len[i], cmp_equal) - p[i];p[i][len[i]] = p[i][0];FOR(k, 0, len[i]){point t = getmid(p[i][k], p[i][k+1], cir[i]);a[l].p = t;a[l++].id = i;FOR(j, 0, n)if( xy(disp2p_sqr(t, cir[j].c), cir[j].r*cir[j].r) ){a[l].p = t;a[l++].id = j;}}}sort(a, a+l, cmp_NODE);l = unique(a, a+l, cmp_NODE_equal) - a;memset(see, false, sizeof(see));FOR(i, 0, l)if( check(a[i], n) )see[a[i].id] = true;FOR(i, 0, n){if( see[i] ){sum++;continue;}if( !inst[i] && check_cover_all(cir[i], n) )sum++;}return sum;}int main(){int n;while( ~scanf("%d", &n) && n ){FOR(i, 0, n){cir[i].get();cir[i].id = i;}int ans = solve( n );printf("%d\n", ans);}return 0;}