hdu 4063 Aircraft

相信在比赛的时候看这题的大家，都很无语吧~~！！

我是卡了整场了。。。中途看了其他题几眼，又回来这题了。。。

我第一个就是看的这个题（因为题目短，而且看着比较顺眼。。。）。。。然后就走向了一条不归路。。。泪。。。

第一反应，这不是裸的最短路么，很快敲了个，直接用floyd了，比较懒，顺利WA。。

然后想到了，可以不经过圆心的，那不就是经过交点喽？？？

然后套模板，圆相交模板。。。算出来交点，然后建图。。。用SPFA。。。顺利WA。。。

后来回来看这个题，想到了，如下图这个trick，谈不上trick吧，就是完全可以不经过任何交点到达另外一个点。这么一来，任意两个交点（包括圆心，以及俩圆的交点）都可能存在这样的路。

判断是个问题，后来小白说，可以一段一段判断，想了下，不难实现，就写了。。。结果WA了。。。结果比赛结束这题还是没能过。。。

后来找大牛要了几组数据，发现有个小于等于写成小于了，改了后，就A掉了。。

其中这个判断线段是否完全被圆形覆盖应该是比较困扰的，可以求出线段与圆所有交点，然后对这些交点排序，然后判断相邻俩点之间的线段是否被一个圆覆盖，这个的话可以判断这俩交点之间的中点是否被一个圆覆盖。

总体的时间复杂度应该是N^5。。到N^6之间。。

#include <set>#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 = 1000;const double eps = 1e-6;const double inf = 1e50;struct point{double x, y;void get(){scanf("%lf%lf", &x, &y);}};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 == ydouble 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 ) );}bool operator==(point a, point b){return dd(a.x, b.x) && dd(a.y, b.y);}typedef struct NODE{int to;double len;NODE *next;}NODE;NODE *p[MAX],node[MAX*MAX];struct circle{point c;double r;};circle c[MAX];int cou;void init(){cou = 0;memset(p, 0, sizeof(p));}void Add(int from,int to,double len){node[cou].next = p[from];node[cou].to = to;node[cou].len = len;p[from] = &node[cou++];}queue<int> q;double SPFA_List(int from,int to,int n){while( !q.empty() ) q.pop();double dis[MAX];bool inq[MAX];for(int i=0; i<n; i++)dis[i] = inf;memset(inq,false,sizeof(inq));dis[from] = 0;q.push(from);inq[from] = 1;while( !q.empty() ){int now = q.front();q.pop();inq[now] = false;NODE *head = p[now];while( head != NULL ){int v = head->to;double len = head->len;if( dy(dis[v], dis[now] + len) ){dis[v] = dis[now] + len;if( !inq[v] ){inq[v] = true;q.push(v);}}head = head->next;}}if( dd(dis[to], inf) ) return -1;return dis[to];}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;}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);}bool c2c_tangent(point a,double r1,point b,double r2){if( dd(disp2p(a,b),r1+r2) || dd(disp2p(a,b),fabs(r1-r2)) )return true;return false;}point c2c_tangent_p(point a,double r1,point b,double r2){point t;if( dd(disp2p(a,b),r1 + r2)  ){t.x = (r1*b.x + r2*a.x)/(r1 + r2);t.y = (r1*b.y + r2*a.y)/(r1 + r2);return t;}t.x = (r1*b.x - r2*a.x)/(r1 - r2);t.y = (r1*b.y - r2*a.y)/(r1 - r2);return t;}point g[MAX];bool f[MAX][MAX];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 disp2l(point a,point l1,point l2){return fabs( crossProduct(a,l1,l2) )/disp2p(l1,l2);}bool onSegment(point a, point b, point c){if( dd(crossProduct(a,b,c),0.0) && dyd(c.x,min(a.x,b.x)) && xyd(c.x,max(a.x,b.x)) && dyd(c.y,min(a.y,b.y)) && xyd(c.y,max(a.y,b.y)) )return true;return false;}bool cmp(point a, point b){if( dd(a.x, b.x) )return xy(a.y, b.y);return xy(a.x, b.x);}point tp[MAX];bool check(int cnt, int n){FOR(i, 1, cnt){point tt;tt.x = (tp[i].x + tp[i-1].x)/2;tt.y = (tp[i].y + tp[i-1].y)/2;bool f = false;FOR(k, 0, n)if( xyd(disp2p(c[k].c, tt), c[k].r) ){f = true;break;}if( !f )return false;}return true;}double solve(int n){int l = 0;int s = 0, t = n-1;FOR(i, 0, n)g[l++] = c[i].c;FOR(i, 0, n){FOR(k, i+1, n){if( dy(disp2p(c[i].c, c[k].c), c[i].r + c[k].r) || c[i].c == c[k].c )continue;if( c2c_tangent(c[i].c, c[i].r, c[k].c, c[k].r) ){point tt = c2c_tangent_p(c[i].c, c[i].r, c[k].c, c[k].r);g[l++] = tt;continue;}point t1, t2;c2c_inst_p(c[i].c, c[i].r, c[k].c, c[k].r, t1, t2);g[l++] = t1;g[l++] = t2;}}memset(f, false, sizeof(f));int tmp[MAX];FOR(i, 0, n){int cnt = 0;FOR(k, 0, l)if( xyd(disp2p(c[i].c, g[k]), c[i].r) )tmp[cnt++] = k;FOR(k, 0, cnt)FOR(j, k+1, cnt){int x = tmp[k], y = tmp[j];if( f[x][y] ) continue;double dis = disp2p(g[x], g[y]);Add(x, y, dis);Add(y, x, dis);f[x][y] = f[y][x] = true;}}FOR(i, 0, l)FOR(k, i+1, l){if( f[i][k] ) continue;int cnt = 0;tp[cnt++] = g[i];tp[cnt++] = g[k];FOR(j, 0, n)if( xyd(disp2l(c[j].c, g[i], g[k]),c[j].r) ){point t1, t2;l2c_inst_p(c[j].c, c[j].r, g[i], g[k], t1, t2);if( onSegment(g[i], g[k], t1) )tp[cnt++] = t1;if( onSegment(g[i], g[k], t2) )tp[cnt++] = t2;}sort(tp, tp+cnt, cmp);if( check(cnt, n) ){Add(i, k, disp2p(g[i], g[k]));Add(k, i, disp2p(g[k], g[i]));}}return SPFA_List(s, t, l);}int main(){int ncases, n, ind = 1;scanf("%d", &ncases);while( ncases-- ){scanf("%d", &n);init();FOR(i, 0, n){c[i].c.get();scanf("%lf", &c[i].r);}double ans = solve(n);printf("Case %d: ",ind++);if( ans < -eps )puts("No such path.");elseprintf("%.4lf\n", ans);}return 0;}