106 lines
2.3 KiB
C++
106 lines
2.3 KiB
C++
//weiwancheng
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#include<iostream>
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#include<cstdio>
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#include<cmath>
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#include<cstring>
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using namespace std;
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const double inf=1e100,EPS=1e-6;
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const int N=101;
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double l,now;
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int n;
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struct edge
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{
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double x[2],y[2];
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} e[N];
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//big bug:double->int
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double length(double x1,double y1,double x2,double y2)
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{
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double x=x1-x2,y=y1-y2;
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return sqrt(x*x+y*y);
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}
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double cross(double x1,double y1,double x2,double y2)
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{
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return x1*y2-x2*y1;
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}
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double same(double x0,double y0,double x3,double y3,double x1,double y1,double x2,double y2)
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{
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double x=x3-x0,y=y3-y0;//y0->x0
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return cross(x,y,x1-x0,y1-y0)*cross(x,y,x2-x0,y2-y0);
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}
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//WA double 判断正负要用
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int sign(double d)
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{
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if(fabs(d)<EPS)
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{
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return 0;
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}
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else return d>0? 1:-1;
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}
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bool ob(int li,int lj,int ri,int rj)
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{
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for(int i=li+1; i<ri; i++)
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{
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for(int j=0; j<=4; j+=2)
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{
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// printf("(%.0lf,%.0lf)(%.0lf,%.0lf)/(%.0lf,%.0lf)(%.0lf,%.0lf):",w[i].x,w[i].y[j],w[i].x,w[i].y[j+1],w[li].x,w[li].y[lj],w[ri].x,w[ri].y[rj]);
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// printf("[%.2lf/%.2lf]\n",same(w[i].x,w[i].y[j],w[i].x,w[i].y[j+1],w[li].x,w[li].y[lj],w[ri].x,w[ri].y[rj]),same(w[li].x,w[li].y[lj],w[ri].x,w[ri].y[rj],w[i].x,w[i].y[j],w[i].x,w[i].y[j+1]));
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//两个都同方向才是，应该&&，错以为|| ,不是0都是真 两个小于0才是香蕉
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if(sign(same(w[i].x,w[i].y[j],w[i].x,w[i].y[j+1],w[li].x,w[li].y[lj],w[ri].x,w[ri].y[rj]))<0&&sign(same(w[li].x,w[li].y[lj],w[ri].x,w[ri].y[rj],w[i].x,w[i].y[j],w[i].x,w[i].y[j+1]))<0)
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{
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return 0;
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}
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}
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}
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return 1;
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}
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void find(int ii,int jj)
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{
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for(int i=0; i<ii; i++)
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{
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for(int j=1; j<=4; j++)
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{
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l=length(w[ii].x,w[ii].y[jj],w[i].x,w[i].y[j]);
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now=w[i].val[j]+l;//i j ii,jj搞反了
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if(now<w[ii].val[jj]&&ob(i,j,ii,jj))
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{
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w[ii].val[jj]=now;
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}
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// printf("(%.0lf,%.0lf)(%.0lf,%.0lf):length:%.3lf,now:%.3lf，val:%.3lf\n",w[ii].x,w[ii].y[jj],w[i].x,w[i].y[j],l,now,w[ii].val[jj]);
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if(i==0) break;
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}
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}
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}
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void doit()
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{
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for(int i=1; i<=n; i++)
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{
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for(int j=1; j<=4; j++)
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{
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find(i,j);
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if(i==n) break;
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}
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}
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}
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int main()
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{
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scanf("%d",&t);
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while(t--)
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{
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scanf("%d",&n);
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memset(&e,0,sizeof(edge)*(n+1));
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for(int i=1; i<=n; i++)
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{
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for(int j=0; j<2; j++)// 又是j
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{
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scanf("%lf",&e[i].x[j],&e[i].y[j]);
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}
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}
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w[++n].x=10;
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w[n].y[1]=5;
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w[n].val[1]=inf;// bug 没赋值
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doit();
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printf("%.2lf\n",w[n].val[1]);
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}
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}
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