Zoj 1721 The Doors dijstra + computational ry

Practice: Well, it is very similar to the fzu question to find the point in the shortest path. The points on both sides of the door, the start point, and the end point are all drawn. Then, obtain the accessibility of each vertex Based on the graph.

#include<cstdio>#include<cmath>#define eps 1e8const int LMT=102;int wn,n;double gra[LMT][LMT];struct wall{double y1,y2,x;}w[LMT];struct point{double x,y;}p[LMT];inline double max(double a,double b){return a>b?a:b;}inline double min(double a,double b){return a<b?a:b;}inline double length(point a,point b){return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}inline int min(int a,int b){return a<b?a:b;}void init(void){int i,j;for(i=0;i<LMT;i++)for(j=0;j<LMT;j++)gra[i][j]=eps;wn=0;n=0;}bool can(point a,point b){int i;if(a.x==b.x){for(i=0;i<wn&&w[i].x!=a.x;i++);for(;i<wn&&w[i].x==a.x;i++)if(a.y<=w[i].y1&&w[i].y2<=b.y)return 0;return 1;}double k=(b.y-a.y)/(b.x-a.x);for( i=0;i<wn&&w[i].x<=a.x;i++);for(;i<wn&&w[i].x<b.x;i++)if(b.y-k*(b.x-w[i].x)<=w[i].y2&&b.y-k*(b.x-w[i].x)>=w[i].y1)return 0;return 1;}int main(){int doors,i,j,k;double x,y1,y2,y3,y4;while(~scanf("%d",&doors)&&doors!=-1){init();p[n].x=0;p[n].y=5;n++;for( i=0;i<doors;i++){scanf("%lf%lf%lf%lf%lf",&x,&y1,&y2,&y3,&y4);p[n].x=x;p[n].y=y1;n++;p[n].x=x;p[n].y=y2;n++;p[n].x=x;p[n].y=y3;n++;p[n].x=x;p[n].y=y4;n++;w[wn].x=x;w[wn].y1=0;w[wn].y2=y1;wn++;w[wn].x=x;w[wn].y1=y2;w[wn].y2=y3;wn++;w[wn].x=x;w[wn].y1=y4;w[wn].y2=10;wn++;}p[n].x=10;p[n].y=5;n++;for(i=0;i<n;i++){for( j=i+1;j<n;j++)if(can(p[i],p[j]))gra[i][j]=gra[j][i]=length(p[i],p[j]);gra[i][i]=0;}        for(k=0;k<n;k++)for( i=0;i<n;i++)if(gra[i][k]!=eps)for(j=0;j<n;j++)if(gra[k][j]!=eps)   gra[i][j]=min(gra[i][j],gra[i][k]+gra[k][j]);printf("%.2lf\n",gra[0][n-1]);}return 0;}