poj 1556 The Doors 计算几何+图论

题目链接：http://poj.org/problem?id=1556

这是一道很好的计算几何和图论的综合题，贵在建图，哪些点之间可以连线，只有当两点之间没有墙时才可以，怎么判断呢 只用把墙的x坐标带入两点的直线中看y坐标是否在墙上就可以了。

 

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  1 #include <iostream>
  2 #include <cstring>
  3 #include <cstdio>
  4 #include <cmath>
  5 using namespace std;
  6 const int MAX=20;
  7 const int MMAX=80;
  8 const double inf = 10000000.0;
  9 struct ss{
 10     double x;
 11     double P[5];
 12 }wall[MAX];
 13 double map[MMAX][MMAX];
 14 double bx = 0.0,by = 5.0;
 15 double ex = 10.0,ey = 5.0;
 16 int m;
 17 double Distance(double x1,double y1,double x2,double y2)//计算两点的距离
 18 {
 19     return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
 20 }
 21 double yy(double x1,double y1,double x2,double y2,double x)//计算y点的坐标  
 22 {
 23     return ((y2-y1)/(x2-x1)*x+(x2*y1-x1*y2)/(x2-x1));
 24 }
 25 
 26 int Con(double x1,double y1,double x2,double y2,int i,int k)
 27 {
 28     int j;
 29     double y;
 30     if(i>k)return 1;
 31     for (j=i;j<=k;j++)
 32     {
 33         y=yy(x1,y1,x2,y2,wall[j].x);
 34         if((y>=wall[j].P[1]&&y<=wall[j].P[2])||(y>=wall[j].P[3]&&y<=wall[j].P[4]));
 35         else return 0;           
 36     }
 37     return 1;
 38 }
 39 
 40 void Buildmap()//建图
 41 {
 42     double len;
 43     int i,j,k,p;
 44     if (Con(bx,by,ex,ey,1,m))
 45     {
 46         len=Distance(bx,by,ex,ey);
 47         map[0][m*4+1]=len;
 48     }
 49     
 50     for (i=1;i<=m;i++)//起点建图
 51         for (k=1;k<=4;k++)
 52             if (Con(bx,by,wall[i].x,wall[i].P[k],1,i-1))
 53             {
 54                 len=Distance(bx,by,wall[i].x,wall[i].P[k]);
 55                 map[0][(i-1)*4+k]=len;
 56             }
 57             
 58      for (i=1;i<m;i++)//中间点建图
 59         for (k=1;k<=4;k++)
 60             for (j=i+1;j<=m;j++)
 61                 for (p=1;p<=4;p++)
 62                     if (Con(wall[i].x,wall[i].P[k],wall[j].x,wall[j].P[p],i+1,j-1))
 63                         {
 64                             len=Distance(wall[i].x,wall[i].P[k],wall[j].x,wall[j].P[p]);
 65                             map[(i-1)*4+k][(j-1)*4+p]=len;
 66                         }
 67                             
 68         for (i=1;i<=m;i++)//终点建图
 69             for (k=1;k<=4;k++)
 70                     if (Con(ex,ey,wall[i].x,wall[i].P[k],i+1,m))
 71                         {
 72                             len=Distance(ex,ey,wall[i].x,wall[i].P[k]);
 73                             map[(i-1)*4+k][m*4+1]=len;
 74                         }
 75 }
 76 
 77 double Dijkstra(int s,int t,int n)
 78 {
 79     double dis[MMAX];
 80     fill(dis,dis+MMAX,inf);
 81     int i,k;
 82     bool used[MMAX];
 83     memset(used,false,sizeof(used));
 84     dis[s] = 0.0; used[s] = true;
 85     int now = s;
 86     for(i=0; i<n; i++)
 87     {
 88         for(k=0; k<=n; k++)
 89             if( map[now][k] >= 0.0 && dis[k] > dis[now] + map[now][k] )
 90                 dis[k] = dis[now] + map[now][k];
 91             double min = inf;
 92             for(k=0; k<=n; k++)
 93                 if( !used[k] && dis[k] < min )
 94                     min = dis[now = k];
 95                 used[now] = 1;
 96     }
 97     return dis[t];
 98 }
 99 void Dijkstra(int s,int n)
100 {
101     int i,j,v,vist[MMAX*4];
102     double dist[MMAX*4],ans;
103     memset(vist,0,sizeof(vist));
104     for (i=1;i<=MMAX;i++)
105         dist[i]=inf;
106     vist[s]=1;
107     v=s;
108     for (i=0;i<n;i++)
109     {
110         for (j=0;j<=n;j++)
111             if(dist[j]>dist[v]+map[v][j]&&map[v][j]>=0.0)
112             {
113                 dist[j]=dist[v]+map[v][j];
114             }        
115             ans=inf;
116             for (j=0;j<=n;j++)
117                 if(!vist[j]&&dist[j]<ans)
118                 {
119                     ans=dist[j];
120                     v=j;
121                 }
122                 vist[v]=1;
123     }
124 }
125 
126 int main()
127 {
128     int i,j,n;
129     while (scanf("%d",&m)!=EOF)
130     {
131         if(m==-1)break;
132         for (i=0;i<MMAX;i++)
133             for(j=0;j<MMAX;j++)
134             {
135                 map[i][j]=-1.0;
136             }
137             for (i=1;i<=m;i++)
138             {
139                 scanf("%lf",&wall[i].x);
140                 for (j=1;j<=4;j++)
141                 {
142                     scanf("%lf",&wall[i].P[j]);
143                 }
144             }
145             Buildmap();
146             n=m*4+1;
147             double ans=Dijkstra(0,n,n);
148             printf("%.2lf\n",ans);  
149     }
150     return 0;
151 }

 

转载于:https://www.cnblogs.com/zhaoguanqin/archive/2012/04/03/2430903.html