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 }
