题目大意：在10*10的空间里，中间有几堵平行于竖直方向的墙，每个墙由两扇门，求从（0，5）-->(10,5)
的最短路径
思路：把门边沿的点两点建立一条边（中间无墙挡住），和起点，终点同样建边。。最后就成最短路。。
网上大部分Dijkstra，其实数据很小，floyed足以搞定。。
PS：这道题数据很弱，但是弱菜下标写错了。。调了非常久。。。还有看到网上有的写了接近300行代码吓尿了。。
写的很搓的代码：
1 /* 2 Time:2013-03-25 23:56:19 3 State： Accepted 4 */ 5 6 #include<iostream> 7 #include<cstring> 8 #include<string> 9 #include<cstdlib> 10 #include<cstdio> 11 #include<fstream> 12 #include<cmath> 13 #include<algorithm> 14 #define INF 1000000 15 #define ee 1e-8 16 using namespace std; 17 int n, tot; 18 double a[200][6] , map[202][202] ; 19 int pos[200][6], p1, p2 , st , en; 20 21 22 void init(){ 23 memset(a, 0, sizeof(a)); 24 tot = 0; 25 for (int i = 0; i<= 200; ++i) 26 for (int j = 0; j <= 200; ++j) 27 map[i][j] = INF; 28 for (int i = 1; i <= n; ++i){ 29 scanf("%lf%lf%lf%lf%lf",&a[i][0], &a[i][1] , 30 &a[i][2] , &a[i][3], &a[i][4] ); 31 for (int j = 1; j <= 4; ++j) pos[i][j] = i + j * n - n; 32 } 33 } 34 35 double work_x(double x2, double y2, double x1 , double y1){ 36 return x1*y2 - x2*y1; 37 } 38 39 void work(double x1 , double y1, double x2 , double y2){ 40 double x , y3 , y4; 41 bool flag; 42 for (int i = st; i <= en ; ++i){ 43 flag = 0; 44 x = a[i][0]; 45 y3 = a[i][1]; 46 y4 = a[i][2]; 47 if (work_x(x1 - x , y1 - y3, 0, y4-y3)*work_x(x2 -x , y2 - y3, 0, y4-y3) < 0 48 && work_x(x-x1 , y3-y1, x2 - x1, y2-y1)*work_x(x-x1 , y4-y1, x2-x1, y2-y1) < 0) flag = 1; 49 y3 = a[i][3]; 50 y4 = a[i][4]; 51 if (work_x(x1 -x , y1 - y3, 0, y4-y3)*work_x(x2 -x , y2 - y3, 0, y4-y3) < 0 52 && work_x(x-x1 , y3-y1, x2 - x1, y2-y1)*work_x(x-x1 , y4-y1, x2-x1, y2-y1) < 0) flag = 1; 53 if (!flag) return; 54 55 } 56 map[p2][p1] = map[p1][p2] = sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2)); 57 } 58 59 void solve(){ 60 tot = n * 4 + 1; 61 for (int i = 1; i <= n; ++i) 62 for (int k = 1; k <= 4; ++k){ 63 /* for (int k1 = k + 1; k1 <= 4; ++k1) 64 map[pos[i][k1]][pos[i][k]] = 65 map[pos[i][k]][pos[i][k1]] = a[i][k1] - a[i][k]; */ 66 p1 = pos[i][k]; 67 p2 = tot; 68 en = n; 69 st = i + 1; 70 work(a[i][0],a[i][k],10,5); 71 p1 = 0; 72 p2 = pos[i][k]; 73 en = i - 1; 74 st = 1; 75 work(0,5,a[i][0],a[i][k]); 76 for (int j = 1; j < i; ++j) 77 for (int l = 1; l <= 4 ; ++l){ 78 st = j + 1; 79 p1 = pos[j][l]; 80 work(a[j][0], a[j][l], a[i][0], a[i][k]); 81 } 82 83 } 84 p1 =0; 85 p2 = tot; 86 en = n; 87 st = 1; 88 work(0,5,10,5); 89 /* for (int i = 0; i <= tot; ++i) 90 for (int j =i + 1; j <=tot; ++j) 91 if (map[i][j]!=INF) printf("map[%d][%d] = %lf\n",i , j ,map[i][j]);*/ 92 for (int k = 0; k <= tot; ++k) 93 for (int i = 0; i <= tot; ++i) 94 for (int j = 0; j <= tot; ++j){ 95 if (i == j || j == k || k == i) continue; 96 map[i][j] = min(map[i][j], map[i][k] + map[k][j]); 97 } 98 printf("%0.2f\n",map[0][tot]); 99 100 101 } 102 103 int main(){ 104 freopen("poj1556.in","r",stdin); 105 freopen("poj1556.out","w",stdout); 106 while (scanf("%d", &n) != EOF && n != -1){ 107 init(); 108 solve(); 109 } 110 fclose(stdin); fclose(stdout); 111 }