标签：
Description
You‘re in space.
You want to get home.
There are asteroids.
You don‘t want to hit them.Input
Input to this problem will consist of a (non-empty) series of up to 100 data sets. Each data set will be formatted according to the following description, and there will be no blank lines separating data sets.
A single data set has 5 components:
Start line - A single line, "START N", where 1 <= N <= 10.
Slice list - A series of N slices. Each slice is an N x N matrix representing a horizontal slice through the asteroid field. Each position in the matrix will be one of two values:
‘O‘ - (the letter "oh") Empty space
‘X‘ - (upper-case) Asteroid present
Starting Position - A single line, "A B C", denoting the <A,B,C> coordinates of your craft‘s starting position. The coordinate values will be integers separated by individual spaces.
Target Position - A single line, "D E F", denoting the <D,E,F> coordinates of your target‘s position. The coordinate values will be integers separated by individual spaces.
End line - A single line, "END"
The origin of the coordinate system is <0,0,0>. Therefore, each component of each coordinate vector will be an integer between 0 and N-1, inclusive.
The first coordinate in a set indicates the column. Left column = 0.
The second coordinate in a set indicates the row. Top row = 0.
The third coordinate in a set indicates the slice. First slice = 0.
Both the Starting Position and the Target Position will be in empty space.Output
For each data set, there will be exactly one output set, and there will be no blank lines separating output sets.
A single output set consists of a single line. If a route exists, the line will be in the format "X Y", where X is the same as N from the corresponding input data set and Y is the least number of moves necessary to get your ship from the starting position to the target position. If there is no route from the starting position to the target position, the line will be "NO ROUTE" instead.
A move can only be in one of the six basic directions: up, down, left, right, forward, back. Phrased more precisely, a move will either increment or decrement a single component of your current position vector by 1.Sample Input
START 1 O 0 0 0 0 0 0 END START 3 XXX XXX XXX OOO OOO OOO XXX XXX XXX 0 0 1 2 2 1 END START 5 OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO XXXXX XXXXX XXXXX XXXXX XXXXX OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO 0 0 0 4 4 4 ENDSample Output
1 0 3 4 NO ROUTE
1 /* 2 By:OhYee 3 Github:OhYee 4 Email:oyohyee@oyohyee.com 5 Blog:http://www.cnblogs.com/ohyee/ 6 7 かしこいかわいい？ 8 エリーチカ！ 9 要写出来Хорошо的代码哦~ 10 */ 11 12 #include <cstdio> 13 #include <algorithm> 14 #include <cstring> 15 #include <cmath> 16 #include <string> 17 #include <iostream> 18 #include <vector> 19 #include <list> 20 #include <queue> 21 #include <stack> 22 #include <map> 23 using namespace std; 24 25 //DEBUG MODE 26 #define debug 0 27 28 //循环 29 #define REP(n) for(int o=0;o<n;o++) 30 31 const int maxn = 11; 32 int n; 33 int dis[maxn][maxn][maxn]; 34 char Map[maxn][maxn][maxn]; 35 36 const int delta[6][3] = {{1,0,0},{-1,0,0},{0,1,0},{0,-1,0},{0,0,1},{0,0,-1}}; 37 38 struct point { 39 int x,y,z; 40 point() { 41 x = y = z = -1; 42 } 43 point(int a,int b,int c) { 44 x = a; 45 y = b; 46 z = c; 47 } 48 bool operator == (const point &rhs)const { 49 return ((x == rhs.x) && (y == rhs.y) && (z == rhs.z)); 50 } 51 }; 52 53 inline int read_int() { 54 char c; 55 int ans = 0; 56 while (c = getchar(),!(c >= ‘0‘ && c <= ‘9‘)); 57 while (c >= ‘0‘&&c <= ‘9‘) { 58 ans *= 10; 59 ans += (int)c - ‘0‘; 60 c = getchar(); 61 } 62 return ans; 63 } 64 65 int BFS(point s,point v) { 66 if (s == v) 67 return 0; 68 69 memset(dis,-1,sizeof(dis)); 70 queue<point> Q; 71 72 Q.push(s); 73 dis[s.x][s.y][s.z] = 0; 74 75 while (!Q.empty()) { 76 int x = Q.front().x; 77 int y = Q.front().y; 78 int z = Q.front().z; 79 Q.pop(); 80 81 REP(6) { 82 int xx = x + delta[o][0]; 83 int yy = y + delta[o][1]; 84 int zz = z + delta[o][2]; 85 86 //非法路径 87 if (xx < 0 || xx >= n || yy < 0 || yy >= n || zz < 0 || zz >= n) 88 continue; 89 //墙 90 if (Map[xx][yy][zz] == ‘X‘) 91 continue; 92 93 //尚未访问过 94 if (dis[xx][yy][zz] == -1) { 95 dis[xx][yy][zz] = dis[x][y][z] + 1; 96 //到达终点 97 if (point(xx,yy,zz) == v) 98 return dis[xx][yy][zz]; 99 Q.push(point(xx,yy,zz)); 100 } 101 } 102 } 103 return -1; 104 } 105 106 bool Do() { 107 char c; 108 if (scanf("\n%c",&c) == EOF) 109 return false; 110 n = read_int(); 111 //printf(" (%d) \n",n); 112 113 for (int k = 0;k < n;k++)//块 114 for (int i = 0;i < n;i++)//行 115 scanf("%s",Map[k][i]); 116 117 int s1,s2,s3,v1,v2,v3; 118 s1 = read_int(); 119 s2 = read_int(); 120 s3 = read_int(); 121 v1 = read_int(); 122 v2 = read_int(); 123 v3 = read_int(); 124 //scanf("%d%d%d",&s1,&s2,&s3); 125 //scanf("%d%d%d",&v1,&v2,&v3); 126 point s = point(s3,s1,s2); 127 point v = point(v3,v1,v2); 128 129 130 131 132 int ans = BFS(s,v); 133 134 if (ans == -1) 135 printf("NO ROUTE\n"); 136 else 137 printf("%d %d\n",n,ans); 138 139 scanf("%*s"); 140 141 return true; 142 } 143 144 int main() { 145 while (Do()); 146 return 0; 147 }
标签：
原文地址：http://www.cnblogs.com/ohyee/p/5428109.html
