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题意,给出一个N,这是这个三空间的大小,
然后给出所有面的状况O为空地,X为墙,再给出起始点的三维坐
标和终点的坐标,输出到达的步数
三维的BFS,很裸的题,不过一定要注意这一题给的坐标x表示列,y表示行,z表示层,
实际输入的时候调试一下看看给的坐标是不是跟样例在图中所指的对象一样就行了。
http://blog.csdn.net/iaccepted/article/details/23277717
bfs讲得好http://iprai.hust.edu.cn/icl2002/algorithm/algorithm/commonalg/graph/traversal/bfs.htm
bfs
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cstring> 4 #include <cmath> 5 #include <queue> 6 #include <climits> 7 8 using namespace std; 9 10 const int MAX = 11; 11 const int INF = INT_MAX -3; 12 const int dirx[6]= {1,-1,0,0,0,0},diry[6]= {0,0,1,-1,0,0},dirz[6]= {0,0,0,0,1,-1}; 13 14 typedef struct Point 15 { 16 int x,y,z; 17 }; 18 19 char map[MAX][MAX][MAX]; 20 int dist[MAX][MAX][MAX]; 21 int sx,sy,sz,dx,dy,dz; 22 int minx,n; 23 24 Point pp; 25 26 queue<Point> que; 27 28 int bfs(int x,int y,int z) 29 { 30 pp.x = x; 31 pp.y = y; 32 pp.z = z; 33 que.push(pp); 34 int i,tx,ty,tz; 35 while(!que.empty()) 36 { 37 pp = que.front(); 38 x = pp.x; 39 y = pp.y; 40 z = pp.z; 41 que.pop(); 42 if(x==dx && y==dy && z==dz)break; 43 for(i=0; i<6; ++i) 44 { 45 tx = x+dirx[i]; 46 ty = y+diry[i]; 47 tz = z+dirz[i]; 48 if(tx<0 || ty<0 || tz<0 || tx>=n || ty>=n || tz>=n)continue; 49 if(dist[tz][tx][ty]!=-1)continue; 50 if(map[tz][tx][ty]==‘X‘)continue; 51 dist[tz][tx][ty] = dist[z][x][y]+1;//十分巧妙的计算结果 52 pp.x = tx; 53 pp.y = ty; 54 pp.z = tz; 55 que.push(pp); 56 } 57 } 58 return dist[dz][dx][dy]; 59 } 60 61 int main() 62 { 63 //freopen("in.txt","r",stdin); 64 char str[10]; 65 int i,j,k; 66 while(scanf("%s %d%*c",str,&n)!=EOF) 67 { 68 for(k=0; k<n; ++k) 69 { 70 for(i=0; i<n; ++i) 71 { 72 for(j=0; j<n; ++j) 73 { 74 map[k][i][j]=getchar(); 75 dist[k][i][j] = -1; 76 } 77 getchar(); 78 } 79 } 80 scanf("%d %d %d",&sy,&sx,&sz); 81 scanf("%d %d %d",&dy,&dx,&dz); 82 scanf("%s",str); 83 dist[sz][sx][sy] = 0; 84 minx = bfs(sx,sy,sz); 85 if(minx==-1) 86 { 87 printf("NO ROUTE\n"); 88 } 89 else 90 { 91 printf("%d %d\n",n,minx); 92 } 93 /*while(!que.empty())释放队列,可有可无 94 { 95 que.pop(); 96 }*/ 97 } 98 return 0; 99 } 100 /* 101 #include <iostream> 102 #define MAX 10000 103 using namespace std; 104 struct node 105 { 106 int j, i, k, step; 107 } que[MAX]; 108 char s[10]; 109 int n; 110 int sj, si, sk; 111 int ej, ei, ek; 112 char map[11][11][11]; 113 int dir[6][3]= {{0,1,0},{0,-1,0},{1,0,0},{-1,0,0},{0,0,1},{0,0,-1}}; 114 int ok; 115 116 bool inmap(int j ,int i, int k) 117 { 118 return j >= 0 && j < n && i >= 0 && i <n && k >= 0 && k < n; 119 } 120 int bfs() 121 { 122 int front = 0, rear = 1; 123 int ti, tj, tk; 124 que[0].i = si; 125 que[0].j = sj; 126 que[0].k = sk; 127 que[0].step = 0; 128 map[si][sj][sk] = ‘X‘; 129 while(front < rear) 130 { 131 if(que[front].i == ei && que[front].j == ej && que[front].k == ek) 132 { 133 printf("%d %d\n",n, que[front].step); 134 return 1; 135 break; 136 } 137 for(int d = 0; d < 6; d++) 138 { 139 ti = que[front].i + dir[d][1]; 140 tj = que[front].j + dir[d][0]; 141 tk = que[front].k + dir[d][2]; 142 if(inmap(ti, tj, tk) && map[ti][tj][tk] == ‘O‘) 143 { 144 map[ti][tj][tk] = ‘X‘; 145 que[rear].j = tj; 146 que[rear].i = ti; 147 que[rear].k = tk; 148 que[rear++].step = que[front].step + 1; 149 } 150 } 151 front++; 152 } 153 return 0; 154 } 155 int main() 156 { 157 while(scanf("%s%d", s, &n) != EOF) 158 { 159 int i, j, k; 160 ok = 0; 161 memset(map, 0, sizeof(map)); 162 for(i = 0; i < n; i ++) 163 for(j = 0; j < n; j++) 164 for(k = 0; k < n; k++) 165 { 166 cin >>map[j][k][i]; 167 } 168 scanf("%d%d%d%d%d%d", &si, &sj, &sk, &ei, &ej, &ek); 169 scanf("%s", s); 170 ok = bfs(); 171 if(!ok) 172 printf("NO ROUTE\n"); 173 } 174 return 0; 175 } 176 */
dfs
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cstring> 4 #include <cmath> 5 #include <climits> 6 7 const int MAX = 11; 8 const int INF = INT_MAX ;//最大整数 9 const int dirx[6]={1,-1,0,0,0,0},diry[6]={0,0,1,-1,0,0},dirz[6]={0,0,0,0,1,-1}; 10 11 char map[MAX][MAX][MAX]; 12 int dist[MAX][MAX][MAX]; 13 int sx,sy,sz,dx,dy,dz; 14 int minx,n; 15 16 void dfs(int x,int y,int z,int cnt){ 17 if(x<0 || y<0 || z<0 || x>=n || y>=n || z>=n)return; 18 if(map[z][x][y]==‘X‘)return; 19 if(x==dx && y==dy && z==dz){ 20 if(cnt<minx)minx = cnt; 21 return; 22 } 23 if(dist[z][x][y]<=cnt)return; 24 else dist[z][x][y] = cnt; 25 int i,tx,ty,tz; 26 for(i=0;i<6;++i){ 27 tx = x+dirx[i]; 28 ty = y+diry[i]; 29 tz = z+dirz[i]; 30 dfs(tx,ty,tz,cnt+1); 31 } 32 } 33 34 int main(){ 35 //freopen("in.txt","r",stdin); 36 char str[10]; 37 int i,j,k,cnt; 38 while(scanf("%s %d%*c",str,&n)!=EOF){ 39 for(k=0;k<n;++k){ 40 for(i=0;i<n;++i){ 41 for(j=0;j<n;++j){ 42 map[k][i][j]=getchar(); 43 dist[k][i][j]=INF; 44 } 45 getchar(); 46 } 47 } 48 scanf("%d %d %d",&sy,&sx,&sz); 49 scanf("%d %d %d",&dy,&dx,&dz); 50 scanf("%s",str); 51 cnt = 0; 52 minx = INT_MAX; 53 dfs(sx,sy,sz,cnt); 54 if(minx==INT_MAX){ 55 printf("NO ROUTE\n"); 56 }else{ 57 printf("%d %d\n",n,minx); 58 } 59 } 60 61 62 return 0; 63 }
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原文地址:http://www.cnblogs.com/SSYYGAM/p/4508927.html