hdu1240 三维搜索 bfs dfs都可以

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

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <queue>
#include <climits>

using namespace std;

const int MAX = 11;
const int INF = INT_MAX    -3;
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};

typedef struct Point
{
    int x,y,z;
};

char map[MAX][MAX][MAX];
int dist[MAX][MAX][MAX];
int sx,sy,sz,dx,dy,dz;
int minx,n;

Point pp;

queue<Point> que;

int bfs(int x,int y,int z)
{
    pp.x = x;
    pp.y = y;
    pp.z = z;
    que.push(pp);
    int i,tx,ty,tz;
    while(!que.empty())
    {
        pp = que.front();
        x = pp.x;
        y = pp.y;
        z = pp.z;
        que.pop();
        if(x==dx && y==dy && z==dz)break;
        for(i=0; i<6; ++i)
        {
            tx = x+dirx[i];
            ty = y+diry[i];
            tz = z+dirz[i];
            if(tx<0 || ty<0 || tz<0 || tx>=n || ty>=n || tz>=n)continue;
            if(dist[tz][tx][ty]!=-1)continue;
            if(map[tz][tx][ty]=='X')continue;
            dist[tz][tx][ty] = dist[z][x][y]+1;//十分巧妙的计算结果
            pp.x = tx;
            pp.y = ty;
            pp.z = tz;
            que.push(pp);
        }
    }
    return dist[dz][dx][dy];
}

int main()
{
    //freopen("in.txt","r",stdin);
    char str[10];
    int i,j,k;
    while(scanf("%s %d%*c",str,&n)!=EOF)
    {
        for(k=0; k<n; ++k)
        {
            for(i=0; i<n; ++i)
            {
                for(j=0; j<n; ++j)
                {
                    map[k][i][j]=getchar();
                    dist[k][i][j] = -1;
                }
                getchar();
            }
        }
        scanf("%d %d %d",&sy,&sx,&sz);
        scanf("%d %d %d",&dy,&dx,&dz);
        scanf("%s",str);
        dist[sz][sx][sy] = 0;
        minx = bfs(sx,sy,sz);
        if(minx==-1)
        {
            printf("NO ROUTE
");
        }
        else
        {
            printf("%d %d
",n,minx);
        }
        /*while(!que.empty())释放队列，可有可无
        {
            que.pop();
        }*/
    }
    return 0;
}
/*
#include <iostream>
#define MAX 10000
using namespace std;
struct node
{
    int j, i, k, step;
} que[MAX];
char s[10];
int n;
int sj, si, sk;
int ej, ei, ek;
char map[11][11][11];
int dir[6][3]= {{0,1,0},{0,-1,0},{1,0,0},{-1,0,0},{0,0,1},{0,0,-1}};
int ok;

bool inmap(int j ,int i, int k)
{
    return j >= 0 && j < n && i >= 0 && i <n && k >= 0 && k < n;
}
int bfs()
{
    int front = 0, rear = 1;
    int ti, tj, tk;
    que[0].i = si;
    que[0].j = sj;
    que[0].k = sk;
    que[0].step = 0;
    map[si][sj][sk] = 'X';
    while(front < rear)
    {
        if(que[front].i == ei && que[front].j == ej && que[front].k == ek)
        {
            printf("%d %d
",n, que[front].step);
            return 1;
            break;
        }
        for(int d = 0; d < 6; d++)
        {
            ti = que[front].i + dir[d][1];
            tj = que[front].j + dir[d][0];
            tk = que[front].k + dir[d][2];
            if(inmap(ti, tj, tk) && map[ti][tj][tk] == 'O')
            {
                map[ti][tj][tk] = 'X';
                que[rear].j = tj;
                que[rear].i = ti;
                que[rear].k = tk;
                que[rear++].step = que[front].step + 1;
            }
        }
        front++;
    }
    return 0;
}
int main()
{
    while(scanf("%s%d", s, &n) != EOF)
    {
        int i, j, k;
        ok = 0;
        memset(map, 0, sizeof(map));
        for(i = 0; i < n; i ++)
            for(j = 0; j < n; j++)
                for(k = 0; k < n; k++)
                {
                    cin >>map[j][k][i];
                }
        scanf("%d%d%d%d%d%d", &si, &sj, &sk, &ei, &ej, &ek);
        scanf("%s", s);
        ok = bfs();
        if(!ok)
            printf("NO ROUTE
");
    }
    return 0;
}
*/

dfs

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <climits>

const int MAX = 11;
const int INF = INT_MAX    ;//最大整数
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};

char map[MAX][MAX][MAX];
int dist[MAX][MAX][MAX];
int sx,sy,sz,dx,dy,dz;
int minx,n;

void dfs(int x,int y,int z,int cnt){
    if(x<0 || y<0 || z<0 || x>=n || y>=n || z>=n)return;
    if(map[z][x][y]=='X')return;
    if(x==dx && y==dy && z==dz){
        if(cnt<minx)minx = cnt;
        return;
    }
    if(dist[z][x][y]<=cnt)return;
    else dist[z][x][y] = cnt;
    int i,tx,ty,tz;
    for(i=0;i<6;++i){
        tx = x+dirx[i];
        ty = y+diry[i];
        tz = z+dirz[i];
        dfs(tx,ty,tz,cnt+1);
    }
}

int main(){
    //freopen("in.txt","r",stdin);
    char str[10];
    int i,j,k,cnt;
    while(scanf("%s %d%*c",str,&n)!=EOF){
        for(k=0;k<n;++k){
            for(i=0;i<n;++i){
                for(j=0;j<n;++j){
                    map[k][i][j]=getchar();
                    dist[k][i][j]=INF;
                }
                getchar();
            }
        }
        scanf("%d %d %d",&sy,&sx,&sz);
        scanf("%d %d %d",&dy,&dx,&dz);
        scanf("%s",str);
        cnt = 0;
        minx = INT_MAX;
        dfs(sx,sy,sz,cnt);
        if(minx==INT_MAX){
            printf("NO ROUTE
");
        }else{
            printf("%d %d
",n,minx);
        }
    }


    return 0;
}