HDOJ 2102 A计划

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题意：从一个三维迷宫中的(0,0,0)到P点的最短时间，每个时刻只能移动一格。

链接：http://acm.hdu.edu.cn/showproblem.php?pid=2102

思路：基础bfs

注意点：两层对应位置都为传送器时，相当于两边都是墙

以下为AC代码：

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <vector>#include <deque>#include <list>#include <cctype>#include <algorithm>#include <climits>#include <queue>#include <stack>#include <cmath>#include <map>#include <set>#include <iomanip>#include <cstdlib>#include <ctime>#define ll long long#define ull unsigned long long#define all(x) (x).begin(), (x).end()#define clr(a, v) memset( a , v , sizeof(a) )#define pb push_back#define RDI(a) scanf ( "%d", &a )#define RDII(a, b) scanf ( "%d%d", &a, &b )#define RDIII(a, b, c) scanf ( "%d%d%d", &a, &b, &c );#define RDS(s) scanf ( "%s", s );#define PIL(a) printf ( "%d\n", a );#define PIIL(a,b) printf ( "%d %d\n", a, b );#define PIIIL(a,b,c) printf ( "%d %d %d\n", a, b, c );#define REP(i,m,n) for ( int i = m; i <= n; i ++ )#define DEP(i,m,n) for ( int i = m; i >= n; i -- )#define REPI(i,m,n,k) for ( int i = m; i <= n; i += k )#define DEPI(i,m,n,k) for ( int i = m; i >= n; i -= k )#define PSL(s) printf ( "%s\n", s );#define read(f) freopen(f, "r", stdin)#define write(f) freopen(f, "w", stdout)using namespace std;const double pi = acos(-1);const double eps = 1e-10;const int dir[4][2] = { 1,0, -1,0, 0,1, 0,-1 };struct node{    int x, y, z, cnt;    node(){}    node( int _x, int _y, int _z, int _cnt ) : x(_x), y(_y), z(_z), cnt(_cnt) {}};int m, n, t;char str[2][15][15];bool vis[2][15][15];void init(){    clr ( str, '*' );    RDIII ( m, n, t );    REP ( k, 0, 1 ){        REP ( i, 1, m ){            //getchar();            RDS ( & str[k][i][1] );            str[k][i][n+1] = '*';        }    }    clr ( vis, 0 );}int bfs ( int x, int y, int z ){    vis[x][y][z] = 1;    queue<node> q;    q.push ( node ( x, y, z, 0 ) );    while ( ! q.empty() ){        node tmp = q.front();        if ( str[tmp.x][tmp.y][tmp.z] == 'P' )return tmp.cnt;        q.pop();        REP ( i, 0, 3 ){            int xi = tmp.x;            int yi = tmp.y + dir[i][0];            int zi = tmp.z + dir[i][1];            int cnt = tmp.cnt + 1;            if ( yi <= 0 || yi > m || zi <= 0 || zi > n || str[xi][yi][zi] == '*' )continue;            if ( str[xi][yi][zi] == '#' && str[xi^1][yi][zi] == '#' )continue;            if ( str[xi][yi][zi] == '#' && str[xi^1][yi][zi] == '*' )continue;            if ( vis[xi][yi][zi] == 0 ){                if ( str[xi][yi][zi] == '#' && vis[xi^1][yi][zi] == 0 ){                    vis[xi^1][yi][zi] = vis[xi][yi][zi] = 1;                    q.push ( node ( xi^1, yi, zi, cnt ) );                }                else if ( str[xi][yi][zi] != '#' ){                    vis[xi][yi][zi] = 1;                    q.push ( node ( xi, yi, zi, cnt ) );                }            }        }    }    return t + 1;}int main(){    int ncase;    RDI ( ncase );    while ( ncase -- ){        init();        printf ( "%s\n", ( bfs ( 0, 1, 1)  <= t ) ? "YES" : "NO" );    }    return 0;}

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