胜利大逃亡

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4785    Accepted Submission(s): 1454

代码如下:

/*

MiYu原创, 转帖请注明 : 转载自 ______________白白の屋

http://www.cnblog.com/MiYu

Author By : MiYu

Test      : 1

Program   : HDU1253

*/

#include <iostream>

#include <queue>

using namespace std;

int TLE[56][56][56];

const int d[6][3] = { { 0,0,1 },{ 0,0,-1 },{ 0,-1,0 },{ 0,1,0 },{ 1,0,0 },{ -1,0,0 } };

int A, B, C, T, m;

typedef struct pos{

pos(){ x=y=z=n=0; }

void setPos ( int a,int b,int c, int count ){ x=a;y=b;z=c;n=count; }

bool isEnd () { if ( x==1&&y==1&&z==1 )return true;return false; }

int x,y,z;

int n;

}pos;

pos t,p;

#define CMP(A,B) (A.n < B.n)

typedef class Heap {

public:

pos h[70000 * 2];

int n, p, c;

Heap() {

n = 0;

}

void inline push(pos e) {

for (p = ++n; p > 1 && CMP(e,h[p>>1]); h[p] = h[p>>1], p >>= 1)

;

h[p] = e;

}

int inline pop(pos &e) {

if (!n)

return 0;

for (e = h[p = 1], c = 2; c < n

&& CMP(h[c += (CMP(h[c + 1],h[c]) && c < n - 1)], h[n]);

h[p] = h[c], p = c, c <<= 1)

;

h[p] = h[n--];

return 1;

}

}Heap;

Heap WA;

int RE ()

{

if ( A+B+C-2 > T || TLE[A][B][C] == 0 )

return -1;

t.setPos ( A,B,C,0 );

TLE[A][B][C] = 0;

WA.push ( t );

while ( WA.pop(t) ){

if ( t.x+t.y+t.z-2 > T-t.n )

continue;

if ( t.isEnd() )

return t.n;

for ( int i = 0; i != 6; ++ i ){

int x = t.x+d[i][0], y=t.y+d[i][1], z=t.z+d[i][2];

if ( TLE[ x ][ y ][ z ] != 0 ){

TLE[ x ][ y ][ z ] = 0;

p.setPos ( x, y, z, t.n+1 );

if ( p.isEnd() && p.n <= T )

return p.n;

if ( p.n <= T )

WA.push ( p );

}

}

}

return -1;

}

inline bool scan_d(int &num)

{

char in;bool IsN=false;

in=getchar();

if(in==EOF) return false;

while(in!='-'&&(in<'0'||in>'9')) in=getchar();

if(in=='-'){ IsN=true;num=0;}

else num=in-'0';

while(in=getchar(),in>='0'&&in<='9'){

num*=10,num+=in-'0';

}

if(IsN) num=-num;

return true;

}

int main ()

{

int K;

scan_d(K);

while ( K -- ){

while ( WA.pop(p) ) ;

memset ( TLE, 0 , sizeof ( TLE ) );

scan_d(A); scan_d(B); scan_d(C); scan_d(T);

for ( int i = 1; i <= A; ++ i ){

for ( int j = 1; j <= B; ++ j ){

for ( int k = 1; k <= C; ++ k ){

scan_d(m);

TLE[i][j][k] = m == 1 ? 0 : 1;

}

}

}

TLE[1][1][1] = 1;

cout << RE () << endl;

}

//system( "pause" );

return 0;

}