Hdu-1026-Ignatius and the Princess I

A n * M graph can take one step up, down, or down in one second. If there is a number, it will take time to stay in the grid, outputs the path from the start point (0, 0) to the end point (N-1, M-1) each second.

 

--> It will select the priority queue to do the job. Basically, it's okay.

[Cpp] # include <iostream>
# Include <cstdio>
# Include <cstring>
# Include <queue>
 
Using namespace std;
 
Const int maxn = 100 + 10;
Struct node
{
Int x;
Int y;
Int step;
Int fax;
Int fay;
Node (){}
Node (int xx, int yy): x (xx), y (yy ){}
Bool operator <(const node & e) const
{
Return step> e. step;
}
Void init (int I, int j)
{
X = I;
Y = j;
Step = fax = fay =-1;
}
} No [maxn] [maxn];
Char MAP [maxn] [maxn];
Int N, M, cnt;
Int dx [] = {-1, 1, 0, 0 };
Int dy [] = {0, 0,-1, 1 };
 
Bool bfs ()
{
Priority_queue <node> pq;
Pq. push (no [0] [0]);
Node temp;
While (! Pq. empty ())
{
Temp = pq. top ();
Pq. pop ();
For (int I = 0; I <4; I ++)
{
Int newx = temp. x + dx [I];
Int newy = temp. y + dy [I];
If (newx> = 0 & newx <N & newy> = 0 & newy <M & MAP [newx] [newy]! = 'X' & no [newx] [newy]. step =-1)
{
No [newx] [newy]. step = temp. step + 1;
If (MAP [newx] [newy]! = '.') No [newx] [newy]. step + = MAP [newx] [newy]-'0 ';
No [newx] [newy]. fax = temp. x;
No [newx] [newy]. fay = temp. y;
Pq. push (no [newx] [newy]);
If (newx = N-1 & newy = M-1) return 1;
}
}
}
Return 0;
}
 
Void print (node u)
{
If (u. fax = 0 & u. fay = 0)
{
Printf ("It takes % d seconds to reach the target position, let me show you the way. \ n", no [N-1] [M-1]. step );
Printf ("% ds :( 0, 0)-> (% d, % d) \ n", cnt ++, u. x, u. y );
For (int I = 0; I <MAP [u. x] [u. y]-'0'; I ++) printf ("% ds: FIGHT AT (% d, % d) \ n", cnt ++, u. x, u. y );
Return;
}
Print (no [u. fax] [u. fay]);
Printf ("% ds :( % d, % d)-> (% d, % d) \ n", cnt ++, u. fax, u. fay, u. x, u. y );
For (int I = 0; I <MAP [u. x] [u. y]-'0'; I ++) printf ("% ds: FIGHT AT (% d, % d) \ n", cnt ++, u. x, u. y );
}
 
Int main ()
{
Int I, j;
While (scanf ("% d", & N, & M) = 2)
{
For (I = 0; I <N; I ++)
For (j = 0; j <M; j ++)
{
Cin> MAP [I] [j];
No [I] [j]. init (I, j );
}
 
No [0] [0]. step = 0;
Cnt = 1;
If (bfs () print (no [N-1] [M-1]);
Else printf ("God please help our poor hero. \ n ");
Printf ("FINISH \ n ");
}
Return 0;
}

# Include <iostream>
# Include <cstdio>
# Include <cstring>
# Include <queue>

Using namespace std;

Const int maxn = 100 + 10;
Struct node
{
Int x;
Int y;
Int step;
Int fax;
Int fay;
Node (){}
Node (int xx, int yy): x (xx), y (yy ){}
Bool operator <(const node & e) const
{
Return step> e. step;
}
Void init (int I, int j)
{
X = I;
Y = j;
Step = fax = fay =-1;
}
} No [maxn] [maxn];
Char MAP [maxn] [maxn];
Int N, M, cnt;
Int dx [] = {-1, 1, 0, 0 };
Int dy [] = {0, 0,-1, 1 };

Bool bfs ()
{
Priority_queue <node> pq;
Pq. push (no [0] [0]);
Node temp;
While (! Pq. empty ())
{
Temp = pq. top ();
Pq. pop ();
For (int I = 0; I <4; I ++)
{
Int newx = temp. x + dx [I];
Int newy = temp. y + dy [I];
If (newx> = 0 & newx <N & newy> = 0 & newy <M & MAP [newx] [newy]! = 'X' & no [newx] [newy]. step =-1)
{
No [newx] [newy]. step = temp. step + 1;
If (MAP [newx] [newy]! = '.') No [newx] [newy]. step + = MAP [newx] [newy]-'0 ';
No [newx] [newy]. fax = temp. x;
No [newx] [newy]. fay = temp. y;
Pq. push (no [newx] [newy]);
If (newx = N-1 & newy = M-1) return 1;
}
}
}
Return 0;
}

Void print (node u)
{
If (u. fax = 0 & u. fay = 0)
{
Printf ("It takes % d seconds to reach the target position, let me show you the way. \ n", no [N-1] [M-1]. step );
Printf ("% ds :( 0, 0)-> (% d, % d) \ n", cnt ++, u. x, u. y );
For (int I = 0; I <MAP [u. x] [u. y]-'0'; I ++) printf ("% ds: FIGHT AT (% d, % d) \ n", cnt ++, u. x, u. y );
Return;
}
Print (no [u. fax] [u. fay]);
Printf ("% ds :( % d, % d)-> (% d, % d) \ n", cnt ++, u. fax, u. fay, u. x, u. y );
For (int I = 0; I <MAP [u. x] [u. y]-'0'; I ++) printf ("% ds: FIGHT AT (% d, % d) \ n", cnt ++, u. x, u. y );
}

Int main ()
{
Int I, j;
While (scanf ("% d", & N, & M) = 2)
{
For (I = 0; I <N; I ++)
For (j = 0; j <M; j ++)
{
Cin> MAP [I] [j];
No [I] [j]. init (I, j );
}

No [0] [0]. step = 0;
Cnt = 1;
If (bfs () print (no [N-1] [M-1]);
Else printf ("God please help our poor hero. \ n ");
Printf ("FINISH \ n ");
}
Return 0;
}