1644: [Usaco2007 oct]obstacle Course Obstacle Training session time Limit:5 Sec Memory limit:64 MB
submit:383 solved:196
[Submit] [Status] [Discuss] Description
Consider an n x N (1 <= n <= 100) square ranch with 1 squares. Some squares are not for cows to set foot on, they are labeled ' X '. For example:
. . B x.
. x x A.
.... x.
. X...
. . X. .
Bessie found herself exactly at point A, and she wanted to lick salt from the salt block at B. Slow and clumsy animals, such as cows, hate to turn. Still, of course they will turn when necessary. For a given ranch, please calculate the minimum number of turns from A to B. At the beginning, Bessie could make a face in either direction. Bessie knew that she must be able to arrive.
Input
Line 1th: An integer N row
2..N + 1: line i+1 has N characters ('. ', ' x ', ' A ', ' B '), indicating the state of each point.
Output
Line 1: An integer with a minimum number of turns.
Sample Input3
. XA
...
Bx.
Sample Output2
HINT Source
Silver
The problem: (Hansbug: Too happy, this question I amuse the ha haha wnjxyk: sweat = =)
In fact, although this problem is also a bfs water problem, but compared with the general maze, this question has a pit--for a point first to traverse to is not necessarily optimal, in fact, even if the current optimal also because of the problem in the subject because of the direction, so the follow-up is not necessarily optimal, and if the simple irrigation method, The true optimal value cannot be updated because it has been traversed.
For example, the 5th point in the data:
33
........ xa.....x.......x .....
... x...........x.........x ....
........... x.xx.....x.x .....
....... x.....x.........x .....
..... x................x.x .....
... x.. X....x.........xx .....
................ x......x ...
... x.....x....................x.
...... x..........................x.
X................................
.................................
..... x...x. X....................
................. ........
.............. x.........x ...
........... x...x....x...x ....
...... xx.......x. X..........
............ x----------
................... xx. xxx, .....
.. x...............x.x .......
. x....x...............x....x....x
.. X......xx.......x.x.......x ....
........... x...xx.....x.x. X
. x.............x........x...x.x.
... x.x............. .......
................. x......x
..... x.......x............xx...xx
..... x....x.....xx.x.....xx.x.
....... x...x........x.x ... Xxb
. x.x.x.............x...........x.
X.............x................x.
....... x.x.....x.....x.....x ...
..... x..............xx.....x.
......... xx......x.x.....x.
(2,10) position the optimal value is obviously 1, and no matter how the traversal is 1
But (3,10) because it involves the direction of the origin (2,10), it is possible to make a 2 optimal solution, but in fact the positive solution is 1
That's all I have to say.
1/**************************************************************2Problem:16443 User:hansbug4 language:pascal5 result:accepted6Time: +Ms7Memory:1728KB8****************************************************************/9 Ten ConstDd:Array[1..4,1..2] ofLongint= ((1,0),(-1,0),(0,1),(0,-1)); One var A I,j,k,l,m,n,x0,y0,x1,y1,f,r,x,y:longint; -B:Array[0..205,0..205] ofLongint; -A:Array[0..205,0..205] ofLongint; theD:Array[0..100000,0..2] ofLongint; - Ch:char; - functionmin (x,y:longint): Longint; - begin + ifX<y ThenMin:=xElsemin:=y; - End; + functionMax (x,y:longint): Longint; A begin at ifX>y ThenMax:=xElsemax:=y; - End; - - begin - READLN (n); -Fillchar (b,sizeof (b),0); inFillchar (A,sizeof (a),0); - fori:=0 ton+1 Do to begin +b[0, i]:=1; -b[n+1, i]:=1; theB[i,0]:=1; *b[i,n+1]:=1; $ End;Panax Notoginseng fori:=1 toN Do - begin the forj:=1 toN Do + begin A read (CH); the CaseUpCase (CH) of + '.': b[i,j]:=0; - 'A':begin $x0:=i;y0:=J; $b[i,j]:=1; - End; - 'B':begin thex1:=i;y1:=J; -b[i,j]:=0;Wuyi End; the 'X': b[i,j]:=1; - End; Wu End; - Readln; About End; $f:=1; r:=2;d [1,1]:=x0;d[1,2]:=y0; -a[x0,y0]:=1; a[x0,y0]:=1; a[x0,y0]:=1; a[x0,y0]:=1; - whileF<r Do - begin A fori:=1 to 4 Do + begin theX:=d[f,1];y:=d[f,2]; - whileTrue Do $ begin theX:=x+dd[i,1]; theY:=y+dd[i,2]; the ifb[x,y]=1 ThenBreak ; theL:=a[d[f,1],d[f,2]]+1; - if(L>=a[x,y]) and(a[x,y]>0) Thencontinue; ina[x,y]:=l; the if(X=X1) and(y=y1) Then the begin AboutWriteln (a[x,y]-2); the Readln; the Halt; the End; +D[r,1]:=x;d[r,2]:=y; - Inc (R); the End;Bayi the End; the Inc (f); - End; - End.
1644: [Usaco2007 oct]obstacle Course Obstacle Training Course