. Inputthe input consists of multiple test cases. The first line of all test case contains three integers n, m, and T (1 ' X ': a block of wall, which the doggie cannot enter;' S ': The start point of the doggie;' D ': the Door; Or'. ': an empty block.The input is terminated with three 0 ' s. This test is a not-to-be processed. Outputfor each test case, print on one line "YES" if the doggie can survive, or "NO" otherwise. Sample Input4 4 5S. X... X... Xd.... 3 4 5S. X... X.... D0 0 0Sample Outpu
Title Link: http://poj.org/problem?id=1475A set of test data:7 3###. T.. S.#B #.........Results:Problem-solving ideas: First to determine whether there is a box around the empty space, if there is, to determine if the person can reach the side of the box, if possible, put the box into the empty place, and then continueAC Code:1 //problem-solving ideas: First to determine whether there is a box around the empty space, if there is, to determine if the person can reach the side of the box, if possi
~ 180, rotate an deg angle on the x axis
RotateY
Deg
Deg range-180 ~ 180, rotate an deg angle on the y axis
RotateZ
Deg
Deg range-180 ~ 180, rotate an deg angle on the Z axis
Rotate3d
(X, y, z, deg)
Same as transform-function rotate3d
Scaling:
Method
Parameters
Description
Scale
Sx, [sy]
If one parameter is set, both the x axis and the y axis are scaled to the
; + Point Next,now; -now.x=sx;now.y=Sy; +now.step=0; Anow.dir=-1; at Q.push (now); - while(!q.empty ()) - { -now=Q.front (); - Q.pop (); - if(now.x==exnow.y==ey) in return 1; - for(i=0;i4; i++) to { +next.x=now.x+Dx[i]; -next.y=now.y+Dy[i]; thenext.step=Now.step; *Next.dir=i; $ if(next.dir!=now.dirnow.dir!=-1)//change the direction to add one, but the first step is not a turning pointPanax Notog
]. val [I] {+ + T [d]. num [I]; // here is the statistics on the number of elements from l to I entering the left child. This is the data mainly used by the tree.T [d + 1]. val [+ zn] = t [d]. val [I];} Else if (t [d]. val [I]> midd) // enter the right child{T [d + 1]. val [++ yn] = t [d]. val [I];} Else{If (samed {++ Samed;++ T [d]. num [I];T [d + 1]. val [+ zn] = t [d]. val [I];} Else // you have children.T [d + 1]. val [++ yn] = t [d]. val [I];}}Build (l, mid, d + 1); // create left and right
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