Hdoj title 3309 Roll The Cube (BFS)

Roll the CubeTime limit:3000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 502 Accepted Submission (s): 181


Problem DescriptionThis is a simple game. The goal of the game is to roll from balls to both holes each.
' B '--ball
' H '--hole
'. '--Land
' * '--wall
Remember when a ball rolls to a hole, they (the ball and the hole) disappeared, which is, ' H ' + ' B ' = '. '.
Now is controlling and balls at the same time. Up, down, left, right---once one of these keys are pressed, balls exist roll to that direction, for example, you press Ed up, and the balls both roll up.
A Ball would stay where it is if the next point was a wall, and balls can ' t be overlap.
Your code should give the Minimun times you press the keys to achieve the goal.
Inputfirst there ' s an integer T (t<=100) indicating the case number.
Then T-blocks, each block has a integers n, m (n, m <=) indicating size of the map.
Then n lines each with M characters.
There ' ll always be balls (B) and both holes (H) in a map.
The boundary of the map is always walls (*).
Outputthe minimum times you press to achieve the goal.
Tell me "Sorry, sir, my poor program fails to get a answer." If you can never achieve the goal.
Sample Input

3****b**b**h**h****4 4*****bb**hh*****4 4*****bh**hb*****5 6*******. bb***. h*h**. *.*******

Sample Output

312Sorry, sir, my poor program fails to get a answer.

Authormadfrog
Sourcehdoj Monthly contest–2010.02.06
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#include <stdio.h> #include <string.h> #include <iostream> #include <queue>using namespace std; int N,m,vis[25][25][25][25],sx[2],sy[2];char map[25][25];int dx[4]={0,1,0,-1};int dy[4]={1,0,-1,0};struct s{int x[2] , y[2],step,b[2],h[2];//friend bool operator < (s a,s b)//{//return a.step>b.step;//}}a,temp;int BFS () {memset (Vis , 0,sizeof (Vis)); a.x[0]=sx[0],a.x[1]=sx[1];a.y[0]=sy[0],a.y[1]=sy[1];a.b[0]=a.b[1]=a.h[0]=a.h[1]=0;vis[sx[0]][ Sy[0]][sx[1]][sy[1]]=1;a.step=0;//priority_queue<struct s>q;queue<struct S>q;q.push (a); while (! Q.empty ()) {int i,j;//a=q.top (); A=q.front (); Q.pop (); for (i=0;i<4;i++) {temp=a;for (j=0;j<2;j++) {if (Temp.b[j]) Continue;temp.x[j]=a.x[j]+dx[i];temp.y[j]=a.y[j]+dy[i];if (map[temp.x[j]][temp.y[j]]== ' * ') {temp.x[j]=a.x[j]; TEMP.Y[J]=A.Y[J];}} if (Vis[temp.x[0]][temp.y[0]][temp.x[1]][temp.y[1]]) continue;if (temp.x[0]==temp.x[1]&&temp.y[0]==temp.y [1]&&temp.b[0]+temp.b[1]==0] Continue;vis[temp.x[0]][temp.y[0]][temp.x[1]][temp.y[1]]=1;temp.step=a.step+1;int flag=1;for (j=0;j<2;j++) {int now=map[temp.x[j]][temp.y[j]];if (now<2&&! Temp.h[now]) {temp.h[now]=1;temp.b[j]=1;} if (!temp.b[j]) flag=0;} if (flag) return Temp.step;q.push (temp);}} return-1;} int main () {int t;scanf ("%d", &t), while (t--) {int i,j;scanf ("%d%d", &n,&m); int Cnt=0,cot=0;for (i=0;i<n; i++) {scanf ("%s", Map[i]), for (j=0;j<m;j++) {if (map[i][j]== ' H ') {map[i][j]=cnt++;} if (map[i][j]== ' B ') {sx[cot]=i;sy[cot]=j;cot++;}}} int Ans=bfs (); if (ans==-1) {printf ("Sorry, sir, my poor program fails to get an answer.\n");} elseprintf ("%d\n", ans);}}

Hdoj title 3309 Roll The Cube (BFS)