HDOJ 3309 Roll The Cube （BFS）

Roll The Cube

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 558    Accepted Submission(s): 215

Problem Description

This is a simple game.The goal of the game is to roll two balls to two holes each.
'B' -- ball
'H' -- hole
'.' -- land
'*' -- wall
Remember when a ball rolls into a hole, they(the ball and the hole) disappeared, that is , 'H' + 'B' = '.'.
Now you are controlling two balls at the same time.Up, down , left , right --- once one of these keys is pressed, balls exist roll to that direction, for example , you pressed up , two balls both roll up. 
A ball will stay where it is if its next point is a wall, and balls can't be overlap.
Your code should give the minimun times you press the keys to achieve the goal.

 

Input

First there's an integer T(T<=100) indicating the case number.
Then T blocks , each block has two integers n , m (n , m <= 22) indicating size of the map.
Then n lines each with m characters.
There'll always be two balls(B) and two holes(H) in a map.
The boundary of the map is always walls(*).

 

Output

The minimum times you press to achieve the goal.
Tell me "Sorry , sir , my poor program fails to get an answer." if you can never achieve the goal.

 

Sample Input

46 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 an answer.

 

#include <cstdio>#include <cstring>#include <queue>#define maxn 25#define INF 0x3f3f3f3fusing namespace std;int n, m, ans;int sx[2], sy[2];int dx[] = {-1, 1, 0, 0};int dy[] = {0, 0, -1, 1};bool vis[maxn][maxn][maxn][maxn];char mp[maxn][maxn];char s[maxn];struct Node{    int x[2], y[2], step;    int b[2], h[2];}cur, now;queue<Node> q;bool bfs(){    int i, j, t, flag;    memset(vis, 0, sizeof(vis));    while (!q.empty())        q.pop();    cur.x[0] = sx[0], cur.y[0] = sy[0];    cur.x[1] = sx[1], cur.y[1] = sy[1];    cur.h[0] = cur.h[1] = 0;    cur.b[0] = cur.b[1] = 0;    cur.step = 0;    vis[sx[0]][s[0]][sx[1]][sy[1]] = 1;    q.push(cur);    while (!q.empty()){        now = q.front();        q.pop();        for (i = 0; i < 4; i++){            cur = now;            for (j = 0; j < 2; j++){                if (cur.b[j]) continue;                cur.x[j] += dx[i];                cur.y[j] += dy[i];                if (mp[cur.x[j]][cur.y[j]] == '*'){                    cur.x[j] -= dx[i];                    cur.y[j] -= dy[i];                }            }            if (vis[cur.x[0]][cur.y[0]][cur.x[1]][cur.y[1]] || cur.x[0] == cur.x[1] && cur.y[0] == cur.y[1] && cur.b[0] + cur.b[1] == 0)                continue;            cur.step++;            vis[cur.x[0]][cur.y[0]][cur.x[1]][cur.y[1]] = 1;            flag = 1;            for (j = 0; j < 2; j++){                t = mp[cur.x[j]][cur.y[j]];                if (t < 2 && !cur.h[t])                    cur.b[j] = 1, cur.h[t] = 1;                if (!cur.b[j]) flag = 0;            }            if (flag){                ans = cur.step;                return true;            }            q.push(cur);        }    }    return false;}int main(){    int i, j, t, cnt1, cnt2;    scanf("%d", &t);    while (t--){        scanf("%d%d", &n, &m);        cnt1 = cnt2 = 0;        for (i = 1; i <= n; i++){            scanf("%s", s);            for (j = 1; j <= m; j++){                mp[i][j] = s[j - 1];                if (mp[i][j] == 'H')                    mp[i][j] = cnt1++;                else if (mp[i][j] == 'B')                    sx[cnt2] = i, sy[cnt2] = j, cnt2++;            }        }        if (bfs())            printf("%d\n", ans);        else            printf("Sorry, sir, my poor program fails to get an answer.\n");    }    return 0;}