Saturday, November 28, 2015

图BFS boj1534 Obstacle Course-jiangwen127-ChinaUnix博客


图BFS boj1534 Obstacle Course-jiangwen127-ChinaUnix博客
You are working on the team assisting with programming for the Mars rover. To conserve energy, the rover needs to find optimal paths across the rugged terrain to get from its starting location to its final location. The following is the first approximation for the problem.

N x N square matrices contain the expenses for traversing each individual cell. For each of them, your task is to find the minimum-cost traversal from the top left cell [0][0] to the bottom right cell [N - 1][N - 1]. Legal moves are up, down, left, and right; that is, either the row index changes by one or the column index changes by one, but not both.



Input
Each problem is specified by a single integer between 2 and 125 giving the number of rows and columns in the N x N square matrix. The file is terminated by the case N = 0.

Following the specification of N you will find N lines, each containing N numbers. These numbers will be given as single digits, zero through nine, separated by single blanks.



Output
Each problem set will be numbered (beginning at one) and will generate a single line giving the problem set and the expense of the minimum-cost path from the top left to the bottom right corner, exactly as shown in the sample output (with only a single space after "Problem" and after the colon).


Sample Input

3
5 5 4
3 9 1
3 2 7
5
3 7 2 0 1
2 8 0 9 1
1 2 1 8 1
9 8 9 2 0
3 6 5 1 5
7
9 0 5 1 1 5 3
4 1 2 1 6 5 3
0 7 6 1 6 8 5
1 1 7 8 3 2 3
9 4 0 7 6 4 1
5 8 3 2 4 8 3
7 4 8 4 8 3 4
0



Sample Output

Problem 1: 20
Problem 2: 19
Problem 3: 36



Hint
This problem was used also in the Southern California region with the title Rover Obstacles

题意描述:
从左上到右下的路径中,寻找这样一条路径,它经过的格子中的数字之和最小。移动方向可以上、下、左、右。

解题思路:
1.之前做过类似的题,不过移动方向限制在右和下,这种情况下使用dp来做。在最先看到这个题以后也考虑过用dp来做,不过貌似行不通,因为一个经过的点可以再一次被遍历到。
2.换思路,暴力搜索,BFS,过之.
3.搜索的过程中,当到达A点时,判断它周围的4个点,如果从A点到达它们可以减少路径的长度,那么更新它周围的这个点的最小路径(这是关键!!!!!!),然后将它入队列,在取出它之后,更新它周围的点,如此往复,知道所有点的最小路径值都不能再减小为止.
4.输出右下点的最小值即可.

#define INT_MAX 0x7fffffff

typedef struct _node
{
    int x, y;
    int steps;
}ST_NODE;

/*min_path[i][j]记录当前从[0][0]点到[i][j]点的最小路径*/
int map[150][150], min_path[150][150];
deque Que;
/*小技巧,可以循环处理4个方向上的相邻节点*/
int dir_x[4] = {-1, 1, 0 , 0};
int dir_y[4] = {0 , 0, -1, 1};

void input(int size)
{
    int i, j;
    for (i=0 ; i        for (j=0 ; j            scanf("%d", &map[i][j]);
}

void initialize(int size)
{
    int i, j;
    Que.clear();

    for (i=0 ; i        for (j=0 ; j            min_path[i][j] = INT_MAX;
}

void bfs(int size)
{
    int i, x_next, y_next, x_cur, y_cur, steps;
    ST_NODE node;
    min_path[0][0] = map[0][0];
    node.x = node.y = 0;
    node.steps = map[0][0];
    Que.push_back(node);
    while (! Que.empty())
    {
        node = Que.front();
        Que.pop_front();
        x_cur = node.x;
        y_cur = node.y;
        steps = node.steps;
        for (i=0 ; i<4 ; i++)
        {
            x_next = x_cur + dir_x[i];
            y_next = y_cur + dir_y[i];
            if (x_next >= 0 && x_next < size && y_next >= 0 && y_next < size)
            {
                /*如果通过[x_cur][y_cur]到[x_next][y_next]能够减少min_path[x_next][y_next]的值,那么更新
                        之,并将x_next, y_next入队列,以用于下次出队时更新它周围的点的最小路径值
                 */
                if (steps + map[x_next][y_next] < min_path[x_next][y_next])
                {
                    min_path[x_next][y_next] = steps + map[x_next][y_next];
                    node.x = x_next;
                    node.y = y_next;
                    node.steps = min_path[x_next][y_next];
                    Que.push_back(node);
                }
            }
        }
    }
}

int main(int argc, char *argv[])
{
    int N, k = 1;

    while (scanf("%d", &N))
    {
        if (0 == N)
            break;
        input(N);
        initialize(N);
        bfs(N);
        printf("Problem %d: %d\n", k++, min_path[N - 1][N - 1]);
    }
}

http://www.1point3acres.com/bbs/thread-145290-1-1.html
一个matrix, 1代表机器人,0代表通路,x代表路障,求到所有机器人距离和最小的一点
需要注意的是机器人本身也是通路的一部分,可以通过他到达其他机器人,甚至最小点
也可以和机器人在同一个点上。
另外如果被路障围起来的点不能被算进来

面试时先介绍一下自己,问几个行为问题,然后讲题目思路,这样就差不多过了半小时,后面写程序时间有点紧,
所以介绍时尽量简短一点。
说思路的时候,说了bfs, dfs,面试官说bfs好,又问内存用多少,答O(m*n), 要求节省内存,但我没有想出来不用queue做bfs,
现在想内存最多是O(m + n)应该已经很好了。不过后来又用了一个存距离和判断访问过的二维数组,所以应该还是需要O(m*n), 
不知道有什么办法节省。后来面试官没有深究。
基本思路就是找到一个机器人位置,从这点开始bfs, 把所有点到这个机器人的距离加到dis数组,最后扫一遍看看谁距离最短就返回
这点。
.1point3acres缃�
/*//problem: robot merge point
//input:
//robot: 1
//obstacle: X
[
0   0   0   M   1
0   1   X   0   0
0   X   0   0   0
0   0   0   1   0
0   0   0   0   0
//output:
//best merge point: M
3 + 1 + 3 = 7
. visit 1point3acres.com for more.
Definition: Best merge point: minimal sum of path num between robots and merge point*/
  1. public class Solution{
  2.         class Point{
  3.                 int x;
  4.                 int y;
  5.                 public Point(int x, int y){. 鍥磋鎴戜滑@1point 3 acres
  6.                         this.x = x;
  7.                     this.y = y;
  8.                 }
  9.         }. more info on 1point3acres.com
  10.         Point bestMergePoint(char[][] mat){
  11.          int m = mat.length;. 涓€浜�-涓夊垎-鍦帮紝鐙鍙戝竷
  12.          if (m == 0) return null;
  13.          int n  = mat[0].length;
  14.         
  15.          // bfs each point that is 1. visit 1point3acres.com for more.
  16.          int[][] dis = new int[m][n];
  17.          for (int i = 0; i < m; i++){
  18.              for (int j = 0; j < n; j++){
  19.                  if (mat[i][j] == '1') bfs(mat, i, j, dis);
  20.              }
  21.          }
  22.          // count number-google 1point3acres
  23.          int min = Integer.MAX_VALUE;-google 1point3acres
  24.          Point ret = null;
  25.          for (int i = 0; i < m; i++){
  26.              for (int j = 0; j < n; j++){
  27.                      // if this point is not visited by a robot
  28.                      if (dis[i][j] == 0 && mat[i][j] == '0') continue;.鏈枃鍘熷垱鑷�1point3acres璁哄潧
  29.                  if (mat[i][j] != 'X' && dis[i][j] < min){
  30.                      min = dis[i][j];. 1point 3acres 璁哄潧
  31.                      ret = new Point(i, j);
  32.                  }
  33.              }
  34.          }. 涓€浜�-涓夊垎-鍦帮紝鐙鍙戝竷
  35.          return ret;
  36.         }

  37.         //bfs matrix to mark the distance from x and y
  38.         void bfs(char[][] mat, int x, int y, int[][] dis){
  39.                  Queue<Point> path = new LinkedList();
  40.                  int[] dx = {1,-1,0,0};
  41.                  int[] dy = {0,0,1,-1};. 鐣欏鐢宠璁哄潧-涓€浜╀笁鍒嗗湴
  42.                  int m = mat.length;
  43.                  int n = mat[0].length;
  44.                  boolean[][] visited = new boolean[m][n];
  45.                  int ct = 0;
  46.                  System.out.println("Started at: " + x + " " + y);
  47.                  int di = 0;
  48.                  path.offer(new Point(x, y));
  49.                  visited[x][y] = true;
  50.                  while (!path.isEmpty()){
  51.                      int count = path.size();
  52.                      for (int i = 0; i < count; i++){
  53.                          Point cur = path.poll();
  54.                          ct++;
  55.                          // System.out.println(cur.x + " " + cur.y);
  56.                          // visited[cur.x][cur.y] = true;
  57.                          // update cur.value
  58.                          dis[cur.x][cur.y] += di;
  59.                          // check its neighbors
  60.                          for (int j = 0; j < 4; j++){
  61.                              int nx = cur.x + dx[j];
  62.                              int ny = cur.y + dy[j];
  63.                              if (nx >= 0 && nx < m && ny >= 0 && ny < m && mat[nx][ny] != 'X' && !visited[nx][ny]){. 涓€浜�-涓夊垎-鍦帮紝鐙鍙戝竷
  64.                                  path.offer(new Point(nx, ny));. Waral 鍗氬鏈夋洿澶氭枃绔�,
  65.                                  visited[nx][ny] = true;
  66.                              }
  67.                          }
  68.                      }
  69.                      di++;
  70.                  }
  71.                  System.out.println("total points added to queue: " + ct);
  72.         }
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