## Saturday, November 28, 2015

### 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

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]);
}
}

.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.                 }
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.          }
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.         }