[TopCoder]SRM 646 Div2 The Grid Div Two | 书影博客
John is standing at the origin of an infinite two-dimensional grid. He is going to move along this grid. During each second he can either stay where he is or he can move by one unit in one of the four cardinal directions (north, south, east, or west). Some of the grid points are blocked. John is not allowed to move to a blocked grid point.
You are given the coordinates of the blocked grid points as tuple (integer)s x and y. For each valid i, the grid point that is x[i] units east and y[i] units north of the origin is blocked. You are also given an integer k. Compute and return the maximal possible x-coordinate of a point John can reach in k seconds.
Constraints
- x will contain between 0 and 47 elements, inclusive.
- x and y will contain the same number of elements.
- Each element of x will be between -1,000 and 1,000, inclusive.
- Each element of y will be between -1,000 and 1,000, inclusive.
- All pairs (x[i], y[i]) will be distinct.
- Each pair (x[i], y[i]) will be different from (0, 0).
- k will be between 1 and 1,000, inclusive.
给定禁止通行的网格坐标点，同时给定一个整数k，John至多行走k秒，求John可以到达点的x坐标的最大值。
约束条件如上。
剪枝策略：由于禁止通行的点至多只有47个，因此可以排除y坐标>25或者<-25的点，以及x坐标<-25的点。
障碍物的极限情况分别为：
x轴极限情况：
障碍物由原点出发，上下包围x轴向左延伸至-47 / 2处，y坐标为±1。
y轴极限情况：
障碍物平行于y轴依次排列。
def find(self, x, y, k): HORIZON, MAX = 1024, 2048 dx = (1, -1, 0, 0) dy = (0, 0, 1, -1) visited = [[0] * MAX for i in range(MAX)] #visited[row][col] size = len(x) for i in range(size): visited[x[i] + HORIZON][y[i] + HORIZON] = True sp = (0, 0, k) visited[sp[0]][sp[1]] = True queue = [sp] ans = 0 while queue: #BFS point = queue.pop(0) if point[2] == 0: continue for j in range(4): np = (point[0] + dx[j], point[1] + dy[j], point[2] - 1) if np[0] < -25 or np[1] > 25 or np[1] < -25: #Pruning continue if np[2] and np[0] + np[2] <= ans or visited[np[0] + HORIZON][np[1] + HORIZON]: continue visited[np[0] + HORIZON][np[1] + HORIZON] = True queue.append(np) ans = max(ans, np[0]) return ans
Read full article from [TopCoder]SRM 646 Div2 The Grid Div Two | 书影博客
John is standing at the origin of an infinite two-dimensional grid. He is going to move along this grid. During each second he can either stay where he is or he can move by one unit in one of the four cardinal directions (north, south, east, or west). Some of the grid points are blocked. John is not allowed to move to a blocked grid point.
You are given the coordinates of the blocked grid points as tuple (integer)s x and y. For each valid i, the grid point that is x[i] units east and y[i] units north of the origin is blocked. You are also given an integer k. Compute and return the maximal possible x-coordinate of a point John can reach in k seconds.
Constraints
- x will contain between 0 and 47 elements, inclusive.
- x and y will contain the same number of elements.
- Each element of x will be between -1,000 and 1,000, inclusive.
- Each element of y will be between -1,000 and 1,000, inclusive.
- All pairs (x[i], y[i]) will be distinct.
- Each pair (x[i], y[i]) will be different from (0, 0).
- k will be between 1 and 1,000, inclusive.
题目大意：
John站在无限二维坐标网格的原点(0,0)。他沿着网格行走，每秒任选坐标系4个方向（东、南、西、北）之一移动1个单位。网格中的一些点禁止通行。给定禁止通行的网格坐标点，同时给定一个整数k，John至多行走k秒，求John可以到达点的x坐标的最大值。
约束条件如上。
解题思路：
宽度优先搜索（BFS） + 剪枝（Pruning）剪枝策略：由于禁止通行的点至多只有47个，因此可以排除y坐标>25或者<-25的点，以及x坐标<-25的点。
障碍物的极限情况分别为：
x轴极限情况：
障碍物由原点出发，上下包围x轴向左延伸至-47 / 2处，y坐标为±1。
y轴极限情况：
障碍物平行于y轴依次排列。
注意：采用Python解题必须剪枝才能通过系统测试，否则会超时。使用C++和Java采用朴素的BFS也可通过测试。
def find(self, x, y, k): HORIZON, MAX = 1024, 2048 dx = (1, -1, 0, 0) dy = (0, 0, 1, -1) visited = [[0] * MAX for i in range(MAX)] #visited[row][col] size = len(x) for i in range(size): visited[x[i] + HORIZON][y[i] + HORIZON] = True sp = (0, 0, k) visited[sp[0]][sp[1]] = True queue = [sp] ans = 0 while queue: #BFS point = queue.pop(0) if point[2] == 0: continue for j in range(4): np = (point[0] + dx[j], point[1] + dy[j], point[2] - 1) if np[0] < -25 or np[1] > 25 or np[1] < -25: #Pruning continue if np[2] and np[0] + np[2] <= ans or visited[np[0] + HORIZON][np[1] + HORIZON]: continue visited[np[0] + HORIZON][np[1] + HORIZON] = True queue.append(np) ans = max(ans, np[0]) return ans
Read full article from [TopCoder]SRM 646 Div2 The Grid Div Two | 书影博客
