7.数据结构---图（遍历）

posted @ 2019-10-15 10:47 nxf_rabbit75 阅读(443) 评论(0) 编辑收藏举报

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一、深度遍历

1.员工的重要性 leetcode 690

思路：深度遍历

"""
# Employee info
class Employee:
    def __init__(self, id, importance, subordinates):
        # It's the unique id of each node.
        # unique id of this employee
        self.id = id
        # the importance value of this employee
        self.importance = importance
        # the id of direct subordinates
        self.subordinates = subordinates
"""
class Solution:
    def getImportance(self, employees, id):
        """
        :type employees: Employee
        :type id: int
        :rtype: int
        """
        self.res = 0
        emp_infos = dict()
        for employee in employees:
            emp_infos[employee.id] = [employee.importance,employee.subordinates]
          
        def dfs(subs):
            for sub in subs:
                self.res += emp_infos[sub][0]
                dfs(emp_infos[sub][1])
                      
        dfs([id])
        return self.res
        

2.图像渲染 leetcode 773

以中心点上下左右扩散

思路：深度搜索，遍历中心点的上下左右四个点，然后到达一个点，再继续遍历其到上下左右到节点，直到碰到边界，就返回

class Solution:
    def floodFill(self, image: List[List[int]], sr: int, sc: int, newColor: int) -> List[List[int]]:
        self.recode = []
        self.m = len(image)
        self.n = len(image[0])
        def dfs(sr,sc,image):
            directions = [[-1,0],[1,0],[0,-1],[0,1]]#上下左右
              
            for d in directions:
                new_sr = sr+d[0]
                new_sc = sc+d[1]
                if new_sr >= 0 and new_sr < self.m and new_sc >= 0 and new_sc < self.n:#未越界
                    if image[new_sr][new_sc] == image[sr][sc] and [new_sr,new_sc] not in self.recode:
                        print(new_sr,new_sc)
                        self.recode.append([new_sr,new_sc])
                        dfs(new_sr,new_sc,image)
              
        dfs(sr,sc,image)
        image[sr][sc] = newColor
        for src in self.recode:
            n_sr,n_sc = src[0],src[1]
            image[n_sr][n_sc] = newColor
        return image

3.岛屿的个数 leetcode200

思路：深度搜索

class Solution(object):
    dx = [1,0,-1,0]
    dy = [0,-1,0,1]
    def numIslands(self, grid):
        ans = 0
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j] == '1':
                    self.dfs(grid,i,j)
                    ans +=1
        return ans
    def dfs(self,grid,i,j):
        grid[i][j] = '.'
        for k in range(4):
            m = i+self.dx[k]
            n = j+self.dy[k]
            if m>-1 and m<len(grid) and n >-1 and n<len(grid[0]) and grid[m][n] == '1':
                self.dfs(grid,m,n)
        return

4.两点间路径条数

思路：深度遍历

# row, col = map(int, input().split())
# graph = []
# for _ in range(row):
#     graph.append(list(map(int, input().split())))
# print(graph)
  
# x1, y1, x2, y2 = map(int, input().split())
  
dirs = [(-1, 0), (1, 0), (0, 1), (0, -1)]#上,下,右,左
M = 10 ** 9
res = [0]
  
graph = [[0, 1, 0, 0, 0], [0, 2, 3, 0, 0], [0, 0, 4, 5, 0], [0, 0, 7, 6, 0]]
row = 4
col = 5
x1, y1, x2, y2 = 0, 1, 3, 2
  
def dfs(x1, y1, visited):
    if x1 == x2 and y1 == y2:
        res[0] += 1
        return
    for i, j in dirs:#遍历4个方向,判断下一步哪步可行,然后从这步开始,继续深度遍历,直到到达终点
        tmp_x = i + x1
        tmp_y = j + y1
        if 0 <= tmp_x < row and 0 <= tmp_y < col and graph[tmp_x][tmp_y] > graph[x1][y1] \
                and (tmp_x, tmp_y) not in visited:
            dfs(tmp_x, tmp_y, visited | {(tmp_x, tmp_y)})
    print (visited)
dfs(x1, y1, {(x1, y1)})
  
print(res[0] % M)

二、广度遍历

（一）图

图结构：节点、边
分类：有向图、无向图
特殊的图：二叉树（二叉搜索树）、普通树（并查集）、堆
连通图与非连通图

　　连通图（Connected Graphs）指图内任意两个节点间，总能找到一条路径连接它们，否则，为非连通图（Disconnected Graphs）。也就是说，如果图中包含岛（Island）,则是非连通图。如果岛内的节点都是连通的，这些岛就被成为一个部件（Component，有时也叫 Cluster）。