dfs采用的是栈，bfs采用的是队列。
sorted(L.items(),key=lambda x:(x[0],x[1],x[2],x[3]))

拓扑排序

深度优先搜索方法

import collections
graph = {
    'A':['F','E','C'],
    'B':['A','C'],
    'C':[],
    'D':['F'],
    'E':[],
    'F':['E','G'],
    'G':['F']
}
n=len(graph)
visted={key:0 for key in graph.keys()}
ans=[]
def dfs(u):
    if visted[u]==2:
        return
    visted[u]=1
    for v in graph[u]:
        if visted[v]==0:
            dfs(v)
            
    visted[u]=2
    ans.append(u)
for key in graph.keys():
    dfs(key)
print(ans)
# ['E', 'G', 'F', 'C', 'A', 'B', 'D']

广度优先搜索方法

import collections
graph = {
    'A':['F','E','C'],
    'B':['A','C'],
    'C':[],
    'D':['F'],
    'E':[],
    'F':['E','G'],
    'G':['F']
}
n=len(graph)
visted={key:0 for key in graph.keys()}
queue=[]
ans=[]
def bfs(u):
    if visted[u]:return
    queue.append(u)
    visted[u]=1
    while queue:
        node=queue.pop(0)
        ans.append(node)
        for v in graph[node]:
            if visted[v]==0:
                visted[v]=1
                queue.append(v)
                
                
for key in graph.keys():
    bfs(key)
# bfs('A')
# print(ans,visted)
# bfs('B')
# print(ans,visted)
# ['E', 'G', 'F', 'C', 'A', 'B', 'D']
print(ans)
# ['A', 'F', 'E', 'C', 'G', 'B', 'D']

无向无权图

graph={
    1:[2,5],
    2:[1,3,4],
    3:[2],
    4:[2],
    5:[1,6],
    6:[5,7],
    7:[6]
}
n=len(graph)+1
# visted=[0]*n
visted=[0, 0, 0, 0, 0, 0, 0,0]
def dfs(u):
    ans=1
    visted[u]=1
    for v in graph[u]:
        if visted[v]==0:
            vh=dfs(v)
#             print(u,v,vh)
            ans=max(ans,vh+1)
    return ans

dfs(2) #5

无向有权图

有向无权图 利用广度优先搜索算法

graph={
    1:[2,4],
    2:[4,5],
    3:[1,6],
    4:[3,5,6,7],
    5:[7],
    6:[],
    7:[6],
}
def shortest(root,graph):
    n=len(graph)+1
    path=[0]*n
    visit=[0]*n
    dist=[float('inf')]*n
    queue=[root]
    visit[root]=1
    dist[root]=0
    while queue:
        ver=queue.pop()
        for nex in graph[ver]:
            if visit[nex]==0:
                path[nex]=ver
                visit[nex]=1
                dist[nex]=dist[ver]+1
                queue.append(nex)
    print(dist,path,visit)
                
shortest(3,graph)

有向有权图 带排序的广度优先算法/dijkstra

import heapq
graph1={
    1:[(2,2),(1,4)],
    2:[(3,4),(10,5)],
    3:[(4,1),(5,6)],
    4:[(2,3),(2,5),(8,6),(4,7)],
    5:[(1,7)],
    6:[],
    7:[(1,6)],
}
def shortest2(root,graph):
    n=len(graph)+1
    path=[0]*n
    visit=[0]*n
    dist=[float('inf')]*n
    
    dist[root]=0
    queue=[(0,root)]
    heapq.heapify(queue)
    while queue:
        dis,ver=heapq.heappop(queue)
        visit[ver]=1
        for nex in graph[ver]:
            if visit[nex[1]]==0:
                heapq.heappush(queue,nex)
                if dist[ver]+nex[0]<dist[nex[1]]:
                    dist[nex[1]]=dist[ver]+nex[0]
                    path[nex[1]]=ver
    print(dist,path,visit)
    
shortest2(3,graph1)
#[inf, 4, 6, 0, 5, 7, 5, 8] [0, 3, 1, 0, 1, 4, 3, 5] [0, 1, 1, 1, 1, 1, 1, 1]

最小生成树

prims算法

树是连通 无向 无环图。n个结点一定有n-1条边。
一个普通的生成树

最小生成树

Kruskal’s Algorithm

graph=[
    [3,1,4],[3,4,2],[3,6,5],[1,4,1],[4,6,8],[1,2,2],[6,7,1],[2,4,3],[4,7,4],[2,5,10],[5,7,6],[4,5,7]
]
def kruskal(graph,n):
    find=[i for i in range(n+1)]
    graph.sort(key=lambda x:x[2])
    def parent(x):
        if find[x]==x:
            return x
        else:
            return find[x]
    
    ans=[]
    while len(ans)<n-1:
        x,y,w = graph.pop(0)
        if parent(x)!=parent(y):
            ans.append([x,y,w])
            find[x]=y
    return ans
    
kruskal(graph,7)    
# [[1, 4, 1], [6, 7, 1], [3, 4, 2], [1, 2, 2], [2, 4, 3], [3, 1, 4]]

最小割 min-cut

二分图 Bipartite Graph 队列

例题1 所有可能的路径

https://leetcode.cn/problems/all-paths-from-source-to-target/description/

class Solution:
    def allPathsSourceTarget(self, graph: List[List[int]]) -> List[List[int]]:
        path = [0]
        ans=[]

        def dfs(graph,index):
            if index==len(graph)-1:
                ans.append(path[:])
                return 
            for node in graph[index]:
                path.append(node)
                dfs(graph,node)
                path.pop()
        dfs(graph,0)
        return ans

class Solution:
    def allPathsSourceTarget(self, graph: List[List[int]]) -> List[List[int]]:
        from collections import deque
        ans=[]
        q=deque([[0],])
        while q:
            path = q.popleft()
            if path[-1]==len(graph)-1:
                ans.append(path)
                continue
            for v in graph[path[-1]]:
                q.append(path+[v])
        return ans 

例题2 岛屿数量

https://leetcode.cn/problems/number-of-islands/description/

class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        n=len(grid)
        m=len(grid[0])
        ans=0

        def wipe(x,y):
            dir=[
            lambda x,y:[x+1,y],#下
            lambda x,y:[x-1,y],#上
            lambda x,y:[x,y+1],#右
            lambda x,y:[x,y-1],#左
            ]
            stack=[(x,y)]
            while stack:
                x,y=stack.pop()
                grid[x][y]='0'
                for i in range(4):
                    nxt_x,nxt_y=dir[i](x,y)
                    if nxt_x>=0 and nxt_x <n and nxt_y<m and nxt_y>=0:
                        if grid[nxt_x][nxt_y]=='1':
                            stack.append((nxt_x,nxt_y))
                        
        for i in range(n):
            for j in range(m):
                if grid[i][j]=='1':
                    ans+=1
                    wipe(i,j)
        return ans

例题3 岛屿最大面积

https://leetcode.cn/problems/max-area-of-island/

class Solution:
    def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
        n=len(grid)
        m=len(grid[0])
        ans=0
        for i in range(n):
            for j in range(m):
                if grid[i][j]==1:
                    grid[i][j]=0
                    tmp=1
                    stack=[(i,j)]
                    dir=[
                    lambda x,y:[x+1,y],
                    lambda x,y:[x-1,y],
                    lambda x,y:[x,y+1],
                    lambda x,y:[x,y-1],
                    ]
                    while stack:
                        x,y=stack.pop()
                        grid[x][y]=0
                        for k in range(4):
                            nx,ny = dir[k](x,y)
                            if nx>=0 and nx<n and ny>=0 and ny<m and grid[nx][ny]==1:
                                tmp+=1
                                grid[nx][ny]=0
                                stack.append((nx,ny))
                    ans = max(ans,tmp)
        return ans

例题4 飞地的数量

https://leetcode.cn/problems/number-of-enclaves/

class Solution:
    def numEnclaves(self, grid: List[List[int]]) -> int:
        n=len(grid)
        m=len(grid[0])
        ans=0
        def dfs(x,y):
            for dx,dy in (0,1),(1,0),(-1,0),(0,-1):
                nx = x+dx
                ny =y+dy
                if 0<= nx <n and 0<= ny <m and grid[nx][ny]==1:
                    grid[nx][ny]=0
                    dfs(nx,ny)

        for i in [0,n-1] :
            for j in range(m):
                if grid[i][j]==1:
                    ans+=1
                    grid[i][j]=0
                    dfs(i,j)


        for i in range(n):
            for j in [0,m-1]:
                if grid[i][j]==1:
                    ans+=1
                    grid[i][j]=0
                    dfs(i,j)
        return sum(sum(g) for g in grid)

class Solution {
    public static int[][] dirs={{-1,0},{1,0},{0,1},{0,-1}};
    private int n,m;
    private boolean[][] visted;

    public int numEnclaves(int[][] grid) {
        n = grid.length;
        m=grid[0].length;
        visted = new boolean[n][m];
        for(int i=0;i<n;i++){
            dfs(grid,i,0);
            dfs(grid,i,m-1);
        }
        for(int j=0;j<m;j++){
            dfs(grid,0,j);
            dfs(grid,n-1,j);
        }
        int ans=0;
        for(int i=0;i<n;i++){
            for(int j =0;j<m;j++){
                if(grid[i][j]==1 && !visted[i][j]){
                    ans+=1;
                }
            }
        }
        return ans;
    }

    public void dfs(int[][]grid,int x,int y){
        if (x<0 || x>=n || y<0 || y>=m || grid[x][y]==0 || visted[x][y]){
            return;
        }
        visted[x][y]=true;
        for(int[] dir:dirs){
            dfs(grid,x+dir[0],y+dir[1]);
        }
    }
}

例题5 被围绕的区域

https://leetcode.cn/problems/surrounded-regions/description/

class Solution:
    def solve(self, board: List[List[str]]) -> None:
        """
        Do not return anything, modify board in-place instead.
        """
        n=len(board)
        m=len(board[0])
        visted=[[0]*m for _ in range(n)]

        def dfs(x,y):
            if 0<=x<n and 0<=y<m and board[x][y]=='O' and not visted[x][y]:
                visted[x][y]=1
                for dx,dy in (0,1),(0,-1),(1,0),(-1,0):
                    dfs(x+dx,y+dy)
            else:
                return

        for i in range(n):
            dfs(i,0)
            dfs(i,m-1)

        for j in range(m):
            dfs(0,j)
            dfs(n-1,j)
        
        for i in range(n):
            for j in range(m):
                if board[i][j]=='O' and visted[i][j]==0 :
                    board[i][j]='X'

例题6 太平洋大西洋水流问题

https://leetcode.cn/problems/pacific-atlantic-water-flow/description/

class Solution:
    def pacificAtlantic(self, heights: List[List[int]]) -> List[List[int]]:
        n=len(heights)
        m=len(heights[0])
        pacific =[[0]*m for _ in range(n)]
        atlantic=[[0]*m for _ in range(n)]
        ans=[]
        def dfs(x,y,grid):
                grid[x][y]=1
                for dx,dy in (0,1),(0,-1),(1,0),(-1,0):
                    nx=x+dx
                    ny=y+dy
                    if 0<=nx<n and 0<=ny<m and heights[nx][ny]>=heights[x][y] and grid[nx][ny]==0:
                        dfs(nx,ny,grid)

        for i in range(n):
            for j in range(m):
                if j==0 or i==0:
                    dfs(i,j,pacific)
                if i==n-1 or j==m-1:
                    atlantic[i][j]=1
                    dfs(i,j,atlantic)
        for i in range(n):
            for j in range(m):
                if pacific[i][j]==1 and atlantic[i][j]==1:
                    ans.append([i,j])
        return ans

例题7 钥匙和房间

https://leetcode.cn/problems/keys-and-rooms/

class Solution:
    def canVisitAllRooms(self, rooms: List[List[int]]) -> bool:
        n=len(rooms)
        visted=[0]*n
        def dfs(x):
            if visted[x]:
                return
            visted[x]=1
            for nx in rooms[x]:
                dfs(nx)

        dfs(0)
        return sum(visted)==n

例题8 寻找图中是否存在路径

https://leetcode.cn/problems/find-if-path-exists-in-graph/description/

class Solution:
    def validPath(self, n: int, edges: List[List[int]], source: int, destination: int) -> bool:
            parent=[i for i in range(n)]
            def find(x):
                if x==parent[x]:
                    return x
                parent[x]=find(parent[x])
                return find(parent[x])
            def union(x,y):
                px=find(x)
                py=find(y)
                if px!=py:
                    parent[px]=py

            for u,v in edges:
                if find(u)!=find(v):
                    union(u,v)
            return find(source)==find(destination)

例题9 冗余连接

https://leetcode.cn/problems/redundant-connection/description/

class Solution:
    def findRedundantConnection(self, edges: List[List[int]]) -> List[int]:
        n=len(edges)
        parent=[i for i in range(n+1)]
        def find(x):
            if x==parent[x]:
                return x
            parent[x]=find(parent[x])
            return find(parent[x])
        def union(x,y):
            px=find(x)
            py=find(y)
            if px !=py:
                parent[px]=py
        for u,v in edges:
            if find(u)==find(v):
                return [u,v]
            else:
                union(u,v)

例题10 课程表 拓扑排序

https://leetcode.cn/problems/course-schedule/description/

import collections
class Solution:
    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
        edges=collections.defaultdict(list)
        visted=[0]*numCourses
        restult=[]
        valid = True
        for u,v in prerequisites:
            edges[v].append(u)
        
        def dfs(u):
            nonlocal valid
            visted[u]=1
            for v in edges[u]:
                if visted[v]==0:
                    dfs(v)
                    if not valid:
                        return
                elif visted[v]==1:
                    valid=False
                    return
            visted[u]=2
            restult.append(u)

        for i in range(numCourses):
            if valid and not visted[i]:
                dfs(i)
        return valid

import collections
class Solution:
    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
        edges=collections.defaultdict(list)
        indeg=[0]*numCourses

        for u,v in prerequisites:
            edges[v].append(u)
            indeg[u]+=1
        
        queue = [u for u in range(numCourses) if indeg[u]==0]
        visted =0
        while queue:
            u = queue.pop(0)
            visted+=1
            for v in edges[u]:
                indeg[v]-=1
                if indeg[v]==0:
                    queue.append(v)

        return visted==numCourses

例题11 单词接龙

https://leetcode.cn/problems/word-ladder/description/

class Solution:
    def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
        wordList = set(wordList)
        if endWord not in wordList:
            return 0
        q = deque([(beginWord, 1)])
        while q:
            cur, step = q.popleft()
            for i, x in enumerate(cur):
                for y in [chr(ord('a')+i) for i in range(26)]:
                    if y != x:
                        nxt = cur[:i] + y + cur[i+1:]
                        if nxt == endWord:
                            return step + 1
                        if nxt in wordList:
                            wordList.remove(nxt)
                            q.append((nxt, step+1))
        return 0

例题12 最小高度树

https://leetcode.cn/problems/minimum-height-trees/description/

import collections
class Solution:
    def findMinHeightTrees(self, n: int, edges: List[List[int]]) -> List[int]:
        graph=collections.defaultdict(list)
        res=[]
        for u,v in edges:
            graph[u].append(v)
            graph[v].append(u)
        
        def dfs(u):
            ans=1
            visted[u]=1
            for v in graph[u]:
                if visted[v]==0:
                    vh=dfs(v)
        #             print(u,v,vh)
                    ans=max(ans,vh+1)
            return ans

        visted=[0]*n
        res.append([0,dfs(0)])
        for i in range(1,n):
            visted=[0]*n
            tmp=dfs(i)
            if tmp<res[-1][-1]:
                res=[[i,tmp]]
            elif tmp==res[-1][-1]:
                res.append([i,tmp])


        return [u for u,v in res]

例题13 省份数量

https://leetcode.cn/problems/number-of-provinces/description/
1.利用并查集

class Solution:
    def findCircleNum(self, isConnected: List[List[int]]) -> int:
        n=len(isConnected)
        par=[i for i in range(n+1)]
        def find(x):
            if x==par[x]:
                return x
            par[x]=find(par[x])
            return find(par[x])

        for i in range(n):
            for j in range(n):
                if i!=j and isConnected[i][j]==1:
                    pi=find(i+1)
                    pj= find(j+1)
                    par[pi]=pj
        for i in range(n+1):
            par[i]=find(i)
        return len(set(par))-1

class Solution:
    def findCircleNum(self, isConnected: List[List[int]]) -> int:
        n=len(isConnected)
        visted=[0]*n
        ans=0
        def dfs(x):
            visted[x]=1
            for j in range(n):
                if isConnected[x][j]==1 and visted[j]==0:
                    dfs(j)

        for i in range(n):
            if visted[i]==0:
                dfs(i)
                ans+=1
        return ans

例题14 判断二分图

https://leetcode.cn/problems/is-graph-bipartite/

class Solution:
    def __init__(self):
        self.ans = True

    def isBipartite(self, graph: List[List[int]]) -> bool:
        n=len(graph)
        par=[i for i in range(n)]
        visted=[0]*n
        def dfs(x,flag):
            if not self.ans :return self.ans
            par[x]=flag 
            visted[x]=1
        #     print(x,visted,dic)
            for nx in graph[x]:
                if visted[nx]==1:
                    # print(x,nx,flag,par[nx],par[nx] == flag,dic,visted)
                    if par[nx] == flag:
                        self.ans = False
                else:
                    dfs(nx,1-flag)
            return self.ans
        
        res=True
        for i in range(n):
            if visted[i]==0:
                res = res and dfs(i,0)

        return res
      

class Solution:
    def isBipartite(self, graph: List[List[int]]) -> bool:
        n=len(graph)
        visted=[0]*n
        color=[0]*n
        valid = True

        def dfs(x,c):
            nonlocal valid
            color[x]=c
            visted[x]=1
            for nx in graph[x]:
                if visted[nx]==0:
                    dfs(nx,1-c)
                    if not valid:
                        return
                elif color[nx]!=1-c:
                    valid=False
                    return
        
        for i in range(n):
            if visted[i]==0:
                dfs(i,0)
                if not valid:
                    break

        return valid

class Solution:
    def isBipartite(self, graph: List[List[int]]) -> bool:
        n=len(graph)
        visted=[0]*n
        color=[0]*n

        for i in range(n):
            if visted[i]==0:  #初始化
                q=collections.deque([i])
                color[i]=0
                visted[i]=1
                while q:
                    x=q.popleft()
                    for nx in graph[x]:
                        if visted[nx]==0:
                            color[nx]=1-color[x]
                            q.append(nx)
                            visted[nx]=1
                        elif color[nx]==color[x]:
                            return False
        return True

class Solution {
    public boolean isBipartite(int[][] graph) {
        int n=graph.length;
        int[] color = new int[n];
        int[] visted = new int[n];
        Arrays.fill(visted,0);

        for(int i=0;i<n;i++){
            if(visted[i]==0){
                Queue<Integer> queue =new LinkedList<Integer>();
                queue.offer(i);
                color[i]=0;
                visted[i]=1;
                while(!queue.isEmpty()){
                    int x = queue.poll();
                    for(int nx :graph[x]){
                        if(visted[nx]==0){
                            color[nx]=1-color[x];
                            visted[nx]=1;
                            queue.offer(nx);
                        }else if (color[nx]==color[x]) return false;
                    }
                }
            }
        }
        return true;

    }
}

class Solution {
    private static int[] color;
    private static int[] visted;
    private static boolean valid;

    public boolean isBipartite(int[][] graph) {
        int n = graph.length;
        color = new int[n];
        visted= new int[n];
        valid = true;
        Arrays.fill(visted,0);
        for(int i=0;i<n;i++){
            if(visted[i]==0){
                dfs(i,0,graph);
                if(!valid){
                    break;
                }
            }
        }
        return valid;
    }

    public void dfs(int x,int c,int[][] graph){
        visted[x]=1;
        color[x]=c;
        for(int nx:graph[x]){
            if (visted[nx]==0){
                dfs(nx,1-c,graph);
                if(!valid)return;
            }else if(color[nx]==c){
                valid=false;
                return;
            }
        }
    }
}