7. Professional wrestlers can be divided into two types: the "baby-faced" ("good guy") type and the "high heels" ("bad guy") type. It is possible for any pair of professional wrestlers to be in contention. Suppose there are n professional wrestlers and there is a linked list of r pairs of wrestlers that give the competitive relationship. Give an algorithm with a time of O(n+r) to determine if you can classify some wrestlers as "baby-faced" and the rest as "heels", so that all competition is only between baby-faced and high-heeled fighters. If such a division is possible, the algorithm should also generate such a division. If you want to write code, use go.
Wenxin's words:
To solve this problem, we can use the concept of bipartite graph in graph theory. A bipartite graph is a special kind of graph in which vertices can be divided into two non-overlapping subsets, so that the two vertices connected by each edge of the graph belong to these two different subsets. In this problem, the wrestler can be considered as the vertex of the graph, and the competitive relationship can be seen as the edge of the graph. We want to determine whether the rivalry between a given wrestler can form a dichotomous diagram and give a way to divide it.
You can use depth-first search (DFS) or breadth-first search (BFS) to detect and construct dichotomous plots. Here I will use DFS to solve the problem and give an implementation of Go.
Go language implementation
package main
import(
"fmt"
)
typeGraphstruct{
vertices int
adjList [][]int
colors []int// 0: uncolored, 1: first color, -1: second color
}
func NewGraph(n int)*Graph{
return&Graph{
vertices: n,
adjList:make([][]int, n),
colors:make([]int, n),
}
}
func (g *Graph)AddEdge(v, w int){
g.adjList[v]=append(g.adjList[v], w)
g.adjList[w]=append(g.adjList[w], v)// 因为是无向图
}
func (g *Graph) isBipartite()bool{
for i :=0; i < g.vertices; i++{
if g.colors[i]==0&&!g.dfs(i,1){
returnfalse
}
}
returntrue
}
func (g *Graph) dfs(v, color int)bool{
g.colors[v]= color
for _, neighbor :=range g.adjList[v]{
if g.colors[neighbor]== color {
// 发现同色顶点，不是二分图
returnfalse
}
if g.colors[neighbor]==0&&!g.dfs(neighbor,-color){
returnfalse
}
}
returntrue
}
func main(){
n :=5// 假设有5个摔跤手
g :=NewGraph(n)
// 假设有以下竞争关系：(0, 1), (1, 2), (3, 4)
g.AddEdge(0,1)
g.AddEdge(1,2)
g.AddEdge(3,4)
if g.isBipartite(){
fmt.Println("可以划分为二分图")
for i, color :=range g.colors {
if color ==1{
fmt.Printf("摔跤手 %d 是 '娃娃脸' 型\n", i)
}elseif color ==-1{
fmt.Printf("摔跤手 %d 是 '高跟鞋' 型\n", i)
}
}
}else{
fmt.Println("无法划分为二分图")
}
} illustrate
- 1. Graph structure: Contains the number of vertices, adjacency tables, and a color array to store the color markers during DFS.
- 2. AddEdge 方法：向图中添加无向边。
- 3. isBipartite method: Detect whether the plot is a bipartite plot and color mark it by DFS.
- 4. DFS Method: Perform a depth-first search and try to color each vertex. If vertices of the same color are found to be connected, false is returned.
- 5. main function: build the graph, call isBipartite, and print the result.
The above code will check if a given wrestler rivalry can form a dichotomous graph, and output the corresponding partition result.
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iFLYTEK Spark:
package main
import(
"fmt"
)
typeWrestlerstruct{
id int
color int// 0 for "娃娃脸", 1 for "高跟鞋"
}
func canPartition(n int, edges [][]int)bool{
wrestlers :=make([]Wrestler, n)
for i :=range wrestlers {
wrestlers[i]=Wrestler{id: i, color:-1}
}
// 构建邻接表表示竞争关系
adjList :=make([][]int, n)
for _, edge :=range edges {
u, v := edge[0], edge[1]
adjList[u]=append(adjList[u], v)
adjList[v]=append(adjList[v], u)
}
// 使用深度优先搜索进行染色
var dfs func(int, int)bool
dfs =func(node int, color int)bool{
if wrestlers[node].color !=-1{
return wrestlers[node].color == color
}
wrestlers[node].color = color
for _, neighbor :=range adjList[node]{
if!dfs(neighbor,1-color){
returnfalse
}
}
returntrue
}
// 遍历所有节点，尝试染色
for i :=0; i < n; i++{
if wrestlers[i].color ==-1&&!dfs(i,0){
returnfalse
}
}
returntrue
}
func main(){
n :=5
edges :=[][]int{{0,1},{1,2},{3,4}}
result := canPartition(n, edges)
fmt.Println("是否可以划分：", result)
} 360 Smart Brain:
This problem can be solved by a bipartite graph judgment algorithm in graph theory. We can think of each wrestler as a node in the graph, and rivalry as an edge in the graph. If there is a way to divide nodes into two sets such that there are no edges connected to each other within the same set, then the graph is a dichotomous graph.
Here's an example of Go code that uses depth-first search (DFS) to determine and generate bipartite graph partitions:
package main
import"fmt"
typeWrestlerstruct{
id int
isBabyface bool// true for "娃娃脸", false for "高跟鞋"
}
typeEdgestruct{
from, to int
}
func isBipartite(wrestlers []Wrestler, edges []Edge)(bool,map[int]bool){
n :=len(wrestlers)
color :=make([]bool, n)// color[i]表示第i个摔跤手的类型
for i :=range color {
color[i]=false// 初始化所有摔跤手为"高跟鞋"
}
// 构建邻接表
adjList :=make([][]int, n)
for _, edge :=range edges {
adjList[edge.from]=append(adjList[edge.from], edge.to)
adjList[edge.to]=append(adjList[edge.to], edge.from)
}
var dfs func(int, bool)bool
dfs =func(node int, c bool)bool{
if color[node]!=false{// 如果已经着色
return color[node]== c // 检查颜色是否一致
}
color[node]= c // 着色
for _, neighbor :=range adjList[node]{
if!dfs(neighbor,!c){// 对邻居进行DFS，颜色相反
returnfalse
}
}
returntrue
}
for i :=range wrestlers {
if color[i]==false&&!dfs(i,true){// 从每个未着色的节点开始DFS
returnfalse,nil
}
}
returntrue, color
}
func main(){
// 假设我们有5个摔跤手，竞争关系如下：
// 0 vs 1, 0 vs 2, 1 vs 3, 2 vs 4
wrestlers :=[]Wrestler{
{id:0},
{id:1},
{id:2},
{id:3},
{id:4},
}
edges :=[]Edge{
{from:0, to:1},
{from:0, to:2},
{from:1, to:3},
{from:2, to:4},
}
isPossible, colors := isBipartite(wrestlers, edges)
if isPossible {
fmt.Println("可以进行划分，划分结果如下：")
for i, color :=range colors {
fmt.Printf("摔跤手%d: %s\n", wrestlers[i].id,map[bool]string{true:"娃娃脸",false:"高跟鞋"}[color])
}
}else{
fmt.Println("无法进行划分。")
}
} This code first defines the Wrestler and Edge structs, which represent wrestlers and rivalries, respectively. Then define an isBipartite function that takes a list of wrestlers and a list of edges as input, returns a boolean value indicating whether a division can be made, and a color mapping table where true represents "baby face" and false represents "heels".
In the main function, we have created a simple example to test the isBipartite function. If it can be divided, it prints out the type of each wrestler; If not, it will output the corresponding information.
Zhipu Qingyan:
This problem can be solved by detecting whether a given competition graph is a dichotomous graph or not. A bipartite graph is a special type of graph in which a vertex set can be divided into two disjoint subsets, such that each edge connects only one vertex of the two subsets. Here's the algorithm implemented in Go:
package main
import(
"fmt"
)
// Graph represents an adjacency list graph
typeGraphstruct{
adjList map[int][]int
}
// NewGraph creates a new graph
func NewGraph(vertices int)*Graph{
return&Graph{
adjList:make(map[int][]int, vertices),
}
}
// AddEdge adds an edge to the graph
func (g *Graph)AddEdge(u, v int){
g.adjList[u]=append(g.adjList[u], v)
g.adjList[v]=append(g.adjList[v], u)
}
// IsBipartite checks if the graph is bipartite and returns the partition if possible
func (g *Graph)IsBipartite()(bool,map[int]bool){
color :=make(map[int]bool)// false for one set, true for another
for vertex, _ :=range g.adjList {
if _, visited := color[vertex];!visited {
if!g.isBipartiteUtil(vertex, color){
returnfalse,nil
}
}
}
returntrue, color
}
// isBipartiteUtil is a utility function to check if a component is bipartite
func (g *Graph) isBipartiteUtil(vertex int, color map[int]bool)bool{
queue :=[]int{vertex}
color[vertex]=false// Start coloring with false
forlen(queue)>0{
u := queue[0]
queue = queue[1:]
for _, v :=range g.adjList[u]{
if _, visited := color[v];!visited {
color[v]=!color[u]
queue =append(queue, v)
}elseif color[v]== color[u]{
returnfalse
}
}
}
returntrue
}
func main(){
// Example usage:
// Create a graph with 4 vertices
graph :=NewGraph(4)
// Add edges (competitive relationships)
graph.AddEdge(0,1)
graph.AddEdge(0,3)
graph.AddEdge(1,2)
graph.AddEdge(2,3)
// Check if the graph is bipartite and get the partition
isBipartite, partition := graph.IsBipartite()
if isBipartite {
fmt.Println("The wrestlers can be divided into two groups:")
for wrestler, isHeel :=range partition {
if isHeel {
fmt.Printf("Wrestler %d is a 'High Heel'\n", wrestler)
}else{
fmt.Printf("Wrestler %d is a 'Baby Face'\n", wrestler)
}
}
}else{
fmt.Println("The wrestlers cannot be divided into two groups with the given competitive relationships.")
}
} This algorithm first creates a graph and then uses a breadth-first search (BFS) to check if the graph is bipartite. If it is, it will return a partition where false is for the "babyface" type player and true is for the "heels" type. The time complexity of the algorithm is O(n+r), where n is the number of wrestlers and r is the number of competitive relationships.
Heavenly Craftsmanship:
This problem can be solved by the concept of dichotomous graphs in graph theory. Considering the wrestler as the vertex in the graph and the competitive relationship as an undirected edge, the problem is transformed into judging whether a given undirected graph is a dichotomous graph and giving a possible dichotomous way. A dichotomous graph is defined as a vertex in a graph that can be divided into two disjoint sets such that each edge of the graph connects vertices in two different sets. This problem can be solved by either depth-first search (DFS) or breadth-first search (BFS) with a time complexity of O(n+r), where n is the number of vertices and r is the number of edges.
Here's the algorithm implemented in Go:
package main
import(
"fmt"
)
typeWrestlerint
typeEdgestruct{
fromWrestler
to Wrestler
}
typeGraphstruct{
edges []Edge
nodes map[Wrestler]bool
}
func NewGraph()*Graph{
return&Graph{
edges:make([]Edge,0),
nodes:make(map[Wrestler]bool),
}
}
func (g *Graph)AddEdge(from, to Wrestler){
g.edges =append(g.edges,Edge{from, to})
g.nodes[from]=true
g.nodes[to]=true
}
func (g *Graph)IsBipartite()(bool,map[Wrestler]bool){
visited :=make(map[Wrestler]bool)
color :=make(map[Wrestler]bool)
var dfs func(Wrestler, bool)bool
dfs =func(node Wrestler, c bool)bool{
if visited[node]{
return color[node]== c
}
visited[node]=true
color[node]= c
for _, edge :=range g.edges {
if edge.from== node {
if!dfs(edge.to,!c){
returnfalse
}
}elseif edge.to == node {
if!dfs(edge.from,!c){
returnfalse
}
}
}
returntrue
}
for node :=range g.nodes {
if!visited[node]{
if!dfs(node,true){
returnfalse,nil
}
}
}
returntrue, color
}
func main(){
g :=NewGraph()
g.AddEdge(1,2)
g.AddEdge(2,3)
g.AddEdge(3,4)
g.AddEdge(4,5)
g.AddEdge(5,1)
isBipartite, coloring := g.IsBipartite()
fmt.Println("Is Bipartite?", isBipartite)
if isBipartite {
fmt.Println("Coloring:", coloring)
}
} This Go code implements a graph structure and provides a way to add edges. It then checks whether the graph can be split into two sets by binding, each containing vertices with no inner edges, through a depth-first search. If it can be bisected, it also returns a map that represents the set to which the vertex belongs.
Note that this algorithm assumes that the input graphs are connected. If the graph is not connected, you may need to run the algorithm separately for each connected component, or modify the algorithm to handle the situation.
chatgpt：
The following is an algorithm based on depth-first search, using an adjacency table representation of a graph. In this algorithm, we divide the wrestlers into two types and ensure that the rivalry only exists between the same type.
package main
import(
"fmt"
)
typeWrestlerstruct{
Idint
FaceTypestring
Visitedbool
}
func canDivideIntoTypes(wrestlers []*Wrestler, relationships [][]int)bool{
graph :=make(map[int][]*Wrestler)
for _, r :=range relationships {
w1 := wrestlers[r[0]]
w2 := wrestlers[r[1]]
if graph[w1.Id]==nil{
graph[w1.Id]=[]*Wrestler{}
}
if graph[w2.Id]==nil{
graph[w2.Id]=[]*Wrestler{}
}
graph[w1.Id]=append(graph[w1.Id], w2)
graph[w2.Id]=append(graph[w2.Id], w1)
}
for _, wrestler :=range wrestlers {
if!wrestler.Visited&&!dfs(wrestler,"Face", graph){
returnfalse
}
}
returntrue
}
func dfs(wrestler *Wrestler, faceType string, graph map[int][]*Wrestler)bool{
wrestler.FaceType= faceType
wrestler.Visited=true
for _, opponent :=range graph[wrestler.Id]{
if opponent.FaceType== faceType {
returnfalse// 竞争关系在同一类型之间，返回false
}
if!opponent.Visited&&!dfs(opponent, getOppositeFaceType(faceType), graph){
returnfalse
}
}
returntrue
}
func getOppositeFaceType(faceType string)string{
if faceType =="Face"{
return"Heel"
}else{
return"Face"
}
}
func main(){
n :=4// 摔跤手数量
wrestlers :=make([]*Wrestler, n)
for i :=0; i < n; i++{
wrestlers[i]=&Wrestler{
Id: i,
FaceType:"",
Visited:false,
}
}
r :=[][]int{{0,1},{1,2},{2,3}}// 竞争关系链表
canDivide := canDivideIntoTypes(wrestlers, r)
if canDivide {
fmt.Println("可以将摔跤手划分为“娃娃脸”型和“高跟鞋”型")
for _, wrestler :=range wrestlers {
fmt.Printf("摔跤手%d：%s\n", wrestler.Id, wrestler.FaceType)
}
}else{
fmt.Println("无法进行划分")
}
} The time complexity of this algorithm is O(n+r), where n is the number of wrestlers and r is the number of competing relationships.
some:
This problem can be solved by the two-color graph theory in graph theory. If a graph is bipartite, then it can be divided into two disjoint subsets, such that each edge of the graph is connected to vertices in two different subsets. In the case of wrestlers, we can think of "baby face" and "high heels" as two different subsets.
Here are the algorithmic steps to solve this problem:
- 1. Represent the wrestler as a node in the graph.
- 2. Construct an adjacency table of a diagram based on a given competing linked list.
- 3. Use depth-first search (DFS) or breadth-first search (BFS) to traverse the graph and try to assign a color (baby face or heels) to each node.
- 4. During the traversal, if you encounter a node that has already been assigned a color, check if the new color conflicts with the previously assigned color. If there is a conflict, it means that this division is not possible.
- 5. If all nodes are successfully assigned colors and there are no conflicts, then the division is possible and the current assignment is a solution.
Here's an example of the code in Go:
package main
import(
"fmt"
)
// 定义图的结构
typeGraphstruct{
adjacencyList map[int][]int
}
// 创建图
func NewGraph(n int)*Graph{
return&Graph{
adjacencyList:make(map[int][]int, n),
}
}
// 添加边
func (g *Graph)AddEdge(u, v int){
g.adjacencyList[u]=append(g.adjacencyList[u], v)
g.adjacencyList[v]=append(g.adjacencyList[v], u)// 无向图
}
// 深度优先搜索，尝试为图着色
func (g *Graph) DFS(node, parent, color int, colors []int)bool{
colors[node]= color
for _, neighbor :=range g.adjacencyList[node]{
if colors[neighbor]==0{
if!g.DFS(neighbor, node,3-color, colors){
returnfalse
}
}elseif colors[neighbor]== colors[node]{
returnfalse
}
}
returntrue
}
// 检查是否可以为图着色
func (g *Graph)IsBipartite()([]int,bool){
n :=len(g.adjacencyList)
colors :=make([]int, n)
for node :=range g.adjacencyList {
if colors[node]==0{
if!g.DFS(node,-1,1, colors){
returnnil,false
}
}
}
return colors,true
}
func main(){
// 示例：创建图并添加边
g :=NewGraph(4)
g.AddEdge(1,2)
g.AddEdge(1,3)
g.AddEdge(2,4)
g.AddEdge(3,4)
colors, isBipartite := g.IsBipartite()
if isBipartite {
fmt.Println("Graph is bipartite and the coloring is:", colors)
}else{
fmt.Println("Graph is not bipartite.")
}
} This code defines a graph structure and implements the ability to add edges and check if the graph is a bipartite graph. The IsBipartite function returns an array of colors that represents the type of each wrestler (babyface or heels), and a boolean value that indicates whether or not a division can be made. If the returned boolean value is true, it means that a division can be made, and each element (1 or 2) in the colors array represents the type of wrestler. If false, it means that the partition cannot be made.
Mixed elements:
First of all, we need to analyze the requirements of the question. The question asks us to divide professional wrestlers into "baby-faced" and "heeled" types, so that all competition exists only between these two types of fighters. We can solve this problem using the Depth-First Search (DFS) algorithm.
The solution is as follows:
- 1. Create an array of colors that store the color of each wrestler (1 for "babyface", -1 for "heels", and 0 for undyed).
- 2. Create an array graph that stores the competitive relationships between wrestlers. Based on the entered linked list, each pair of competing wrestlers is connected in the graph.
- 3. Define a depth-first search function, dfs, with the input parameter being the index of the current wrestler.
- • If the competitor is not dyed, call the DFS function recursively to color the competitor the opposite color to the current wrestler. If it returns false, it means that the partition could not be completed and returns false.
- • If a competitor is dyed and the color is the same as the current wrestler, the division cannot be completed and returns false.
- • If the current wrestler has been dyed, return directly whether it is the same color as the target color.
- • Dye your current wrestler as the target color and go through all of its competitors.
- • After iterating through all competitors, return true.
- 4. Iterate through all the wrestlers, and for the wrestlers that are not dyed, call the dfs function to try to divide them. If false is returned, the division cannot be completed and the result is returned.
- 5. After the traversal is complete, check if there are undyed wrestlers in the color array, if they are, it means that the division cannot be completed, and the result is returned.
- 6. Finally, generate the partition result based on the color array.
The time complexity of the algorithm is O(n+r), where n is the number of wrestlers and r is the number of competing relationships.
The code implementation is as follows:
func dividePlayers(n int, edges [][]int) ([]int,[]int){
color :=make([]int, n)
graph :=make([][]int, n)
for _, e :=range edges {
a, b := e[0]-1, e[1]-1
graph[a]=append(graph[a], b)
graph[b]=append(graph[b], a)
}
var dfs func(int, int)bool
dfs =func(u, c int)bool{
color[u]= c
for _, v :=range graph[u]{
if color[v]== c {
returnfalse
}
if color[v]==0{
if!dfs(v,-c){
returnfalse
}
}
}
returntrue
}
ok :=true
for i :=0; i < n && ok; i++{
if color[i]==0{
ok = dfs(i,1)
}
}
if!ok {
returnnil,nil
}
for i :=0; i < n; i++{
if color[i]==0{
returnnil,nil
}
}
ans1, ans2 :=[]int{},[]int{}
for i, c :=range color {
if c ==1{
ans1 =append(ans1, i+1)
}else{
ans2 =append(ans2, i+1)
}
}
return ans1, ans2
}