给定一个具有N个顶点和E 个边的图。边由U[]和V[] 给出，使得对于每个索引 i， U[i]连接到V[i] 。任务是找到为给定图形着色所需的最少颜色数。
例子
Input: N = 5, M = 6, U[] = { 1, 2, 3, 1, 2, 3 }, V[] = { 3, 3, 4, 4, 5, 5 };
Output: 3
Explanation:
For the above graph node 1, 3, and 5 cannot have the same color. Hence the count is 3.
方法：
我们将保留两个数组count[]和colors[] 。数组count[]将存储每个节点的边数， colors[]将存储每个节点的颜色。将每个顶点的计数初始化为 0，将每个顶点的颜色初始化为 1。
脚步：
- 为给定的边集创建一个邻接表，并保留每个节点的边数
- 遍历所有节点并将该节点插入到没有边的队列Q中。
- 当Q不为空时，请执行以下操作：
- 从队列中弹出元素。
- 对于连接到弹出节点的所有节点：
- 减少弹出节点的当前边数。
- 如果当前的颜色小于或等于其父节点的颜色(弹出的节点)，则更新当前节点的颜色 = 1 + 弹出节点的颜色
- 如果 count 为零，则将此节点插入队列Q 中。
- color []数组中的最大元素将给出为给定图形着色所需的最小颜色数。
下面是上述方法的实现：
C++
// C++ program to find the minimum
// number of colors needed to color
// the graph
#include
using namespace std;
// Function to count the minimum
// number of color required
void minimumColors(int N, int E,
int U[], int V[])
{
// Create array of vectors
// to make adjacency list
vector adj[N];
// Intialise colors array to 1
// and count array to 0
vector count(N, 0);
vector colors(N, 1);
// Create adjacency list of
// a graph
for (int i = 0; i < N; i++) {
adj[V[i] - 1].push_back(U[i] - 1);
count[U[i] - 1]++;
}
// Declare queue Q
queue Q;
// Traverse count[] and insert
// in Q if count[i] = 0;
for (int i = 0; i < N; i++) {
if (count[i] == 0) {
Q.push(i);
}
}
// Traverse queue and update
// the count of colors
// adjacent node
while (!Q.empty()) {
int u = Q.front();
Q.pop();
// Traverse node u
for (auto x : adj[u]) {
count[x]--;
// If count[x] = 0
// insert in Q
if (count[x] == 0) {
Q.push(x);
}
// If colors of child
// node is less than
// parent node, update
// the count by 1
if (colors[x] <= colors[u]) {
colors[x] = 1 + colors[u];
}
}
}
// Stores the minimumColors
// requires to color the graph.
int minColor = -1;
// Find the maximum of colors[]
for (int i = 0; i < N; i++) {
minColor = max(minColor, colors[i]);
}
// Print the minimum no. of
// colors required.
cout << minColor << endl;
}
// Driver function
int main()
{
int N = 5, E = 6;
int U[] = { 1, 2, 3, 1, 2, 3 };
int V[] = { 3, 3, 4, 4, 5, 5 };
minimumColors(N, E, U, V);
return 0;
} Java
// Java program to find the minimum
// number of colors needed to color
// the graph
import java.util.*;
class GFG
{
// Function to count the minimum
// number of color required
static void minimumColors(int N, int E,
int U[], int V[])
{
// Create array of vectors
// to make adjacency list
Vector []adj = new Vector[N];
// Intialise colors array to 1
// and count array to 0
int []count = new int[N];
int []colors = new int[N];
for (int i = 0; i < N; i++) {
adj[i] = new Vector();
colors[i] = 1;
}
// Create adjacency list of
// a graph
for (int i = 0; i < N; i++) {
adj[V[i] - 1].add(U[i] - 1);
count[U[i] - 1]++;
}
// Declare queue Q
Queue Q = new LinkedList<>();
// Traverse count[] and insert
// in Q if count[i] = 0;
for (int i = 0; i < N; i++) {
if (count[i] == 0) {
Q.add(i);
}
}
// Traverse queue and update
// the count of colors
// adjacent node
while (!Q.isEmpty()) {
int u = Q.peek();
Q.remove();
// Traverse node u
for (int x : adj[u]) {
count[x]--;
// If count[x] = 0
// insert in Q
if (count[x] == 0) {
Q.add(x);
}
// If colors of child
// node is less than
// parent node, update
// the count by 1
if (colors[x] <= colors[u]) {
colors[x] = 1 + colors[u];
}
}
}
// Stores the minimumColors
// requires to color the graph.
int minColor = -1;
// Find the maximum of colors[]
for (int i = 0; i < N; i++) {
minColor = Math.max(minColor, colors[i]);
}
// Print the minimum no. of
// colors required.
System.out.print(minColor +"\n");
}
// Driver function
public static void main(String[] args)
{
int N = 5, E = 6;
int U[] = { 1, 2, 3, 1, 2, 3 };
int V[] = { 3, 3, 4, 4, 5, 5 };
minimumColors(N, E, U, V);
}
}
// This code is contributed by 29AjayKumar Python3
# Python3 program to find the minimum
# number of colors needed to color
# the graph
from collections import deque
# Function to count the minimum
# number of color required
def minimumColors(N, E, U, V):
# Create array of vectors
# to make adjacency list
adj = [[] for i in range(N)]
# Intialise colors array to 1
# and count array to 0
count = [0]*N
colors = [1]*(N)
# Create adjacency list of
# a graph
for i in range(N):
adj[V[i] - 1].append(U[i] - 1)
count[U[i] - 1] += 1
# Declare queue Q
Q = deque()
# Traverse count[] and insert
# in Q if count[i] = 0
for i in range(N):
if (count[i] == 0):
Q.append(i)
# Traverse queue and update
# the count of colors
# adjacent node
while len(Q) > 0:
u = Q.popleft()
# Traverse node u
for x in adj[u]:
count[x] -= 1
# If count[x] = 0
# insert in Q
if (count[x] == 0):
Q.append(x)
# If colors of child
# node is less than
# parent node, update
# the count by 1
if (colors[x] <= colors[u]):
colors[x] = 1 + colors[u]
# Stores the minimumColors
# requires to color the graph.
minColor = -1
# Find the maximum of colors[]
for i in range(N):
minColor = max(minColor, colors[i])
# Print the minimum no. of
# colors required.
print(minColor)
# Driver function
N = 5
E = 6
U = [1, 2, 3, 1, 2, 3]
V = [3, 3, 4, 4, 5, 5]
minimumColors(N, E, U, V)
# This code is contributed by mohit kumar 29C#
// C# program to find the minimum
// number of colors needed to color
// the graph
using System;
using System.Collections.Generic;
class GFG
{
// Function to count the minimum
// number of color required
static void minimumColors(int N, int E,
int []U, int []V)
{
// Create array of vectors
// to make adjacency list
List []adj = new List[N];
// Intialise colors array to 1
// and count array to 0
int []count = new int[N];
int []colors = new int[N];
for (int i = 0; i < N; i++) {
adj[i] = new List();
colors[i] = 1;
}
// Create adjacency list of
// a graph
for (int i = 0; i < N; i++) {
adj[V[i] - 1].Add(U[i] - 1);
count[U[i] - 1]++;
}
// Declare queue Q
List Q = new List();
// Traverse []count and insert
// in Q if count[i] = 0;
for (int i = 0; i < N; i++) {
if (count[i] == 0) {
Q.Add(i);
}
}
// Traverse queue and update
// the count of colors
// adjacent node
while (Q.Count!=0) {
int u = Q[0];
Q.RemoveAt(0);
// Traverse node u
foreach (int x in adj[u]) {
count[x]--;
// If count[x] = 0
// insert in Q
if (count[x] == 0) {
Q.Add(x);
}
// If colors of child
// node is less than
// parent node, update
// the count by 1
if (colors[x] <= colors[u]) {
colors[x] = 1 + colors[u];
}
}
}
// Stores the minimumColors
// requires to color the graph.
int minColor = -1;
// Find the maximum of colors[]
for (int i = 0; i < N; i++) {
minColor = Math.Max(minColor, colors[i]);
}
// Print the minimum no. of
// colors required.
Console.Write(minColor +"\n");
}
// Driver function
public static void Main(String[] args)
{
int N = 5, E = 6;
int []U = { 1, 2, 3, 1, 2, 3 };
int []V = { 3, 3, 4, 4, 5, 5 };
minimumColors(N, E, U, V);
}
}
// This code is contributed by 29AjayKumar 输出：
3
时间复杂度： O(N+E)，其中 N = 顶点数，E = 边数
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