给定具有N个顶点和N-1 条边的树，其中顶点编号从0 到 N-1 ，并且树中存在一个顶点V。假设树中的每个顶点都分配了一种颜色，即白色或黑色，并且顶点的相应颜色由数组arr[] 表示。任务是从包含给定顶点V的给定树的任何可能子树中找到白色顶点数量和黑色顶点数量之间的最大差异。
例子：
Input: V = 0,
arr[] = {'b', 'w', 'w', 'w', 'b',
'b', 'b', 'b', 'w'}
Tree:
0 b
/ \
/ \
1 w 2 w
/ / \
/ / \
5 b w 3 4 b
| | |
| | |
7 b b 6 8 w
Output: 2
Explanation:
We can take the subtree
containing the vertex 0
which contains vertices
0, 1, 2, 3 such that
the difference between
the number of white
and the number of black vertices
is maximum which is equal to 2.
Input:
V = 2,
arr[] = {'b', 'b', 'w', 'b'}
Tree:
0 b
/ | \
/ | \
1 2 3
b w b
Output: 1 方法：思路是利用动态规划的概念来解决这个问题。
- 首先，为颜色数组创建一个向量，对于白色，按 1，对于黑色，按 -1。
- 创建一个数组 dp[] 来计算包含顶点 V 的某个子树中白色和黑色顶点数量之间的最大可能差异。
- 现在，使用深度优先搜索遍历遍历树并更新 dp[] 数组中的值。
- 最后， dp[] 数组中的最小值是所需的答案。
下面是上述方法的实现：
C++
// C++ program to find maximum
// difference between count of
// black and white vertices in
// a path containing vertex V
#include
using namespace std;
// Defining the tree class
class tree {
vector dp;
vector > g;
vector c;
public:
// Constructor
tree(int n)
{
dp = vector(n);
g = vector >(n);
c = vector(n);
}
// Function for adding edges
void addEdge(int u, int v)
{
g[u].push_back(v);
g[v].push_back(u);
}
// Function to perform DFS
// on the given tree
void dfs(int v, int p = -1)
{
dp[v] = c[v];
for (auto i : g[v]) {
if (i == p)
continue;
dfs(i, v);
// Returning calculated maximum
// difference between white
// and black for current vertex
dp[v] += max(0, dp[i]);
}
}
// Function that prints the
// maximum difference between
// white and black vertices
void maximumDifference(int v,
char color[],
int n)
{
for (int i = 0; i < n; i++) {
// Condition for white vertex
if (color[i] == 'w')
c[i] = 1;
// Condition for black vertex
else
c[i] = -1;
}
// Calling dfs function for vertex v
dfs(v);
// Printing maximum difference between
// white and black vertices
cout << dp[v] << "\n";
}
};
// Driver code
int main()
{
tree t(9);
t.addEdge(0, 1);
t.addEdge(0, 2);
t.addEdge(2, 3);
t.addEdge(2, 4);
t.addEdge(1, 5);
t.addEdge(3, 6);
t.addEdge(5, 7);
t.addEdge(4, 8);
int V = 0;
char color[] = { 'b', 'w', 'w',
'w', 'b', 'b',
'b', 'b', 'w' };
t.maximumDifference(V, color, 9);
return 0;
} Java
// Java program to find maximum
// difference between count of
// black and white vertices in
// a path containing vertex V
import java.util.*;
// Defining the
// tree class
class GFG{
static int []dp;
static Vector []g;
static int[] c;
// Constructor
GFG(int n)
{
dp = new int[n];
g = new Vector[n];
for (int i = 0; i < g.length; i++)
g[i] = new Vector();
c = new int[n];
}
// Function for adding edges
void addEdge(int u, int v)
{
g[u].add(v);
g[v].add(u);
}
// Function to perform DFS
// on the given tree
static void dfs(int v, int p)
{
dp[v] = c[v];
for (int i : g[v])
{
if (i == p)
continue;
dfs(i, v);
// Returning calculated maximum
// difference between white
// and black for current vertex
dp[v] += Math.max(0, dp[i]);
}
}
// Function that prints the
// maximum difference between
// white and black vertices
void maximumDifference(int v,
char color[],
int n)
{
for (int i = 0; i < n; i++)
{
// Condition for
// white vertex
if (color[i] == 'w')
c[i] = 1;
// Condition for
// black vertex
else
c[i] = -1;
}
// Calling dfs function
// for vertex v
dfs(v, -1);
// Printing maximum difference
// between white and black vertices
System.out.print(dp[v] + "\n");
}
// Driver code
public static void main(String[] args)
{
GFG t = new GFG(9);
t.addEdge(0, 1);
t.addEdge(0, 2);
t.addEdge(2, 3);
t.addEdge(2, 4);
t.addEdge(1, 5);
t.addEdge(3, 6);
t.addEdge(5, 7);
t.addEdge(4, 8);
int V = 0;
char color[] = {'b', 'w', 'w',
'w', 'b', 'b',
'b', 'b', 'w'};
t.maximumDifference(V, color, 9);
}
}
// This code is contributed by 29AjayKumar Python3
# Python3 program to find maximum
# difference between count of
# black and white vertices in
# a path containing vertex V
# Function for adding edges
def addEdge(g, u, v):
g[u].append(v)
g[v].append(u)
# Function to perform DFS
# on the given tree
def dfs(v, p, dp, c, g):
dp[v] = c[v]
for i in g[v]:
if i == p:
continue
dfs(i, v, dp, c, g)
# Returning calculated maximum
# difference between white
# and black for current vertex
dp[v] += max(0, dp[i])
# Function that prints the
# maximum difference between
# white and black vertices
def maximumDifference(v, color, n, dp, c, g):
for i in range(n):
# Condition for white vertex
if(color[i] == 'w'):
c[i] = 1
# Condition for black vertex
else:
c[i] = -1
# Calling dfs function for vertex v
dfs(v, -1, dp, c, g)
# Printing maximum difference between
# white and black vertices
print(dp[v])
# Driver code
n = 9
g = {}
dp = [0] * n
c = [0] * n
for i in range(0, n + 1):
g[i] = []
addEdge(g, 0, 1)
addEdge(g, 0, 2)
addEdge(g, 2, 2)
addEdge(g, 2, 4)
addEdge(g, 1, 5)
addEdge(g, 3, 6)
addEdge(g, 5, 7)
addEdge(g, 4, 8)
V = 0
color = [ 'b', 'w', 'w',
'w', 'b', 'b',
'b', 'b', 'w' ]
maximumDifference(V, color, 9, dp, c, g)
# This code is contributed by avanitrachhadiya2155C#
// C# program to find maximum
// difference between count of
// black and white vertices in
// a path containing vertex V
using System;
using System.Collections.Generic;
// Defining the
// tree class
class GFG{
static int []dp;
static List []g;
static int[] c;
// Constructor
GFG(int n)
{
dp = new int[n];
g = new List[n];
for (int i = 0; i < g.Length; i++)
g[i] = new List();
c = new int[n];
}
// Function for adding edges
void addEdge(int u, int v)
{
g[u].Add(v);
g[v].Add(u);
}
// Function to perform DFS
// on the given tree
static void dfs(int v, int p)
{
dp[v] = c[v];
foreach (int i in g[v])
{
if (i == p)
continue;
dfs(i, v);
// Returning calculated maximum
// difference between white
// and black for current vertex
dp[v] += Math.Max(0, dp[i]);
}
}
// Function that prints the
// maximum difference between
// white and black vertices
void maximumDifference(int v,
char []color,
int n)
{
for (int i = 0; i < n; i++)
{
// Condition for
// white vertex
if (color[i] == 'w')
c[i] = 1;
// Condition for
// black vertex
else
c[i] = -1;
}
// Calling dfs function
// for vertex v
dfs(v, -1);
// Printing maximum difference
// between white and black vertices
Console.Write(dp[v] + "\n");
}
// Driver code
public static void Main(String[] args)
{
GFG t = new GFG(9);
t.addEdge(0, 1);
t.addEdge(0, 2);
t.addEdge(2, 3);
t.addEdge(2, 4);
t.addEdge(1, 5);
t.addEdge(3, 6);
t.addEdge(5, 7);
t.addEdge(4, 8);
int V = 0;
char []color = {'b', 'w', 'w',
'w', 'b', 'b',
'b', 'b', 'w'};
t.maximumDifference(V, color, 9);
}
}
// This code is contributed by Rajput-Ji Javascript
输出：
2时间复杂度： O(N) ，其中 N 是树中的顶点数。
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