给定使用N-1条道路连接的N个城市。在城市[i, i+1] 之间，从 1 到 N-1 的所有i都存在一条边。
任务是建立供水连接。在一个城市设置供水系统，然后通过公路运输将水从该城市输送到其他城市。某些城市被封锁，这意味着水无法通过该特定城市。确定可以供水的最大城市数量。
输入格式：
- 第一行包含一个整数 >strong>N 表示城市的数量。
- 接下来的 N-1 行包含两个空格分隔的整数uv表示之间的道路
城市 u 和 v。 - 下一行包含 N 个空格分隔的整数，如果第i 个城市是 1
阻塞，否则为0。
例子：
Input :
4
1 2
2 3
3 4
0 1 1 0
Output :
2
Explanation : If city 1 is chosen, then water is supplied from
city 1 to 2. If city 4 is chosen, water is supplied from city 4 to 3
hence maximum of 2 cities can be supplied with water.
Input :
7
1 2
2 3
3 4
4 5
5 6
6 7
0 1 1 0 0 0 0
Output :
5
Explanation : If city 1 is chosen than water is supplied from
city 1 to 2 or if city 4 is chosen water is supplied from city 4 to
3, 5, 6 and 7 hence maximum of 5 cities are supplied with water.
方法：
在这篇文章中，讨论了基于 BFS 的解决方案。
我们对每个城市进行广度优先搜索并检查两件事：城市没有被封锁，城市没有被访问。如果这两个条件都为真，那么我们从该城市运行广度优先搜索并计算可以供水的城市数量。
该解决方案也可以使用深度优先搜索来实现。
下面是上述方法的实现：
C++
// C++ program to solve water
// supply problem using BFS
#include
#include
#include
using namespace std;
// Function to perform BFS
int bfsUtil(int v[], bool vis[], vector adj[],
int src)
{
// Mark current source visited
vis[src] = true;
queue q; //Queue for BFS
q.push(src); // Push src to queue
int count = 0;
while (!q.empty()) {
int p = q.front();
for (int i = 0; i < adj[p].size(); i++) {
// When the adjacent city not visited and
// not blocked, push city in the queue.
if (!vis[adj[p][i]] && v[adj[p][i]] == 0) {
count++;
vis[adj[p][i]] = true;
q.push(adj[p][i]);
}
// when the adjacent city is not visited
// but blocked so the blocked city is
// not pushed in queue
else if (!vis[adj[p][i]] && v[adj[p][i]] == 1) {
count++;
}
}
q.pop();
}
return count + 1;
}
// Utility function to perform BFS
int bfs(int N, int v[], vector adj[])
{
bool vis[N + 1];
int max = 1, res;
// marking visited array false
for (int i = 1; i <= N; i++)
vis[i] = false;
// Check for each and every city
for (int i = 1; i <= N; i++) {
// Checks that city is not blocked
// and not visited.
if (v[i] == 0 && !vis[i]) {
res = bfsUtil(v, vis, adj, i);
if (res > max) {
max = res;
}
}
}
return max;
}
// Driver Code
int main()
{
int N = 4; // Denotes the number of cities
vector adj[N + 1];
int v[N + 1];
// Adjacency list denoting road
// between city u and v
adj[1].push_back(2);
adj[2].push_back(1);
adj[2].push_back(3);
adj[3].push_back(2);
adj[3].push_back(4);
adj[4].push_back(3);
// array for storing whether ith
// city is blocked or not
v[1] = 0;
v[2] = 1;
v[3] = 1;
v[4] = 0;
cout< Java
// Java program to solve water
// supply problem using BFS
import java.util.*;
class GFG{
// Function to perform BFS
static int bfsUtil(int v[], boolean vis[],
Vector adj[],
int src)
{
// Mark current source visited
vis[src] = true;
// Queue for BFS
Queue q = new LinkedList<>();
// Push src to queue
q.add(src);
int count = 0;
while (!q.isEmpty())
{
int p = q.peek();
for(int i = 0; i < adj[p].size(); i++)
{
// When the adjacent city not
// visited and not blocked, push
// city in the queue.
if (!vis[adj[p].get(i)] &&
v[adj[p].get(i)] == 0)
{
count++;
vis[adj[p].get(i)] = true;
q.add(adj[p].get(i));
}
// When the adjacent city is not visited
// but blocked so the blocked city is
// not pushed in queue
else if (!vis[adj[p].get(i)] &&
v[adj[p].get(i)] == 1)
{
count++;
}
}
q.remove();
}
return count + 1;
}
// Utility function to perform BFS
static int bfs(int N, int v[],
Vector adj[])
{
boolean []vis = new boolean[N + 1];
int max = 1, res;
// Marking visited array false
for(int i = 1; i <= N; i++)
vis[i] = false;
// Check for each and every city
for(int i = 1; i <= N; i++)
{
// Checks that city is not blocked
// and not visited.
if (v[i] == 0 && !vis[i])
{
res = bfsUtil(v, vis, adj, i);
if (res > max)
{
max = res;
}
}
}
return max;
}
// Driver Code
public static void main(String[] args)
{
// Denotes the number of cities
int N = 4;
@SuppressWarnings("unchecked")
Vector []adj = new Vector[N + 1];
for (int i = 0; i < adj.length; i++)
adj[i] = new Vector();
int []v = new int[N + 1];
// Adjacency list denoting road
// between city u and v
adj[1].add(2);
adj[2].add(1);
adj[2].add(3);
adj[3].add(2);
adj[3].add(4);
adj[4].add(3);
// Array for storing whether ith
// city is blocked or not
v[1] = 0;
v[2] = 1;
v[3] = 1;
v[4] = 0;
System.out.print(bfs(N, v, adj));
}
}
// This code is contributed by Princi Singh Python3
# Python3 program to solve water
# supply problem using BFS
# Function to perform BFS
def bfsUtil(v, vis, adj, src):
# Mark current source visited
vis[src] = True
# Queue for BFS
q = []
# Push src to queue
q.append(src)
count = 0
while (len(q) != 0):
p = q[0]
for i in range(len(adj[p])):
# When the adjacent city not visited and
# not blocked, push city in the queue.
if (vis[adj[p][i]] == False and v[adj[p][i]] == 0):
count += 1
vis[adj[p][i]] = True
q.push(adj[p][i])
# when the adjacent city is not visited
# but blocked so the blocked city is
# not pushed in queue
elif(vis[adj[p][i]] == False and v[adj[p][i]] == 1):
count += 1
q.remove(q[0])
return count + 1
# Utility function to perform BFS
def bfs(N, v, adj):
vis = [ 0 for i in range(N + 1)]
mx = 1
# marking visited array false
for i in range(1, N + 1, 1):
vis[i] = False
# Check for each and every city
for i in range(1, N + 1, 1):
# Checks that city is not blocked
# and not visited.
if (v[i] == 0 and vis[i] == False):
res = bfsUtil(v, vis, adj, i)
if (res > mx):
mx = res
return mx
# Driver Code
if __name__ == '__main__':
N = 4
# Denotes the number of cities
adj = [[] for i in range(N + 1)]
v = [0 for i in range(N + 1)]
# Adjacency list denoting road
# between city u and v
adj[1].append(2)
adj[2].append(1)
adj[2].append(3)
adj[3].append(2)
adj[3].append(4)
adj[4].append(3)
# array for storing whether ith
# city is blocked or not
v[1] = 0
v[2] = 1
v[3] = 1
v[4] = 0
print(bfs(N, v, adj))
# This code is contributed by Bhupendra_SinghC#
// C# program to solve water
// supply problem using BFS
using System;
using System.Collections.Generic;
class GFG{
// Function to perform BFS
static int bfsUtil(int []v, bool []vis,
List []adj,
int src)
{
// Mark current source visited
vis[src] = true;
// Queue for BFS
Queue q = new Queue();
// Push src to queue
q.Enqueue(src);
int count = 0;
while (q.Count != 0)
{
int p = q.Peek();
for(int i = 0; i < adj[p].Count; i++)
{
// When the adjacent city not
// visited and not blocked, push
// city in the queue.
if (!vis[adj[p][i]] &&
v[adj[p][i]] == 0)
{
count++;
vis[adj[p][i]] = true;
q.Enqueue(adj[p][i]);
}
// When the adjacent city is not visited
// but blocked so the blocked city is
// not pushed in queue
else if (!vis[adj[p][i]] &&
v[adj[p][i]] == 1)
{
count++;
}
}
q.Dequeue();
}
return count + 1;
}
// Utility function to perform BFS
static int bfs(int N, int []v,
List []adj)
{
bool []vis = new bool[N + 1];
int max = 1, res;
// Marking visited array false
for(int i = 1; i <= N; i++)
vis[i] = false;
// Check for each and every city
for(int i = 1; i <= N; i++)
{
// Checks that city is not blocked
// and not visited.
if (v[i] == 0 && !vis[i])
{
res = bfsUtil(v, vis, adj, i);
if (res > max)
{
max = res;
}
}
}
return max;
}
// Driver Code
public static void Main(String[] args)
{
// Denotes the number of cities
int N = 4;
List []adj = new List[N + 1];
for (int i = 0; i < adj.Length; i++)
adj[i] = new List();
int []v = new int[N + 1];
// Adjacency list denoting road
// between city u and v
adj[1].Add(2);
adj[2].Add(1);
adj[2].Add(3);
adj[3].Add(2);
adj[3].Add(4);
adj[4].Add(3);
// Array for storing whether ith
// city is blocked or not
v[1] = 0;
v[2] = 1;
v[3] = 1;
v[4] = 0;
Console.Write(bfs(N, v, adj));
}
}
// This code is contributed by Princi Singh 输出：
2
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