给定一个具有 N 个节点和M 个边的连通无向加权图。任务是执行给定的查询并找到最小生成树的权重。查询分为三种类型：
- query(1) -> 求最小生成树的权重。
- query(2, x, y) -> 将节点 x和y之间的边的权重更改为0 。
- query(3, x, y) -> 将节点 x和y之间的边的权重恢复为其原始权重。
例子：
Input:
query(2, 1, 2),
query(1),
query(3, 1, 2),
query(1)
Output:
2
3
Input:
query(1),
query(2, 3, 4),
query(1)
Output :
4
2
方法：让我们在执行任何查询之前先计算初始图的 MST，并让 T 为这个 MST。关键的观察是，在处理查询的任何时候，当前图的 MST 的权重可以通过在此时权重为零的边和 T 的边上运行 Kruskal 算法来计算。因此，保持权重为零的边在一个数据结构，在查询类型 2 和类型 3 后计算最小生成树的权重。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
#define N 2005
// To store vertices, edges
// and the required answer
int n, m, ans;
// To store parent and rank
int par[N], Rank[N];
// To store edges and the edges in MST
vector > > edges, mst;
// To store the edges with weight zero
queue > zeros;
// Function for initialize
void initialize()
{
for (int i = 0; i <= n; i++) {
par[i] = i;
Rank[i] = 0;
}
}
// Function to add edges
void Add_edge(int u, int v, int weight)
{
edges.push_back({ weight, { u, v } });
}
// Utility function to find set of an element i
// (uses path compression technique)
int find(int x)
{
if (par[x] != x)
par[x] = find(par[x]);
return par[x];
}
// Function that performs union of two sets x and y
// (uses union by rank)
void Union(int x, int y)
{
int xroot = find(x);
int yroot = find(y);
if (Rank[xroot] < Rank[yroot])
par[xroot] = yroot;
else if (Rank[xroot] > Rank[yroot])
par[yroot] = xroot;
else {
par[yroot] = xroot;
Rank[xroot]++;
}
}
// Function to compute minimum spanning tree
void compute_MST()
{
// Sort edges in increasing order of weight
sort(edges.begin(), edges.end());
// Go through all the edges
for (int i = 0; i < m; i++) {
int u = find(edges[i].second.first);
int v = find(edges[i].second.second);
if (u == v)
continue;
// Build minimum spanning tree
// and store minimum cost
mst.push_back(edges[i]);
ans += edges[i].first;
Union(u, v);
}
}
// Function to find the cost of minimum
// spanning tree
void Modified_Kruskal(pair x)
{
initialize();
// Make answer zero
ans = 0;
int sz = zeros.size();
// Keep the edges which only have zero weights
// and remove all the other edges
for (int i = 0; i < sz; i++) {
pair Front = zeros.front();
zeros.pop();
if (Front.first == x.first
and Front.second == x.second)
continue;
// Make union between the vertices of
// edges which have weight zero and keep
// them in queue
Union(Front.first, Front.second);
zeros.push(Front);
}
// Find the cost of the minimum spanning tree
for (int i = 0; i < mst.size(); i++) {
int u = find(mst[i].second.first);
int v = find(mst[i].second.second);
if (u == v)
continue;
ans += mst[i].first;
Union(u, v);
}
}
// Function to handle different queries
void query(int type, int u = 0, int v = 0)
{
// Update edge weight to 0
if (type == 2) {
// push edge in zeros
zeros.push({ u, v });
Modified_Kruskal({ -1, -1 });
}
// Restore edge weight to original value
else if (type == 3) {
// push edge in zeros
zeros.push({ u, v });
Modified_Kruskal({ u, v });
}
else
cout << ans << endl;
}
// Driver code
int main()
{
// Number of nodes and edges
n = 4, m = 4;
initialize();
// Add edges
Add_edge(1, 2, 1);
Add_edge(2, 3, 1);
Add_edge(3, 4, 1);
Add_edge(4, 1, 1);
// Build the minimum spanning tree
compute_MST();
// Execute queries
query(2, 1, 2);
query(1);
query(3, 1, 2);
query(1);
return 0;
} Python3
# Python3 implementation of the approach
from collections import deque
N = 2005
# To store vertices, edges
# and the required answer
n, m, ans = 0, 0, 0
# To store parent and rank
par = [0] * N
Rank = [0] * N
# To store edges and the edges in MST
edges, mst = [], []
# To store the edges with weight zero
zeroes = deque()
# Function for initialize
def initialize():
for i in range(n + 1):
par[i] = i
Rank[i] = 0
# Function to add edges
def add_edge(u: int, v: int, weight: int):
edges.append((weight, (u, v)))
# Utility function to find set of an element i
# (uses path compression technique)
def find(x: int) -> int:
if par[x] != x:
par[x] = find(par[x])
return par[x]
# Function that performs union of two sets x and y
# (uses union by rank)
def union(x: int, y: int):
xroot = find(x)
yroot = find(y)
if Rank[xroot] < Rank[yroot]:
par[xroot] = yroot
elif Rank[xroot] > Rank[yroot]:
par[yroot] = xroot
else:
par[yroot] = xroot
Rank[xroot] += 1
# Function to compute minimum spanning tree
def compute_MST():
global ans
# Sort edges in increasing order of weight
edges.sort()
# Go through all the edges
for i in range(m):
u = find(edges[i][1][0])
v = find(edges[i][1][1])
if u == v:
continue
# Build minimum spanning tree
# and store minimum cost
mst.append(edges[i])
ans += edges[i][0]
union(u, v)
# Function to find the cost of minimum
# spanning tree
def modified_kruskal(x):
global ans
initialize()
# Make answer zero
ans = 0
sz = len(zeroes)
# Keep the edges which only have zero weights
# and remove all the other edges
for i in range(sz):
front = zeroes[0]
zeroes.popleft()
if front[0] == x[0] and front[1] == x[1]:
continue
# Make union between the vertices of
# edges which have weight zero and keep
# them in queue
union(front[0], front[1])
zeroes.append(front)
# Find the cost of the minimum spanning tree
for i in range(len(mst)):
u = find(mst[i][1][0])
v = find(mst[i][1][1])
if u == v:
continue
ans += mst[i][0]
union(u, v)
# Function to handle different queries
def query(type: int, u=0, v=0):
global ans
# Update edge weight to 0
if type == 2:
# push edge in zeros
zeroes.append((u, v))
modified_kruskal((-1, -1))
# Restore edge weight to original value
elif type == 3:
# push edge in zeros
zeroes.append((u, v))
modified_kruskal((u, v))
else:
print(ans)
# Driver Code
if __name__ == "__main__":
# Number of nodes and edges
n = 4
m = 4
initialize()
# Add edges
add_edge(1, 2, 1)
add_edge(2, 3, 1)
add_edge(3, 4, 1)
add_edge(4, 1, 1)
# Build the minimum spanning tree
compute_MST()
# Execute queries
query(2, 1, 2)
query(1)
query(3, 1, 2)
query(1)
# This code is contributed by
# sanjeev2552输出：
2
3时间复杂度：每个查询 O(N)，其中 N 是图中节点的总数。
辅助空间： O(N)