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法律研究用 C++ 算法找到三个结点的连接数最多的元组
本教程将讨论一个用来找到三个结点的连接数最多的元组的程序。
为此，我们将获得一棵包含 N 个结点的树。我们的任务是找到这样三个结点，使连接它们的路劲中覆盖的结点数最多。
示例
#include <bits/stdc++.h>
#define ll long long int
#define MAX 100005
using namespace std;
vector<int> nearNode[MAX];
bool isTraversed[MAX];
//storing the required nodes
int maxi = -1, N;
int parent[MAX];
bool vis[MAX];
int startnode, endnode, midNode;
//implementing DFS to search nodes
void performDFS(int u, int count) {
isTraversed[u] = true;
int temp = 0;
for (int i = 0; i < nearNode[u].size(); i++) {
if (!isTraversed[nearNode[u][i]]) {
temp++;
performDFS(nearNode[u][i], count + 1);
}
}
if (temp == 0) {
if (maxi < count) {
maxi = count;
startnode = u;
}
}
}
void performDFS2(int u, int count) {
isTraversed[u] = true;
int temp = 0;
for (int i = 0; i < nearNode[u].size(); i++) {
if (!isTraversed[nearNode[u][i]] && !vis[nearNode[u][i]]) {
temp++;
performDFS2(nearNode[u][i], count + 1);
}
}
if (temp == 0) {
if (maxi < count) {
maxi = count;
midNode = u;
}
}
}
//finding endnote of diameter
void performDFS1(int u, int count) {
isTraversed[u] = true;
int temp = 0;
for (int i = 0; i < nearNode[u].size(); i++) {
if (!isTraversed[nearNode[u][i]]) {
temp++;
parent[nearNode[u][i]] = u;
performDFS1(nearNode[u][i], count + 1);
}
}
if (temp == 0) {
if (maxi < count) {
maxi = count;
endnode = u;
}
}
}
void calcTreeVertices() {
performDFS(1, 0);
for (int i = 0; i <= N; i++)
isTraversed[i] = false;
maxi = -1;
performDFS1(startnode, 0);
for (int i = 0; i <= N; i++)
isTraversed[i] = false;
int x = endnode;
vis[startnode] = true;
while (x != startnode) {
vis[x] = true;
x = parent[x];
}
maxi = -1;
for (int i = 1; i <= N; i++) {
if (vis[i])
performDFS2(i, 0);
}
}
int main() {
N = 4;
nearNode[1].push_back(6);
nearNode[2].push_back(0);
nearNode[1].push_back(7);
nearNode[3].push_back(0);
nearNode[1].push_back(2);
nearNode[4].push_back(0);
calcTreeVertices();
cout << "Nodes: (" << startnode << ", " << endnode << ", " << midNode << ")";
return 0;
}输出
Nodes: (0, 0, 0)
更新于：2020 年 8 月 19 日
64 次浏览
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