查找三元组，以便在C ++中连接这些三元组的节点数最大
在本教程中，我们将讨论一个寻找三元组的程序，以使连接这些三元组的节点数量最大。
为此，我们将提供一个具有N个节点的树。我们的任务是找到一个三元组的节点，以使路径中覆盖的节点最大程度地将它们连接起来。
示例
#include <bits/stdc++.h>
#define ll long long int
#define MAX 100005
using namespace std;
vector<int> nearNode[MAX];
bool isTraversed[MAX];
//存储所需的节点
int maxi = -1, N;
int parent[MAX];
bool vis[MAX];
int startnode, endnode, midNode;
//实现DFS搜索节点
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;
}
}
}
//查找直径的尾注
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;
}输出结果
Nodes: (0, 0, 0)