Write ac++ program with the following requirements: 1. Use a 5x5 boolean two-dimensional array to represent the connectivity of the graphs that need to be traversed. 2. Use depth first algorithm to traverse the graph. 3. Use breadth first algorithm to traverse the graph. 4. The starting position of the traversal is chosen reasonably based on the connectivity of the graph.
时间: 2024-02-06 15:04:12 浏览: 79
Sure, here's a C++ program that meets the requirements you specified:
```c++
#include <iostream>
#include <queue>
#include <stack>
using namespace std;
const int N = 5; // Size of the graph
bool graph[N][N] = { // Connectivity of the graph
{0, 1, 1, 0, 0},
{1, 0, 1, 1, 0},
{1, 1, 0, 1, 1},
{0, 1, 1, 0, 1},
{0, 0, 1, 1, 0}
};
void dfs(int start) {
stack<int> s;
bool visited[N] = {false};
s.push(start);
visited[start] = true;
while (!s.empty()) {
int v = s.top();
s.pop();
cout << v << " ";
for (int i = 0; i < N; i++) {
if (graph[v][i] && !visited[i]) {
s.push(i);
visited[i] = true;
}
}
}
cout << endl;
}
void bfs(int start) {
queue<int> q;
bool visited[N] = {false};
q.push(start);
visited[start] = true;
while (!q.empty()) {
int v = q.front();
q.pop();
cout << v << " ";
for (int i = 0; i < N; i++) {
if (graph[v][i] && !visited[i]) {
q.push(i);
visited[i] = true;
}
}
}
cout << endl;
}
int main() {
// Choose a starting position based on the connectivity of the graph
int start = 0; // For simplicity, we start at node 0
for (int i = 0; i < N; i++) {
if (graph[start][i]) {
start = i;
break;
}
}
cout << "Starting at node " << start << endl;
cout << "Depth-first traversal: ";
dfs(start);
cout << "Breadth-first traversal: ";
bfs(start);
return 0;
}
```
This program uses a 5x5 boolean two-dimensional array to represent the connectivity of the graph that needs to be traversed. It first chooses a starting position based on the connectivity of the graph by finding the first node that is connected to the starting node. Then, it uses depth-first and breadth-first algorithms to traverse the graph starting from the chosen node. Finally, it outputs the results of the traversals.
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