算法训练营|图论第一天 98. 所有可达路径

题目：所有可到达路径

题目链接：

98. 所有可达路径 (kamacoder.com)

解题思路：

邻接矩阵，注意没有result == -1时候特判

#include<bits/stdc++.h>
using namespace std;
vector<vector<int>>result;
vector<int>path;
void dfs(vector<vector<int>>grid, int x, int n) {
	if (x == n) {
		result.push_back(path);
		return;
	}
	for (int i = 1; i <= n; i++) {
		if (grid[x][i] == 1) {
			path.push_back(i);
			dfs(grid, i, n);
			path.pop_back();
		}
	}
}
int main() {
	int n, m;
	cin >> n >> m;
	vector<vector<int>>grid(n + 1, vector<int>(n + 1, 0));
	while (m--) {
		int s, t;
		cin >> s >> t;
		grid[s][t] = 1;
	}
	path.push_back(1);
	dfs(grid, 1, n);
	if (result.size() == 0) cout << -1 << endl;
	for (int i = 0; i < result.size(); i++) {
		for (int j = 0; j < result[i].size() - 1; j++) {
			cout << result[i][j] << ' ';
		}
		cout << result[i][result[i].size() - 1]<<endl;
	}
}

邻接表的写法：

#include<bits/stdc++.h>
using namespace std;
vector<vector<int>>result;
vector<int>path;
void dfs(vector<list<int>>grid, int x, int n) {
	if (x == n) {
		result.push_back(path);
		return;
	}
	for (auto i : grid[x]) {
		path.push_back(i);
		dfs(grid, i, n);
		path.pop_back();
	}
}
int main() {
	int n, m;
	cin >> n >> m;
	vector<list<int>>grid(n + 1);
	while (m--) {
		int s, t;
		cin >> s >> t;
		grid[s].push_back(t);
	}
	path.push_back(1);
	dfs(grid, 1, n);
	if (result.size() == 0) {
		cout << -1 << endl;
	}
	for (auto path : result) {
		for (int i = 0; i < path.size() - 1; i++) {
			cout << path[i] << ' ';
		}
		cout << path[path.size() - 1] << endl;
	}
	return 0;
}