DFS,BFS 练习（深搜，广搜，图，leetcode）

https://leetcode-cn.com/problems/route-between-nodes-lcci/

节点间通路。给定有向图，设计一个算法，找出两个节点之间是否存在一条路径。

示例1:

输入：n = 3, graph = [[0, 1], [0, 2], [1, 2], [1, 2]], start = 0, target = 2
输出：true
示例2:

输入：n = 5, graph = [[0, 1], [0, 2], [0, 4], [0, 4], [0, 1], [1, 3], [1, 4], [1, 3], [2, 3], [3, 4]], start = 0, target = 4
输出 true

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/route-between-nodes-lcci
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DFS 深度优先搜索：（使用dfs函数递归）

class Solution {
public:
    void dfs(vector<vector<int>>& neighbor,vector<int>& visit, int node, int target, int& mark)
    {
        //cout<<"node:"<<node<<endl;
        if(mark==1)
        return ;

        visit[node] = 1;
        for(auto& i:neighbor[node])
        {
            if(i==target)
            {
                mark=1;
                return;
            }
            else
            dfs(neighbor,visit, i, target, mark);
        }
        visit[node] = 0;
        return ;

    }

    bool findWhetherExistsPath(int n, vector<vector<int>>& graph, int start, int target) {
        vector<vector<int>> neighbor(n);
        for(auto& i:graph)
        neighbor[i[0]].push_back(i[1]);

        vector<int> visit(n,0);

        int mark = 0;
    
        dfs(neighbor,visit, start, target, mark);
        

        if(mark==1)
        return true;
        else
        return false;

    }
};

BFS广度优先搜索：（使用队列queue）

class Solution {
public:

    bool findWhetherExistsPath(int n, vector<vector<int>>& graph, int start, int target) {
        vector<vector<int>> neighbor(n);
        for(auto& i:graph)
        neighbor[i[0]].push_back(i[1]);

        vector<int> visit(n,0);

        int mark = 0;
        queue<int> q;
        q.push(start);

        while(!q.empty() && mark==0)
        {
            int node = q.front();
            q.pop();
            for(auto& i:neighbor[node])
            {
                if(i==target)
                {
                    mark=1;
                    break;
                }
                else
                q.push(i);
            }
        }
        
        if(mark==1)
        return true;
        else
        return false;

    }
};