【C++】广度/深度优先算法（bfs+dfs）理解+例题+对比 --- 未消化

例题：
《迷宫问题》
定义一个二维数组：

0 0 1 0 1 //0表示可走，1表示墙
0 1 1 1 0 //只能↑↓←→走，不能斜着走
0 0 0 0 0 
0 1 1 1 0
0 0 0 1 0 //题目保证了输入是一定有解的

求从左上角（0，0）到右下角（4，4）的最短路线。

bfs解题核心逻辑伪代码：

1，将起点推入队列中；
2，将起点标识为已走过；
while（队列非空）{
  3，取队列首节点vt，并从队列中弹出；
  4，探索上面取出得节点的周围是否有没走过的节点vf，如果有将所有能走的vf的parents指向vt，并将vf加入队列
    （如果vf等于终点，说明探索完成，退出循环）。
}
如果队列为空自然跳出，说明无路可达终点。

实际c++实现：

#include<vector>
#include<iostream>
#include<string>
#include<algorithm>
#include<map>
#include<queue>
using namespace std;
struct Node//定义结构体Node
{
    int xx ;//自身处于组内的位置
    int yy;
    bool qiang;//是否是墙
    bool walked;//是否走过
    Node *parents;//指向父节点的指针
};


int main() {
    int id = 0;
    //int xx, yy;
    queue<Node*> bfs;//创建Node指针队列
    vector<vector<Node>> migong;//创建二维迷宫组
    for (int i = 0; i < 5; i++) {
        vector<Node> hang;
        for (int j = 0; j < 5; j++) {
            int ii;
            cin >> ii;
            Node node{ i,j,ii,false};
            hang.push_back(node);
        }

        migong.push_back(hang);
    }
    //输入完毕
    int ax[4] = { -1,1,0,0 };
    int by[4] = { 0,0,1,-1 };
    
    
    bfs.push(&migong[0][0]);//先将起点推进去
    migong[0][0].walked = true;

    Node *vt;//等下指向父节点的指针
    Node *vf;//等下指向父节点引申出的子节点

    while (!bfs.empty()) {
        vt = bfs.front();
        bfs.pop();


        
            if ((*vt).xx >= 1) {//查询左节点是否可以
                vf = &migong[(*vt).xx + ax[0]][(*vt).yy + by[0]];
                if (!(*vf).qiang && !(*vf).walked) {
                    bfs.push(vf);
                    (*vf).walked = true;
                    (*vf).parents = vt;//子节点指向父节点
                    if ((*vf).xx == 4 && (*vf).yy == 4) break;//如果是终点节点，结束寻找，跳出循环。
                }
            }
            if ((*vt).xx <=3  ) {//查询右节点是否可以
                vf = &migong[(*vt).xx + ax[1]][(*vt).yy + by[1]];
                if (!(*vf).qiang && !(*vf).walked) {
                    bfs.push(vf);
                    (*vf).walked = true;
                    (*vf).parents = vt;
                    if ((*vf).xx == 4 && (*vf).yy == 4) break;
                }
            }
            if ((*vt).yy <= 3) {//查询下节点是否可以
                vf = &migong[(*vt).xx + ax[2]][(*vt).yy + by[2]];
                if (!(*vf).qiang && !(*vf).walked) {
                    bfs.push(vf);
                    (*vf).walked = true;
                    (*vf).parents = vt;
                    if ((*vf).xx == 4 && (*vf).yy == 4) break;
                }
            }
            if ((*vt).yy >= 1) {//查询上节点是否可以
                vf = &migong[(*vt).xx + ax[3]][(*vt).yy + by[3]];
                if (!(*vf).qiang && !(*vf).walked) {
                    bfs.push(vf);
                    (*vf).walked = true;
                    (*vf).parents = vt;
                    if ((*vf).xx == 4 && (*vf).yy == 4) break;
                }
            }
        
    }
    

//结束算法，从vf指向的节点开始寻找父节点。
    vector<Node*> fin;
    
    
    while (true) {
        fin.push_back(vf);
        vf = (*vf).parents;
        if ((*vf).xx == 0 && (*vf).yy == 0) {
            fin.push_back(vf);
            break;
        }
        
    }
//输出
    for (int i = fin.size()-1; i >=0;i--) {
        cout << (*fin[i]).xx << "," << (*fin[i]).yy << endl;
    }
    

    return 0;

}


输出示例：
0 0 1 0 1
0 1 1 1 0
0 0 0 0 0
0 1 1 1 0
0 0 0 1 0
0,0
1,0
2,0
2,1
2,2
2,3
2,4
3,4
4,4

dfs解题核心逻辑伪代码：

1，栈初始化
2，获得起点，将起点标识为已走过，将起点入栈
while（栈非空）{
  取栈顶元素vt
  如果vt周围有为走过的节点vf，则：
      将vf改为已走
      vf入栈
  没有能走的节点，vt出栈
}

代码：

#include<iostream>
#include<vector>
#include<list>
#include<algorithm>
#include<queue>
#include<string>
#include<stack>
#include<time.h>
#include <windows.h>
#include<set>

using namespace std;

struct Node
{
    int x;
    int y;
    bool walked;
    int g;
    int f;//f = g+h
    int h;
    Node* parents;
};

int main() {
    
    vector<vector<Node>> migong;//创建二维迷宫组
    for (int i = 0; i < 5; i++) {
        vector<Node> hang;
        for (int j = 0; j < 5; j++) {
            int ii;
            cin >> ii;
            Node node{ i,j,ii };
            hang.push_back(node);
        }

        migong.push_back(hang);
    }
    
    
    /*-----------------------------------dfs----------------------------------------------*/
    vector<vector<Node>> migong2 = migong;

    stack<Node*> f;
    f.push(&migong2[0][0]);
    migong2[0][0].walked = true;
    while (!f.empty()) {
        Node *vt = f.top();
        bool can = true;
        if (vt->x >= 1) {
            Node *vf = &migong2[vt->x - 1][vt->y];
            if (vf->walked == false) {
                vf->parents = vt;
                vf->walked = true;
                if (vf == &migong2[4][4]) {
                    break;
                }
                f.push(vf);
                can = false;
            }
        }
        if (vt->x <=3) {
            Node *vf = &migong2[vt->x + 1][vt->y];
            if (vf->walked == false) {
                vf->parents = vt;
                vf->walked = true;
                if (vf == &migong2[4][4]) {
                    break;
                }
                f.push(vf);
                can = false;
            }
        }
        if (vt->y >= 1) {
            Node *vf = &migong2[vt->x][vt->y - 1];
            if (vf->walked == false) {
                vf->parents = vt;
                vf->walked = true;
                if (vf == &migong2[4][4]) {
                    break;
                }
                f.push(vf);
                can = false;
            }
        }
        if (vt->y <= 3) {
            Node *vf = &migong2[vt->x ][vt->y + 1];
            if (vf->walked == false) {
                vf->parents = vt;
                vf->walked = true;
                if (vf == &migong2[4][4]) {
                    break;
                }
                f.push(vf);
                can = false;
            }
        }

        if (can) {
            f.pop();
        }

    }

    




    vector<Node*> fin2;
    Node*bb = &migong2[4][4];
    while (true) {

        fin2.push_back(aa);
        if (bb == &migong2[0][0]) {
            break;
        }
        bb = bb->parents;

    }

    

    int count2 = 0;
    for (int i = 0; i < 5; i++) {
        for (int j = 0; j < 5; j++) {
            cout << migong2[i][j].walked;
            if (migong2[i][j].walked)count2++;
        }
        cout << endl;
    }
    reverse(fin2.begin(), fin2.end());




    for (int i = 0; i < fin.size(); i++) {
        cout << fin[i]->x <<" "<< fin[i]->y<< endl;
    }

    return 0;
}


输出：
0 1 0 0 0
0 1 1 1 0
0 0 0 0 0
0 1 1 1 0
0 0 0 1 0
11000
11111
11111
11111
00011
0 0
1 0
2 0
2 1
2 2
2 3
2 4
3 4
4 4

两种方法的时间对比以及路径分析：

#include<iostream>
#include<vector>
#include<list>
#include<algorithm>
#include<queue>
#include<string>
#include<stack>
#include<time.h>
#include <windows.h>
#include<set>

using namespace std;

struct Node
{
    int x;
    int y;
    bool walked;
    int g;
    int f;//f = g+h
    int h;
    Node* parents;
};

int main() {
    int qiang = 0;
    vector<vector<Node>> migong;//创建二维迷宫组
    for (int i = 0; i < 5; i++) {
        vector<Node> hang;
        for (int j = 0; j < 5; j++) {
            int ii;
            cin >> ii;
            if (ii) qiang++;
            Node node{ i,j,ii };
            hang.push_back(node);
        }

        migong.push_back(hang);
    }
    
    int a[10002];
    int i = 0;
    double run_time;
    _LARGE_INTEGER time_start;  //开始时间
    _LARGE_INTEGER time_over;   //结束时间
    double dqFreq;      //计时器频率
    LARGE_INTEGER ff;   //计时器频率
    QueryPerformanceFrequency(&ff);
    dqFreq = (double)ff.QuadPart;
    QueryPerformanceCounter(&time_start);

    /*-----------------------------------dfs----------------------------------------------*/
    vector<vector<Node>> migong2 = migong;

    stack<Node*> f;
    f.push(&migong2[0][0]);
    migong2[0][0].walked = true;
    while (!f.empty()) {
        Node *vt = f.top();
        bool can = true;
        if (vt->x >= 1) {
            Node *vf = &migong2[vt->x - 1][vt->y];
            if (vf->walked == false) {
                vf->parents = vt;
                vf->walked = true;
                if (vf == &migong2[4][4]) {
                    break;
                }
                f.push(vf);
                can = false;
            }
        }
        if (vt->x <=3) {
            Node *vf = &migong2[vt->x + 1][vt->y];
            if (vf->walked == false) {
                vf->parents = vt;
                vf->walked = true;
                if (vf == &migong2[4][4]) {
                    break;
                }
                f.push(vf);
                can = false;
            }
        }
        if (vt->y >= 1) {
            Node *vf = &migong2[vt->x][vt->y - 1];
            if (vf->walked == false) {
                vf->parents = vt;
                vf->walked = true;
                if (vf == &migong2[4][4]) {
                    break;
                }
                f.push(vf);
                can = false;
            }
        }
        if (vt->y <= 3) {
            Node *vf = &migong2[vt->x ][vt->y + 1];
            if (vf->walked == false) {
                vf->parents = vt;
                vf->walked = true;
                if (vf == &migong2[4][4]) {
                    break;
                }
                f.push(vf);
                can = false;
            }
        }

        if (can) {
            f.pop();
        }

    }

    QueryPerformanceCounter(&time_over);    //计时结束
    run_time = 1000000 * (time_over.QuadPart - time_start.QuadPart) / dqFreq;
    float time1 = run_time;

    QueryPerformanceFrequency(&ff);
    dqFreq = (double)ff.QuadPart;
    QueryPerformanceCounter(&time_start);

    /*-----------------------------------bfs----------------------------------------------*/
    int ax[4] = { -1,1,0,0 };
    int by[4] = { 0,0,1,-1 };

    queue<Node*> bfs;
    bfs.push(&migong[0][0]);//先将起点推进去
    migong[0][0].walked = true;

    Node *vt;//等下指向父节点的指针
    Node *vf;//等下指向父节点引申出的子节点

    while (!bfs.empty()) {
        vt = bfs.front();
        bfs.pop();



        if ((*vt).x >= 1) {//查询左节点是否可以
            vf = &migong[(*vt).x + ax[0]][(*vt).y + by[0]];
            if (!(*vf).walked && !(*vf).walked) {
                bfs.push(vf);
                (*vf).walked = true;
                (*vf).parents = vt;//子节点指向父节点
                if ((*vf).x == 4 && (*vf).y == 4) break;//如果是终点节点，结束寻找，跳出循环。
            }
        }
        if ((*vt).x <= 3) {//查询右节点是否可以
            vf = &migong[(*vt).x + ax[1]][(*vt).y + by[1]];
            if (!(*vf).walked && !(*vf).walked) {
                bfs.push(vf);
                (*vf).walked = true;
                (*vf).parents = vt;
                if ((*vf).x == 4 && (*vf).y == 4) break;
            }
        }
        if ((*vt).y <= 3) {//查询下节点是否可以
            vf = &migong[(*vt).x + ax[2]][(*vt).y + by[2]];
            if (!(*vf).walked && !(*vf).walked) {
                bfs.push(vf);
                (*vf).walked = true;
                (*vf).parents = vt;
                if ((*vf).x == 4 && (*vf).y == 4) break;
            }
        }
        if ((*vt).y >= 1) {//查询上节点是否可以
            vf = &migong[(*vt).x + ax[3]][(*vt).y + by[3]];
            if (!(*vf).walked && !(*vf).walked) {
                bfs.push(vf);
                (*vf).walked = true;
                (*vf).parents = vt;
                if ((*vf).x == 4 && (*vf).y == 4) break;
            }
        }

    }

    QueryPerformanceCounter(&time_over);    //计时结束
    run_time = 1000000 * (time_over.QuadPart - time_start.QuadPart) / dqFreq;
    float time2 = run_time;
    /*-----------------------------------A*----------------------------------------------*/
    vector<vector<Node>> migong3 = migong;
    set<Node*> openNode;
    set<Node*> closeNode;
    openNode.insert(&migong3[0][0]);

    /*-----------------------------------结束----------------------------------------------*/

    vector<Node*> fin;
    Node*aa = &migong[4][4];
    while (true) {
        
        fin.push_back(aa);
        if (aa == &migong[0][0]) {
            break;
        }
        aa = aa->parents;
        
    }

    vector<Node*> fin2;
    Node*bb = &migong2[4][4];
    while (true) {

        fin2.push_back(bb);
        if (bb == &migong2[0][0]) {
            break;
        }
        bb = bb->parents;

    }

    cout << "bfs运行后矩阵" << endl;

    int count = 0;
    for (int i = 0; i < 5; i++) {
        for (int j = 0; j < 5; j++) {
            cout << migong[i][j].walked;
            if (migong[i][j].walked)count++;
        }
        cout << endl;
    }
    reverse(fin.begin(), fin.end());

    cout << "dfs运行后矩阵" << endl;

    int count2 = 0;
    for (int i = 0; i < 5; i++) {
        for (int j = 0; j < 5; j++) {
            cout << migong2[i][j].walked;
            if (migong2[i][j].walked)count2++;
        }
        cout << endl;
    }
    reverse(fin2.begin(), fin2.end());




    for (int i = 0; i < fin.size(); i++) {
        cout << fin[i]->x <<" "<< fin[i]->y<< endl;
    }

    cout << "Totle Time of dfs : " << time1 << "s" << endl;
    cout << "Totle Time of bfs: " << time2 << "s" << endl;
    cout << "bfs共搜索过的节点数：" << count- qiang << endl;
    cout << "dfs共搜索过的节点数：" << count2- qiang << endl;
    return 0;
    //https://blog.csdn.net/u012878643/article/details/46723375
}

输出示例1：
0 1 0 0 0
0 1 1 1 0
0 0 0 0 0
0 1 1 1 0
0 0 0 1 0
bfs运行后矩阵
11001
11111
11111
11111
11111
dfs运行后矩阵
11000
11111
11111
11111
00011
bfs路径
0 0
1 0
2 0
2 1
2 2
2 3
2 4
3 4
4 4
dfs路径
0 0
1 0
2 0
2 1
2 2
2 3
2 4
3 4
4 4
Totle Time of dfs : 65.5013s
Totle Time of bfs: 67.3427s
bfs共搜索过的节点数：15
dfs共搜索过的节点数：11


输出示例2：
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
bfs运行后矩阵
11111
11111
11111
11111
11111
dfs运行后矩阵
11111
11111
11111
11111
11111
bfs路径
0 0
1 0
2 0
3 0
4 0
4 1
4 2
4 3
4 4
dfs路径
0 0
0 1
0 2
0 3
0 4
1 4
2 4
2 3
2 2
2 1
2 0
3 0
4 0
4 1
4 2
4 3
4 4
Totle Time of dfs : 133.107s
Totle Time of bfs: 131.792s
bfs共搜索过的节点数：25
dfs共搜索过的节点数：25


输出示例3：
0 0 0 0 0
0 1 1 1 0
0 0 0 0 0
0 1 1 1 0
0 0 0 0 0
bfs运行后矩阵
11111
11111
11111
11111
11111
dfs运行后矩阵
11111
11111
11111
11111
11111
bfs路径
0 0
1 0
2 0
3 0
4 0
4 1
4 2
4 3
4 4
dfs路径
0 0
0 1
0 2
0 3
0 4
1 4
2 4
2 3
2 2
2 1
2 0
3 0
4 0
4 1
4 2
4 3
4 4
Totle Time of dfs : 120.217s
Totle Time of bfs: 99.1726s
bfs共搜索过的节点数：19
dfs共搜索过的节点数：19

参考：dfs详解

作者：Buyun0
链接：https://www.jianshu.com/p/1fc63ab1bcc2

posted on 2022-10-04 01:25 bdy 阅读(53) 评论(0) 编辑收藏举报

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