迷宫寻路问题全解

1、深度优先搜索(DFS)+回溯

最基本的板子:

void DFS(int x,int y)
{
    if (x,y都与目标点相同)
    {
        得到一个解;
    }
    else
    {
        for (int i = 1; i <= 四个方向; i++)
            if (满足进一步搜索条件)
            {
                为进一步搜索所需要的状态打上标记;
                DFS(to_x, to_y);
                恢复到打标记前的状态;//也就是回溯一步
            }
    }
}

适用类型①:求可行解数量

https://www.luogu.org/problemnew/show/P1605

#include <iostream>
using namespace std;

//上下左右
int direction[4][2] = { {-1,0},{1,0},{0,-1},{0,1} };

int m[10][10];

int N, M, T, cnt;
int SX, SY, EX, EY;

void DFS(int x, int y) {
    if (x < 1 || x > N || y < 1 || y > M) return;
    if (x == EX && y == EY) {
        cnt++;
        return;
    }
    for (int i = 0; i < 4; i++) {
        if (m[x + direction[i][0]][y + direction[i][1]] != 1) {
            m[x + direction[i][0]][y + direction[i][1]] = 1;
            DFS(x + direction[i][0], y + direction[i][1]);
            m[x + direction[i][0]][y + direction[i][1]] = 0;
        }
    }
}
int main() {
    cin >> N >> M >> T;
    cin >> SX >> SY >> EX >> EY;
    for (int i = 0; i < T; i++) {
        int x, y;
        cin >> x >> y;
        m[x][y] = 1;
    }
    m[SX][SY] = 1;
    DFS(SX, SY);
    cout << cnt << endl;
    return 0;
}
View Code

适用类型②:输出所有可行解

例题:https://www.luogu.org/problemnew/show/P1238

这类题目需要注意的是,要知道搜索前进方向的顺序,比如本题是:上左右下。

如果题目够严谨的话,一定会写出来的,但如果没写的话,只能根据题目样例去判断。

#include <iostream>
using namespace std;

//上左右下
int direction[4][2] = { {-1,0},{0,-1},{0,1},{1,0} };

int m[20][20];
int path[250][2];

int N, M, T, cnt;
int SX, SY, EX, EY;

void DFS(int x, int y,int k) {
    if (x < 1 || x > M || y < 1 || y > N) return;
    if (x == EX && y == EY) {
        cnt++;
        for (int i = 0; i < k; i++) {
            cout << "(" << path[i][0] << "," << path[i][1] << ")";
            if (i != k - 1)cout << "->";
        }
        cout << endl;
        return;
    }

    for (int i = 0; i < 4; i++) {
        int tox = x + direction[i][0], toy = y + direction[i][1];
        if (m[tox][toy] == 1) {
            m[tox][toy] = 0;
            path[k][0] = tox;
            path[k][1] = toy;
            DFS(tox, toy, k + 1);
            m[tox][toy] = 1;
        }
    }
}
int main() {
    cin >> M >> N;
    for (int i = 1; i <= M; i++) {
        for (int j = 1; j <= N; j++) {
            cin >> m[i][j];
        }
    }
    cin >> SX >> SY >> EX >> EY;
    m[SX][SY] = 0;
    path[0][0] = SX, path[0][1] = SY;
    DFS(SX, SY, 1);
    if (cnt == 0)cout << "-1" << endl;
    return 0;
}
View Code

注意:

  • 当 m , n 较大时,无法胜任,撑死在15左右就嗝屁了(还得是迷宫中障碍的位置比较配合的情况,一般大于10,就要慎重考虑该不该用DFS了)。
  • 搜索前进方向的顺序是可能会影响到效率的,如果起点在左上部分,终点在右下部分,理想情况下,优先选择右方向和下方向,可以根据起点和终点的位置关系,调整方向数组的顺序,效率会更高。
  • 找到的第一条解,不一定是最短的,应该说一般都不是。

2、广度优先搜索(BFS)

适用类型①:最短路径的长度

题目链接:走迷宫

#include <iostream>
#include <stdio.h>
#include <queue>
using namespace std;

#define PAIR make_pair
//上下左右
int direction[4][2] = { {-1,0},{1,0},{0,-1},{0,1} };

char m[20][20];
int cnt[250];
int head = 0, tail = 1;
int startx = 0, starty = 1, endx = 9, endy = 8;

void BFS(int x, int y) {
    queue<pair<int, int>>q;
    q.push(PAIR(x, y));
    while (!q.empty()) {
        x = q.front().first; y = q.front().second;
        q.pop();
        for (int i = 0; i < 4; i++) {
            int tox = x + direction[i][0], toy = y + direction[i][1];
            if (m[tox][toy] == '.' && (tox >= startx && tox <= endx && toy >= starty && toy <= endy)) {
                if (tox == endx && toy == endy) {
                    cout << cnt[head] + 1 << endl;//获得当前的层数
                    return;
                }
                //cnt[head]记录当前到了第几层BFS
                cnt[tail++] = cnt[head] + 1;
                m[tox][toy] = '#';
                q.push(PAIR(tox, toy));
            }
        }
        head++;
    }
}
int main() {
    while (true) {
        head = 0, tail = 1;
        for (int i = 0; i < 10; i++) {
            for (int j = 0; j < 10; j++) {
                if (scanf("%c", &m[i][j]) == EOF) return 0;
            }
            getchar();
        }
        m[startx][starty] = '#';
        BFS(startx, starty);
    }
    return 0;
}
View Code

适用类型②:找到最短的一条路径

题目链接:http://poj.org/problem?id=3984

注意:

①:BFS搜到的第一条一定是最短的,但是最短的不一定只有一条。题目未说明有唯一解的,要注意题目对解的要求。

②:如果不止一个解,一般题目会给定按照字典序(上下左右用U D L R表示)、优先向某个方向等要求,输出指定解。

③:BFS没有办法想DFS那样直接把路径存下来;只能把每个点的前驱记下来,这样最后到了终点,得到的路径正好是是反过来的。两种选择,1、自行处理倒过来输出。2、BFS直接倒过来搜,即从终点向起点搜,负负得正

#include <iostream>
#include <queue>
using namespace std;

#define PAIR make_pair
//上下左右
int direction[4][2] = { {-1,0},{1,0},{0,-1},{0,1} };

int m[20][20];
int path[15][15][2];

void BFS(int x, int y) {
    queue<pair<int, int>>q;
    q.push(PAIR(x, y));
    while (!q.empty()) {
        x = q.front().first; y = q.front().second;
        q.pop();
        for (int i = 0; i < 4; i++) {
            int tox = x + direction[i][0], toy = y + direction[i][1];
            if (m[tox][toy] == 0 && (tox >= 0 && tox <= 4 && toy >= 0 && toy <= 4)) {

                if (tox == 0 && toy == 0) {
                    cout << "(" << 0 << ", " << 0 << ")" << endl;
                    while (!(x == 4 && y == 4)) {
                        cout << "(" << x << ", " << y << ")" << endl;
                        int from_x = path[x][y][0];
                        int from_y = path[x][y][1];
                        x = from_x; y = from_y;
                    }
                    cout << "(" << 4 << ", " << 4 << ")" << endl;
                    return;
                }

                m[tox][toy] = 1;
                path[tox][toy][0] = x;
                path[tox][toy][1] = y;
                q.push(PAIR(tox, toy));
            }
        }
    }
}
int main() {
    for (int i = 0; i < 5; i++) {
        for (int j = 0; j < 5; j++) {
            cin >> m[i][j];
        }
    }
    m[4][4] = 0;
    BFS(4, 4);
    return 0;
}
View Code

优化:双向BFS

正向BFS,与反向BFS用不同的值取标记地图,第一次相遇时(某一方发现对方的标记值),一定是一条最短的路径。这个时候,path中记录着两段方向相反的路径,输出的时候需要处理。

同样的,如果题目要求是按照某种顺序、优先某个方向;那么反向的BFS只要反着来就行了。

#include <iostream>
#include <stack>
#include <queue>
using namespace std;

#define PAIR make_pair

int direction[4][2] = { {-1,0},{1,0},{0,-1},{0,1} };

int m[20][20];
int path[15][15][2];
bool flag = false;
int SX = 0, SY = 0, EX = 4, EY = 4;
void move_one_step(queue<pair<int,int>>&q,int sign) {
    int x = q.front().first, y = q.front().second; q.pop();
    for (int i = 0; i < 4; i++) {
        int tox = x + direction[i][0], toy = y + direction[i][1];
        if (m[tox][toy] == 0 && (tox >= SX && tox <= EX && toy >= SX && toy <= EY)) {
            m[tox][toy] = sign;
            path[tox][toy][0] = x;
            path[tox][toy][1] = y;
            q.push(PAIR(tox, toy));
        }
        //发现对方标记值
        else if (m[tox][toy] == -sign) {
            int tempx = tox, tempy = toy;
            /*输出 起点 到相遇点*/
            stack<pair<int, int>>s;
            while (!(tox == 0 && toy == 0)) {
                int t1 = tempx, t2 = tempy;
                s.push(PAIR(tempx, tempy));
                tempx = path[t1][t2][0];
                tempy = path[t1][t2][1];
            }
            while (!s.empty()) {
                int xx = s.top().first, yy = s.top().second;
                s.pop();
                cout << "(" << xx << ", " << yy << ")" << endl;
            }
            /*----------------*/
            /*从相遇点到终点*/
            tempx = x, tempy = y;
            while (!(tempx == 4 && tempy == 4)) {
                int t1 = tempx, t2 = tempy;
                cout << "(" << tempx << ", " << tempy << ")" << endl;;
                tempx = path[t1][t2][0];
                tempy = path[t1][t2][1];
            }
            flag = true;
            return;
        }
    }
}

void BFS(queue<pair<int, int>>&f, queue<pair<int, int>>&r) {
    int i = 1;
    while ((!f.empty() || !r.empty()) && !flag) {
        if (i & 1)
            move_one_step(f, 2);
        else
            move_one_step(r, -2);
        i++;
    }
}
int main() {
    queue<pair<int, int>>f, r;

    for (int i = 0; i < 5; i++) {
        for (int j = 0; j < 5; j++) {
            cin >> m[i][j];
        }
    }
    f.push(PAIR(SX, SY));
    r.push(PAIR(EX, EY));
    m[EX][EY] = -2;
    m[SX][SY] = 2;
    cout << "(" << SX << ", " << SY << ")" << endl;
    BFS(f, r);
    cout << "(" << EX << ", " << EY << ")" << endl;
    return 0;
}
View Code

 3、A*搜索

A*找到的第一个解不一定是最短的,所以A*不能用来去找最短路径(在有多解的情况下);

A*的搜索特性限制,如果用来输出所有可行解,就没有使用的意义了。

所以其实迷宫题并不适合用A*解,除非题目要求比较特殊(只有一条路),仅做参考。

采取

适用类型:找唯一的路径

#include <iostream>
#include <queue>
using namespace std;

#define PAIR make_pair
//上下左右
int direction[4][2] = { {-1,0},{1,0},{0,-1},{0,1} };
int sx = 0, sy = 0, ex = 4, ey = 4;
class Node {
public:
    int x, y, g, h;
    Node(int x, int y,int g){
        this->x = x;
        this->y = y;
        this->g = g;
        h = abs(ex - x) + abs(ey - y);
    }
    bool operator<(Node n)const {
        return n.g + n.h < g + h;
    }
};

int m[20][20];
int path[15][15][2];
int head = 0, tail = 1;
int cnt[255];

void BFS(int x, int y) {
    priority_queue<Node>q;
    //queue<pair<int, int>>q;
    q.push(Node(x, y, cnt[head]));

    while (!q.empty()) {
        x = q.top().x; y = q.top().y;
        q.pop();
        if (x == 0 && y == 0) {
            while (true) {
                cout << "(" << x << ", " << y << ")" << endl;
                if (x == 4 && y == 4)break;
                int from_x = path[x][y][0];
                int from_y = path[x][y][1];
                x = from_x; y = from_y;
            }
            return;
        }
        for (int i = 0; i < 4; i++) {
            int tox = x + direction[i][0], toy = y + direction[i][1];
            if (m[tox][toy] == 0 && (tox >= 0 && tox <= 4 && toy >= 0 && toy <= 4)) {
                cnt[tail++] = cnt[head] + 1;
                m[tox][toy] = 1;
                path[tox][toy][0] = x;
                path[tox][toy][1] = y;
                q.push(Node(tox, toy, cnt[head] + 1));
            }
        }
        head++;
    }
}
int main() {
    for (int i = 0; i < 5; i++) {
        for (int j = 0; j < 5; j++) {
            cin >> m[i][j];
        }
    }
    m[4][4] = 0;
    BFS(4, 4);
    return 0;
}
View Code

 

posted @ 2019-04-01 23:20  czc1999  阅读(1094)  评论(0编辑  收藏  举报