【作业】用栈模拟dfs

题意：一个迷宫，起点到终点的路径，不用递归。

题解：

#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<stdlib.h>
#include<string>
#include<string.h>
#include<stack>
#include<iostream>
#include<map>
using namespace std;
const int maxn = 1e5 + ;
int dir[][] = { ,,,,-,,,- };
struct node {
    int x, y;
    node(int x=, int y=) :x(x), y(y) {}
    bool operator < (const node &q) const { return  x < q.x; }
    bool operator ==(const node &a)const { return x == a.x&&y == a.y; }
};
struct prob {
    int ord;
    node seat;
    int di;
    prob(int x , node y, int z ) :ord(x), seat(y), di(z) {}
};
int mp[][] = {
    { ,,,,,,,,, },
    { ,,,,,,,,, },
    { ,,,,,,,,, },
    { ,,,,,,,,, },
    { ,,,,,,,,, },
    { ,,,,,,,,, },
    { ,,,,,,,,, },
    { ,,,,,,,,, },
    { ,,,,,,,,, },
    { ,,,,,,,,, },
};

stack<prob> S, road;
int vis[][];
map<node, node>p;
int n, m;
bool ok(int x, int y) {
    if (mp[x][y] == || x <  || x >= m || y <  || y >= n || vis[x][y])return ;
    else return ;
}
node nextpos(node n,int i) {
    return node(n.x + dir[i][], n.y + dir[i][]);
}
int main() {

    cin >> n >> m;

    int sr, sc; cin >> sr >> sc;
    int er, ec; cin >> er >> ec;
    node end = node(er, ec);
    node start = node(sr, sc);
    //S.push(0,node(sr, sc),0);
    node now=start;
    int nows=;
    prob e= prob(nows, now, );
    do {
        if (ok(now.x, now.y)) {
            vis[now.x][now.y] = ;
             e = prob(nows, now, );
            S.push(e);
            if (now== end)break;
            now = nextpos(now, );
            nows++;
        }
        else {
            if (!S.empty()) {
                e = S.top();
                S.pop();
                while (e.di ==  && !S.empty()) {
                    vis[e.seat.x][e.seat.y] = ;
                    e = S.top();
                    S.pop();
                }
                if (e.di < ) {
                    e.di++; S.push(e);
                    now = nextpos(now, e.di);
                }///

            }
        }
    } while (!S.empty());

    stack<prob>ans;
    while (!(S.empty()) ){
        ans.push(S.top());
        S.pop();
    }
    while (!(ans.empty())) {
        cout << ans.top().seat.x << ' ' << ans.top().seat.y << endl;
        ans.pop();
    }
    cin >> n;
}

附：之前模仿bfs写的，不知道怎么存路径。。

#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<stdlib.h>
#include<string>
#include<string.h>
#include<stack>
#include<iostream>
#include<map>
using namespace std;
const int maxn = 1e5 + ;
int dir[][] = { ,,,,-,,,- };
struct node {
    int x, y;
    node(int x=, int y=) :x(x), y(y) {}
    bool operator < (const node &q) const { return  x < q.x; }
};
int mp[][] = {
    { ,,,,,,,,, },
    { ,,,,,,,,, },
    { ,,,,,,,,, },
    { ,,,,,,,,, },
    { ,,,,,,,,, },
    { ,,,,,,,,, },
    { ,,,,,,,,, },
    { ,,,,,,,,, },
    { ,,,,,,,,, },
    { ,,,,,,,,, },
};

stack<node> S, road;
int vis[][];
map<node, node>p;
int n, m;
bool illeg(int x, int y) {
    if (mp[x][y] == '' || x <  || x >= m || y <  || y >= n || vis[x][y])return ;
    else return ;
}

int main() {

    cin >> n >> m;

    int sr, sc; cin >> sr >> sc;
    int er, ec; cin >> er >> ec;
    S.push(node(sr, sc));
    while (!S.empty()) {
        node now = S.top(); S.pop();
        road.push(now);
        //if (mp[now.x][now.y] == '1' || now.x < 0 || now.x >= m || now.y < 0 || now.y >= n || vis[now.x][now.y])continue;
        //S.push(now);
        if (now.x == er&&now.y == ec) break;
        for(int i=;i<;i++){
                int dx = now.x + dir[i][]; int dy = now.y + dir[i][];
                if(illeg(dx,dy))continue;
                if (vis[dx][dy])continue;
                S.push(node(dx, dy));
                node x = node(dx, dy);
                p[x] = now;
                vis[dx][dy] = ;
            }
        /*for (int i = 0; i < m; i++){
            for (int j = 0; j < n; j++) {
                cout << vis[i][j];
            }
        cout<<endl;
        }
        cout << endl;*/
    }
    node now=node(er,ec);
    node x = node(sr, sc);
    while (!(now.x==sr&&now.y==sc) ){
        cout << now.x << ' ' << now.y << endl;
        now = p[now];
    }
    cin >> n;
}