迷宫问题 (最短路径保存输出）

定义一个二维数组： 

int maze[5][5] = {

0, 1, 0, 0, 0,

0, 1, 0, 1, 0,

0, 0, 0, 0, 0,

0, 1, 1, 1, 0,

0, 0, 0, 1, 0,

};

它表示一个迷宫，其中的1表示墙壁，0表示可以走的路，只能横着走或竖着走，不能斜着走，要求编程序找出从左上角到右下角的最短路线。

 

Input一个5 × 5的二维数组，表示一个迷宫。数据保证有唯一解。Output左上角到右下角的最短路径，格式如样例所示。Sample Input

0 1 0 0 0
0 1 0 1 0
0 0 0 0 0
0 1 1 1 0
0 0 0 1 0

Sample Output

(0, 0)
(1, 0)
(2, 0)
(2, 1)
(2, 2)
(2, 3)
(2, 4)
(3, 4)
(4, 4)

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <string>
#include <cstring>
#include <cstdlib>
#include <map>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
#include <deque>
using namespace std;
#define ull unsigned long long
#define lli long long
#define pq priority_queue<int>
#define pql priority_queue<ll>
#define pqn priority_queue<node>
#define v vector<int>
#define vl vector<ll>
#define read(x) scanf("%d",&x)
#define lread(x) scanf("%lld",&x);
#define pt(x) printf("%d\n",(x))
#define YES printf("YES\n");
#define NO printf("NO\n");
#define gcd __gcd
#define out(x) cout<<x<<endl;
#define over cout<<endl;
#define rep(j,k) for (int i = (int)(j); i <= (int)(k); i++)
#define input(k,m) for (int i = 1; i <= (int)(k); i++)  for(int j=1;j<=m;j++) {scanf("%d",&a[i][j]) ; }
#define mem(s,t) memset(s,t,sizeof(s))
#define ok return 0;
#define TLE std::ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define mod(x) ((x)%9973)
#define test cout<<"     ++++++      "<<endl;
#define MAXN 0x3f3f3f3f
#define pi acos(-1.0)
//二叉树
#define lson rt<<1, l, m
#define rson rt<<1|1, m+1, r
//线段树
#define ls now<<1
#define rs now<<1|1
int dir[4][2] = {1,0,-1,0,0,1,0,-1}; //单位移动
//int dir[8][2] = {2,1,2,-1,-2,1,-2,-1,1,2,1,-2,-1,2,-1,-2};
int t,n,m,k,x,y,col,ex,ey,ans,cnt;
lli a[202][202],b[202][202];
int vis[11][11];
typedef struct node {    int x,y,px,py;}node;
node dp[5][5],p;

void BFS()
{
    mem(vis,0);
    queue<node>q;
    dp[1][1].x=dp[1][1].y=0;
    q.push(dp[1][1]);
    vis[1][1]=1;
    while(!q.empty())
    {
        p=q.front();
        //cout<<p.x<< " " <<p.y<<" "<<endl;
        q.pop();
        if(p.x==5 && p.y==5) return;
        for(int i=0;i<4;i++)
        {
            int nx=p.x+dir[i][0];
            int ny=p.y+dir[i][1];
            if(nx<0 || ny<0 || nx>=5 || ny>=5) continue;
            if(!a[nx][ny])
            {
                //cout<<nx<<" nx "<<ny<<endl;
                //cout<<p.x<<" p.x++++ "<<p.y<<endl;
                a[nx][ny] = 1;
                dp[nx][ny].x=nx;
                dp[nx][ny].y=ny;
                dp[nx][ny].px=p.x;
                dp[nx][ny].py=p.y;
                q.push(dp[nx][ny]);
            }
        }
    }
}
void outdp(int x,int y)
{
    //cout<<x<<"     ----- "<<y<<endl;
    if(x==0 &&y==0)
    {
        cout<<"("<<dp[1][1].x<<", "<<dp[1][1].y<<")"<<endl;
        return;
    }
    int nx=dp[x][y].px;
    int ny=dp[x][y].py;
    //cout<<nx<<" "<<ny<<endl;
    outdp(nx,ny);
    cout<<"("<<dp[x][y].x<<", "<<dp[x][y].y<<")"<<endl;
}

int main()
{
    for(int i=0;i<5;i++)
        for(int j=0;j<5;j++)
        cin>>a[i][j];
    BFS();
    outdp(4,4);
    ok;
}