POJ3984 迷宮問題
阿新 • • 發佈:2021-01-20
題目連結:https://vjudge.net/problem/POJ-3984
Description
定義一個二維陣列:
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,
};
它表示一個迷宮,其中的1表示牆壁,0表示可以走的路,只能橫著走或豎著走,不能斜著走,要求程式設計序找出從左上角到右下角的最短路線。
Input
一個5 × 5的二維陣列,表示一個迷宮。資料保證有唯一解。
Output
左上角到右下角的最短路徑,格式如樣例所示。
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)
解題思路:題目要求最短路,BFS即可,把一個節點的座標跟它的前一個點的座標封裝成一個類(或者結構體),這樣跑完BFS後,便可以匯出起點到終點的路徑
Cpp Code
#include<iostream> #include<cstdio> #include<cstring> #include<queue> #include<stack> using namespace std; int const maxn = 10; bool vis[maxn][maxn]; int table[maxn][maxn]; struct node{ int x, y, prex, prey; }index[maxn][maxn]; int dx[4] = {0, 0, -1, 1}; int dy[4] = {-1, 1, 0, 0}; void bfs(){ queue<node> q; node cur, nex; index[0][0].prex = index[0][0].prey = -1; vis[0][0] = 1; q.push(index[0][0]); while ((!q.empty())) { cur = q.front(); q.pop(); if(cur.x==4&&cur.y==4){ return; } for (int i = 0; i < 4;i++){ int nx = cur.x + dx[i]; int ny = cur.y + dy[i]; if(nx>=0&&nx<5&&ny>=0&&ny<5&&!vis[nx][ny]&&table[nx][ny]==0){ vis[nx][ny] = 1; index[nx][ny].prex = cur.x; index[nx][ny].prey = cur.y; index[nx][ny].x = nx; index[nx][ny].y = ny; q.push(index[nx][ny]); } } } } int main(){ for (int i = 0; i<5;i++){ for (int j = 0; j < 5;j++){ cin >> table[i][j]; } } bfs(); stack<node> s; node now, nex; now.x = now.y = 4; now.prex=index[4][4].prex; now.prey=index[4][4].prey; s.push(now); while (1) { if(now.prex==-1&&now.prey==-1){ break; } nex.x=now.prex; nex.y=now.prey; nex.prex = index[nex.x][nex.y].prex; nex.prey = index[nex.x][nex.y].prey; s.push(nex); now = nex; /* code */ } while (!s.empty()) { node x = s.top(); s.pop(); cout << "(" <<x.x << ", " <<x.y << ")" << endl; /* code */ } return 0; }
Java Code
import java.util.LinkedList; import java.util.Queue; import java.util.Scanner; import java.util.Stack; public class Main { static int maxn=10; static boolean vis[][]=new boolean[maxn][maxn]; static int table[][]=new int[maxn][maxn]; static int dx[]= {0, 0, -1, 1}; static int dy[]= {-1, 1, 0, 0}; static class node{ int x,y,prex,prey; node(){} node(int a,int b,int c,int d){ x=a; y=b; prex=c; prey=d; } void out() { System.out.println(x+" "+y+" "+prex+" "+prey); } } static node index[][]=new node[maxn][maxn]; static void bfs() { Queue<node> q=new LinkedList<node>(); node cur,nex; index[0][0]=new node(0,0,-1,-1); vis[0][0]=true; q.offer(index[0][0]); while(!q.isEmpty()) { cur=q.poll(); if(cur.x==4&&cur.y==4) { return; } for(int i=0;i<4;i++) { int nx=cur.x+dx[i]; int ny=cur.y+dy[i]; if(nx>=0&&nx<5&&ny>=0&&ny<5&&!vis[nx][ny]&&table[nx][ny]==0) { vis[nx][ny]=true; index[nx][ny]=new node(nx,ny,cur.x,cur.y); q.offer(index[nx][ny]); } } } } public static void main(String[] args) { for(int i=0;i<maxn;i++) { for(int j=0;j<maxn;j++) { index[i][j]=new node(); } } // TODO Auto-generated method stub Scanner sc=new Scanner(System.in); for(int i=0;i<5;i++) { for(int j=0;j<5;j++) { table[i][j]=sc.nextInt(); } } bfs(); Stack<node> s=new Stack<node>(); node now,nex; now=new node(4,4,index[4][4].prex,index[4][4].prey); s.push(now); while(true) { if(now.prex==-1&&now.prey==-1) { break; } nex=new node(now.prex,now.prey,index[now.prex][now.prey].prex,index[now.prex][now.prey].prey); s.push(nex); now=nex; } while(!s.isEmpty()) { node x=s.pop(); System.out.println("("+x.x+", "+x.y+")"); } } }