試題 D: 迷宮

本題總分：10 分

【問題描述】
下圖給出了一個迷宮的平面圖，其中標記為 1 的為障礙，標記為 0 的為可以通行的地方。

010000
000100
001001
110000

迷宮的入口為左上角，出口為右下角，在迷宮中，只能從一個位置走到這 個它的上、下、左、右四個方向之一。
對于上面的迷宮，從入口開始，可以按DRRURRDDDR 的順序通過迷宮， 一共 10 步。其中 D、U、L、R 分別表示向下、向上、向左、向右走。
對于下面這個更復雜的迷宮（30 行 50 列），請找出一種通過迷宮的方式，其使用的步數最少，在步數最少的前提下，請找出字典序最小的一個作為答案。請注意在字典序中D<L<R<U。（如果你把以下文字復制到文本文件中，請務 必檢查復制的內容是否與文檔中的一致。在試題目錄下有一個文件 maze.txt， 內容與下面的文本相同）

01010101001011001001010110010110100100001000101010 00001000100000101010010000100000001001100110100101 01111011010010001000001101001011100011000000010000 01000000001010100011010000101000001010101011001011 00011111000000101000010010100010100000101100000000 11001000110101000010101100011010011010101011110111 00011011010101001001001010000001000101001110000000 10100000101000100110101010111110011000010000111010 00111000001010100001100010000001000101001100001001 11000110100001110010001001010101010101010001101000 00010000100100000101001010101110100010101010000101 11100100101001001000010000010101010100100100010100 00000010000000101011001111010001100000101010100011 10101010011100001000011000010110011110110100001000 10101010100001101010100101000010100000111011101001 10000000101100010000101100101101001011100000000100 10101001000000010100100001000100000100011110101001 00101001010101101001010100011010101101110000110101 11001010000100001100000010100101000001000111000010 00001000110000110101101000000100101001001000011101 10100101000101000000001110110010110101101010100001 00101000010000110101010000100010001001000100010101 10100001000110010001000010101001010101011111010010 00000100101000000110010100101001000001000000000010 11010000001001110111001001000011101001011011101000 00000110100010001000100000001000011101000000110011 10101000101000100010001111100010101001010000001000 10000010100101001010110000000100101010001011101000 00111100001000010000000110111000000001000000001011 10000001100111010111010001000110111010101101111000

【答案提交】
這是一道結果填空的題，你只需要算出結果后提交即可。本題的結果為一 個字符串，包含四種字母 D、U、L、R，在提交答案時只填寫這個字符串，填寫多余的內容將無法得分。

Code

/*^....0^ .1 ^1^.. 011.^ 1.0^ 1 ^ ^0.11 ^ ^..^0. ^ 0^.0 1 .^.1 ^0 .........001^.1 1. .111100....01^00 11^ ^1. .1^1.^ ^0 0^.^ ^0..1.1 1..^1 .0 ^ ^00. ^^0.^^ 0 ^^110.^0 0 ^ ^^^10.01^^ 10 1 1 ^^^1110.101 10 1.1 ^^^1111110010 01 ^^ ^^^1111^1.^ ^^^10 10^ 0^ 1 ^^111^^^0.1^ 1....^11 0 ^^11^^^ 0.. ....1^ ^ ^1. 0^ ^11^^^ ^ 1 111^ ^ 0.10 00 11 ^^^^^ 1 0 1.0^ ^0 ^0 ^^^^ 0 0.0^ 1.0 .^ ^^^^ 1 1 .0^.^ ^^ 0^ ^1 ^^^^ 0. ^.11 ^ 11 1. ^^^ ^ ^ ..^^..^ ^1 ^.^ ^^^ .0 ^.00..^ ^0 01 ^^^ .. 0..^1 .. .1 ^.^ ^^^ 1 ^ ^0001^ 1. 00 0. ^^^ ^.0 ^.1. 0^. ^.^ ^.^ ^^^ ..0.01 .^^. .^ 1001 ^^ ^^^ . 1^. ^ ^. 11 0. 1 ^ ^^ 0.0 ^. 0 ^0 1 ^^^ 0.0.^ 1. 0^ 0 .1 ^^^ ...1 1. 00 . .1 ^^^ ..1 1. ^. 0 .^ ^^ ..0. 1. .^ . 0 ..1 1. 01 . . ^ 0^.^ 00 ^0 1. ^ 1 1.0 00 . ^^^^^^ ..^ 00 01 ..1. 00 10 1 ^^.1 00 ^. ^^^ .1.. 00 .1 1..01 ..1.1 00 1. ..^ 10^ 1^ 00 ^.1 0 1 1.1 00 00 ^ 1 ^. 00 ^.^ 10^ ^^1.1 00 00 10^..^ 1. ^. 1.0 1 ^. 00 00 .^^ ^. ^ 1 00 ^0000^ ^ 011 0 ^. 00.0^ ^00000 1.00.1 11. 1 0 1^^0.01 ^^^ 01.^ ^ 1 1^^ ^.^1 1 0... 1 ^1 1^ ^ .01 ^ 1.. 1.1 ^0.0^ 0 1..01^^100000..0^1 1 ^ 1 ^^1111^ ^^0 ^ ^ 1 1000^.1 ^.^ . 00.. 1.1 0. 01. . 1. .^1. 1 1. ^0^ . ^.1 00 01^.0 001. .^*//* Procedural objectives：Variables required by the program:Procedural thinking：Functions required by the program:Determination algorithm:Determining data structure:*/ /* My dear Max said： "I like you, So the first bunch of sunshine I saw in the morning is you, The first gentle breeze that passed through my ear is you, The first star I see is also you. The world I see is all your shadow."FIGHTING FOR OUR FUTURE！！！ *//* 01010101001011001001010110010110100100001000101010 00001000100000101010010000100000001001100110100101 01111011010010001000001101001011100011000000010000 01000000001010100011010000101000001010101011001011 00011111000000101000010010100010100000101100000000 11001000110101000010101100011010011010101011110111 00011011010101001001001010000001000101001110000000 10100000101000100110101010111110011000010000111010 00111000001010100001100010000001000101001100001001 11000110100001110010001001010101010101010001101000 00010000100100000101001010101110100010101010000101 11100100101001001000010000010101010100100100010100 00000010000000101011001111010001100000101010100011 10101010011100001000011000010110011110110100001000 10101010100001101010100101000010100000111011101001 10000000101100010000101100101101001011100000000100 10101001000000010100100001000100000100011110101001 00101001010101101001010100011010101101110000110101 11001010000100001100000010100101000001000111000010 00001000110000110101101000000100101001001000011101 10100101000101000000001110110010110101101010100001 00101000010000110101010000100010001001000100010101 10100001000110010001000010101001010101011111010010 00000100101000000110010100101001000001000000000010 11010000001001110111001001000011101001011011101000 00000110100010001000100000001000011101000000110011 10101000101000100010001111100010101001010000001000 10000010100101001010110000000100101010001011101000 00111100001000010000000110111000000001000000001011 10000001100111010111010001000110111010101101111000 */ #include <queue> #include <cstdio> #include <cstring> #include <iostream>using namespace std;int m=30,n=50; char map[70][70]; bool vis[70][70]; char dirc[4]={'D','L','R','U'}; int dir[4][2]={{1,0},{0,-1},{0,1},{-1,0}};struct Node{string str;int x,y,step;Node(int xx,int yy,int ss,string s){x=xx;y=yy;step=ss;str=s;} };queue<Node> que;bool check(int x,int y){if(x<0 || x>m-1 || y<0 || y>n-1 || vis[x][y] || map[x][y]=='1')return false;return true; }void bfs(int x,int y){que.push(Node(0,0,0,""));vis[0][0]=true;while(!que.empty()){Node now=que.front();if(now.x==m-1 && now.y==n-1){cout<<now.str<<endl;cout<<now.step<<endl;break;}que.pop();for(int i=0;i<4;i++){int xx=now.x+dir[i][0];int yy=now.y+dir[i][1];if(check(xx,yy)){que.push(Node(xx,yy,now.step+1,now.str+dirc[i]));vis[xx][yy]=true;}}} }int main(){for(int i=0;i<m;i++)scanf("%s",map[i]);bfs(0,0);return 0; }

總結

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