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蓝桥杯 迷宫

發布時間:2023/12/4 编程问答 34 豆豆
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參考代碼:

#include<bits/stdc++.h> using namespace std; typedef long long ll; int e[55][55], vis[55][55]; //e矩陣表示障礙物信息,vis矩陣表示點是否被訪問過 int m = 30, n = 50; struct node{ //定義結構體,用于存放點的信息 int x;int y;string path; //用來表示從起點到該點的最短路徑 }; queue<node> q;bool check(int x, int y) {if(x >= 1 && x <= m && y >= 1 && y <= n && !vis[x][y] && !e[x][y]) return true;else return false; }int main() {char c;for(int i = 1; i <= m; i++){for(int j = 1; j <= n; j++){c = getchar(); e[i][j] = c - '0'; //表示此初是否有障礙 }c = getchar(); //需要用getchar()來讀取一個換行符 }int newx, newy;node top, temp; //定義結構體變量 temp.x = 1; //定義起點 temp.y = 1;temp.path = "";q.push(temp); //將起點放入隊列中 vis[1][1] = true;string str = "DLRU";int cnx[4] = {1, 0, 0, -1}; //4個移動方向的坐標變化量 int cny[4] = {0, -1, 1, 0};while(!q.empty()) //隊列為空時退出循環 {top = q.front(); //讀取隊列第一個變量,進行bfs遍歷 q.pop();for(int i = 0; i < 4; i++) //對于四個方向依次進行遍歷 {newx = top.x + cnx[i];newy = top.y + cny[i];if(check(newx, newy)) //判斷產生的新節點是否超出邊界,是否在障礙物上,是否已被訪問過 {temp.x = newx;temp.y = newy;temp.path = top.path + str[i]; //從起點到當前節點走過的路徑 q.push(temp);vis[newx][newy] = true; //標記當前節點已經被訪問 if(newx == m && newy == n) //到達終點 {cout << temp.path; //輸入路徑 return 0;}}}}return 0; }

輸入數據:

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

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