蛇梯棋问题
我们知道著名的游戏蛇梯棋。在这个游戏中,棋盘上存在一些房间,有房间编号。一些房间通过梯子或蛇相连。当我们遇到梯子时,我们可以爬上一些房间,无需按顺序移动就可以到达目的地附近。类似地,当我们遇到蛇时,它会将我们送到较低的房间,从该房间重新开始旅程。
在这个问题中,我们必须找到从起点到达终点的最小掷骰子次数。
输入和输出
Input: The starting and ending location of the snake and ladders. Snake: From 26 to 0, From 20 to 8, From 16 to 3, From 18 to 6 Ladder From 2 to 21, From 4 to 7, From 10 to 25, from 19 to 28 Output: Min Dice throws required is 3
算法
minDiceThrow(move, cell)
输入:蛇或梯子的跳跃位置,以及总格数。
输出:到达终点格所需的最小掷骰子次数。
Begin initially mark all cell as unvisited define queue q mark the staring vertex as visited for starting vertex the vertex number := 0 and distance := 0 add starting vertex s into q while q is not empty, do qVert := front element of the queue v := vertex number of qVert if v = cell -1, then //when it is last vertex break the loop delete one item from queue for j := v + 1, to v + 6 and j < cell, increase j by 1, do if j is not visited, then newVert.dist := (qVert.dist + 1) mark v as visited if there is snake or ladder, then newVert.vert := move[j] //jump to that location else newVert.vert := j insert newVert into queue done done return qVert.dist End
示例
#include<iostream> #include <queue> using namespace std; struct vertex { int vert; int dist; // Distance of this vertex from source }; int minDiceThrow(int move[], int cell) { bool visited[cell]; for (int i = 0; i < cell; i++) visited[i] = false; //initially all cells are unvisited queue<vertex> q; visited[0] = true; //initially starting from 0 vertex s = {0, 0}; q.push(s); // Enqueue 0'th vertex vertex qVert; while (!q.empty()) { qVert = q.front(); int v = qVert.vert; if (v == cell-1) //when v is the destination vertex break; q.pop(); for (int j=v+1; j<=(v+6) && j<cell; ++j) { //for next 1 to 6 cells if (!visited[j]) { vertex newVert; newVert.dist = (qVert.dist + 1); //initially distance increased by 1 visited[j] = true; if (move[j] != -1) newVert.vert = move[j]; //if jth place have snake or ladder else newVert.vert = j; q.push(newVert); } } } return qVert.dist; //number of minimum dice throw } int main() { int cell = 30; //consider there are 30 cells int moves[cell]; for (int i = 0; i<cell; i++) moves[i] = -1; //initially no snake or ladder are initialized //For ladder in cell i, it jumps to move[i] moves[2] = 21; moves[4] = 7; moves[10] = 25; moves[19] = 28; //For snake in cell i, it jumps to move[i] moves[26] = 0; moves[20] = 8; moves[16] = 3; moves[18] = 6; cout << "Min Dice throws required is " << minDiceThrow(moves, cell); }
输出
Min Dice throws required is 3
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