POJ 2243Knight Moves (bfs)
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Knight Moves
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 13925 Accepted: 7766
Description
A friend of you is doing research on the Traveling Knight Problem (TKP) where you are to find the shortest closed tour of knight moves that visits each square of a given set of n squares on a chessboard exactly once. He thinks that the most difficult part of the problem is determining the smallest number of knight moves between two given squares and that, once you have accomplished this, finding the tour would be easy.
Of course you know that it is vice versa. So you offer him to write a program that solves the "difficult" part.
Your job is to write a program that takes two squares a and b as input and then determines the number of knight moves on a shortest route from a to b.
Of course you know that it is vice versa. So you offer him to write a program that solves the "difficult" part.
Your job is to write a program that takes two squares a and b as input and then determines the number of knight moves on a shortest route from a to b.
Input
The input will contain one or more test cases. Each test case consists of one line containing two squares separated by one space. A square is a string consisting of a letter (a-h) representing the column and a digit (1-8) representing the row on the chessboard.
Output
For each test case, print one line saying "To get from xx to yy takes n knight moves.".
Sample Input
e2 e4a1 b2b2 c3a1 h8a1 h7h8 a1b1 c3f6 f6
Sample Output
To get from e2 to e4 takes 2 knight moves.To get from a1 to b2 takes 4 knight moves.To get from b2 to c3 takes 2 knight moves.To get from a1 to h8 takes 6 knight moves.To get from a1 to h7 takes 5 knight moves.To get from h8 to a1 takes 6 knight moves.To get from b1 to c3 takes 1 knight moves.To get from f6 to f6 takes 0 knight moves.
搜索中基础的题目。用bfs和dfs都可行,这里使用的是bfs。
(实在想吐槽vs的scanf_s,字符输入怎么都报错)
#include <cstdio>#include <cstring>#include <iostream>#include <queue>#include <algorithm>using namespace std;bool visit[9][9];int dir[8][2] = { -1, -2, -2, -1, 1, -2, 2, -1, 1, 2, 2, 1, -1, 2, -2, 1 };struct State{int x, y;int stepCnt;};bool CheckState(State s){if (!visit[s.x][s.y] && s.x >= 1 && s.x <= 8 && s.y >= 1 && s.y <= 8)return true;elsereturn false;}State bfs(State st, State ed){queue<State> q;State now, next;st.stepCnt = 0;q.push(st);visit[st.x][st.y] = true;while (!q.empty()){now = q.front();if (now.x == ed.x && now.y == ed.y){ return now;}for (int i = 0; i < 8; i++){next.x = now.x + dir[i][0];next.y = now.y + dir[i][1];next.stepCnt = now.stepCnt + 1;if (CheckState(next)){q.push(next);visit[next.x][next.y] = true;}}q.pop();}}int main(){ State re, st, ed; char csx, csy, cdx, cdy; while(scanf("%c%c %c%c%*c", &csx, &csy, &cdx, &cdy) != EOF){st.x = csx - 'a' + 1;st.y = csy - '0';ed.x = cdx - 'a' + 1;ed.y = cdy - '0';st.stepCnt = ed.stepCnt = 0;memset(visit, false, 81);re = bfs(st, ed);cout << "To get from " << csx << csy << " to " << cdx << cdy<< " takes " << re.stepCnt << " knight moves." << endl;}return 0;}
“Let the Sun Shine upon this Lord of Cinder!”
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