UVA – 439
Knight Moves
Time Limit:3000MSMemory Limit:Unknown64bit IO Format:%lld & %llu
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Description
A friend of you is doing research on theTraveling Knight Problem (TKP)where you are to find the shortest closed tour of knight moves that visits each square of a given set ofnsquares 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 squaresaandbas input and then determines the number of knight moves on a shortest route fromatob.
Input Specification
The input file 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 Specification
For each test case, print one line saying "To get fromxxtoyytakesnknight moves.".
Sample Inpute2 e4a1 b2b2 c3a1 h8a1 h7h8 a1b1 c3f6 f6Sample OutputTo 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.
Source
Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: Rare Topics :: Rare Problems ::Knight MovesRoot :: Competitive Programming: Increasing the Lower Bound of Programming Contests (Steven & Felix Halim) :: Chapter 4. Graph :: Breadth First Search ::SSSP on Unweighted GraphRoot :: AOAPC I: Beginning Algorithm Contests (Rujia Liu) :: Volume 2. Data Structures ::GraphsRoot :: AOAPC II: Beginning Algorithm Contests (Second Edition) (Rujia Liu) :: Chapter 6. Data Structures ::ExercisesRoot :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Graph :: Single-Source Shortest Paths (SSSP) ::On Unweighted Graph: BFS
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思路：题目意思看了半天才懂，就是类似象棋中的马走日，可以朝八个方向走，然后给出两个点，问从一个点到另一点最短步数为多少，可以用BFS来搜索，，一步一步模拟过去
AC代码：
#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;char s1[5], s2[5];int ans;int vis[10][10];//记忆化搜索 int dir[8][2] = {{-2, -1},{-2, 1},{-1, 2},{1, 2},{2, 1},{2, -1},{1, -2},{-1, -2}};//朝八个方向 struct node {int x, y, count;//count用于记录现在是第几层 bool operator == (const node &a) const {//字符重载，用于快速比较 return x == a.x && y == a.y;} };int bfs(node pos1, node pos2) {//广搜 queue<node> q;node pre, cur;q.push(pos1);vis[pos1.x][pos1.y] = 1;while(!q.empty()) {pre = q.front();q.pop();if(pre == pos2) return pre.count;//找到后返回一个值（BFS层数，第几步） for(int i = 0; i < 8; i++) {node cur;cur.x = pre.x + dir[i][0];cur.y = pre.y + dir[i][1];if(!vis[cur.x][cur.y] && cur.x >= 1 && cur.x <= 8 && cur.y >= 1 && cur.y <= 8) {vis[cur.x][cur.y] = 1;cur.count = pre.count + 1;q.push(cur);}}}return 0;}int main() {while(scanf("%s %s", s1, s2) != EOF) {node pos1, pos2;pos1.y = s1[0] – 'a' + 1;pos1.x = s1[1] – '0';pos1.count = 0;pos2.y = s2[0] – 'a' + 1;pos2.x = s2[1] – '0';pos2.count = 0;memset(vis, 0, sizeof(vis));ans = bfs(pos1, pos2);printf("To get from %s to %s takes %d knight moves.\n", s1, s2, ans);} return 0;}
离开睁眼闭眼看见的城市，逃离身边的纷纷扰扰，
