UVA - 439 - Knight Moves （BFS）

UVA - 439 - Knight Moves （BFS）

UVA - 439

Knight Moves Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu

 

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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.

 

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 from xx to yy takes n knight moves.".

 

Sample Input

 

e2 e4
a1 b2
b2 c3
a1 h8
a1 h7
h8 a1
b1 c3
f6 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.

 

Source

Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: Rare Topics :: Rare Problems :: Knight Moves
Root :: Competitive Programming: Increasing the Lower Bound of Programming Contests (Steven & Felix Halim) :: Chapter 4. Graph :: Breadth First Search :: SSSP on Unweighted Graph
Root :: AOAPC I: Beginning Algorithm Contests (Rujia Liu) :: Volume 2. Data Structures :: Graphs
Root :: AOAPC II: Beginning Algorithm Contests (Second Edition) (Rujia Liu) :: Chapter 6. Data Structures :: Exercises
Root :: 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：題目意思看了半天才懂，就是類似象棋中的馬走日，可以朝八個方向走，然後給出兩個點，問從一個點到另一點最短步數為多少，可以用BFS來搜索，一步一步模擬過去

 

AC代碼：

 

#include 
#include 
#include 
#include 
using namespace std;

char s1[5], s2[5];
int ans;
int vis[10][10];//記憶化搜索 
int dir[8][2] = {{-2, -1},{-2, 1},{-1, 2},{1, 2},{2, 1},{2, -1},{1, -2},{-1, -2}};//朝八個方向 

struct node {
	int x, y, count;		//count用於記錄現在是第幾層 
	bool operator == (const node &a) const {	//字符重載，用於快速比較 
		return x == a.x && y == a.y;
	} 
};

int bfs(node pos1, node pos2) {		//廣搜 
	queue 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;//找到後返回一個值（BFS層數，第幾步） 
		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;
}