POJ 2243 Knight Moves

原题传送门：http://poj.org/problem?id=2243

Knight Moves

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11197		Accepted: 6318

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

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 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

Ulm Local 1996

 

分析：这道题目也是BFS的水题，主要意思就是给你出发坐标，和目标坐标，然后移动路径只能跳日字格，和中国象棋中的马的走法一直，问你最快需要多少步。

 

//BFS 图的遍历 广度优先搜索POJ 2243

#include<cstdio>
#include<queue>
#include<string.h>
using namespace std;

struct Point{
    int x;
    int y;
    int step;
};

const int MAXN = 10;
char a[3];
char b[3];
int stax,stay,endx,endy,step;
bool visit[MAXN][MAXN];
queue<Point> pl;
int dir[8][2] = {{-2,1},{-1,2},{1,2},{2,1},{-2,-1},{-1,-2},{1,-2},{2,-1}};

int BFS(int stx,int sty,int edx,int edy){
    if(stx == edx && sty == edy)return 0;
    Point p;
    p.x = stx;
    p.y = sty;
    p.step = 0;
    pl.push(p);
    visit[stx][sty] = 1;
    while(!pl.empty()){
        Point temp = pl.front();
        pl.pop();
        for(int i=0;i<8;i++){
            Point newP;
            newP.x = temp.x + dir[i][0];
            newP.y = temp.y + dir[i][1];
            newP.step = temp.step + 1;
            if(newP.x == edx && newP.y == edy){
                return newP.step;
            }
            if(newP.x>0&&newP.y>0&&newP.x<=8&&newP.y<=8&&!visit[newP.x][newP.y])
                pl.push(newP);
                visit[newP.x][newP.y] = 1;
        }
    }
}
int main(){
    while(scanf("%s %s",a,b)== 2){
        while(!pl.empty()){
            pl.pop();
        }
        a[2] = '\0';
        b[2] = '\0';
        memset(visit,0,sizeof(visit));
        stax = (int)(a[1]-48);
        stay = (int)(a[0]-96);
        endx = (int)(b[1]-48);
        endy = (int)(b[0]-96);
        int ans = BFS(stax,stay,endx,endy);
        printf("To get from %s to %s takes %d knight moves.\n",a,b,ans);

    }
    return 0;
}