poj 2243 Knight Moves

Description

Input

Output

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.

题意：国际象棋中，给出骑士的起始位置和目标位数，求骑士最少走几步就能到达目标位置。

解题思路：简单bfs，不明白为什么用c++提交TLE，G++提交AC；

#include <iostream>
#include <queue>
#include <string.h>
#include <stdio.h>
using namespace std;
struct point{
	int x,y;
	int step;
};
int a[]={-1, 1,-2, 2,-2, 2,-1, 1};
int b[]={-2,-2,-1,-1, 1, 1, 2, 2};
bool map[10][10];
int bfs(point start,point end){
	queue<point> q;
	start.step=0;
	q.push(start);
	while (q.size()){
		point p=q.front();
		if (p.x==end.x && p.y==end.y){
			return p.step;
		}
		for (int i=0;i<8;i++){	
			if (p.x+a[i]<0 || p.x+a[i]>7 || p.y+b[i]<0 || p.y+b[i]>7 || map[p.x+a[i]][p.y+b[i]]==true)
				continue;	
			point s;
			s.x=p.x+a[i];
			s.y=p.y+b[i];
			s.step=p.step+1;
			q.push(s);
		}
		map[p.x][p.y]=true;
		q.pop();
	}
	return -1;
}
int main(){
	char start[3],end[3];
	while (cin>>start>>end){		
		memset(map,false,sizeof(map));
		point starts,ends;
		starts.x=start[0]-'a';
		starts.y=start[1]-'1';
		ends.x=end[0]-'a';
		ends.y=end[1]-'1';
		cout<<"To get from "<<start<<" to "<<end<<" takes "<<bfs(starts,ends)<<" knight moves."<<endl;
	}
	return 0;
}

poj 2243 Knight Moves

原文：http://blog.csdn.net/codeforcer/article/details/41357999