單向BFS需要搜索N層才能到達(dá)終點(diǎn)柑营，在每個(gè)層需要進(jìn)行的判斷量（即通常的那個(gè)for循環(huán)）為X。那么猎荠，單BFS的運(yùn)算量為：X^N。
如果換成雙BFS蜀备，那么前后各搜索 N/2層关摇，那么總的運(yùn)算量為：2 * ( X ^ ( N/2 ) )。顯然當(dāng)X比較大時(shí)碾阁，在運(yùn)算量上不僅僅不僅僅是減半那么簡單输虱。特別的，如果X=1脂凶，那么雙BFS也就退化成了單向BFS了宪睹，實(shí)際上，此時(shí)也就是可以用DFS來進(jìn)行深搜了蚕钦，而且代碼相對來說更加簡潔亭病。

poj 2243

Knight Moves
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 14679 Accepted: 8226
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.

雙向BFS：

#include <iostream>  
#include <cstdio>  
#include <cstring>  
#include <queue>  
using namespace std;  
  
struct knight  
{  
    int x,y,step;  
};  
  
int dir[8][2]={{-2,-1},{-2,1},{2,-1},{2,1},{-1,-2},{-1,2},{1,-2},{1,2}};  
int sx, sy, ex, ey;  
int visit[8][8];  
int color[8][8];  
int bfs();  
  
int main()  
{  
    int x1, x2;  
    char y1, y2;  
    while(scanf("%c%d %c%d", &y1, &x1, &y2, &x2) != EOF)  
    {  
        getchar();  
        sx = x1 - 1;  
        sy = y1 - 'a';  
        ex = x2 - 1;  
        ey = y2 - 'a';  
        memset(visit, -1, sizeof(visit));  
        memset(color, 0, sizeof(color));  
        int cost = bfs();  
        printf("To get from %c%d to %c%d takes %d knight moves.\n", y1, x1, y2, x2, cost);  
    }  
    return 0;  
}  
  
int bfs()  
{  
    if(sx == ex && sy == ey)  
        return 0;  
    queue<knight> que_front;  
    queue<knight> que_back;  
    knight front, back;  
    front.x = sx; front.y = sy; front.step = 0;  
    back.x = ex; back.y = ey; back.step = 1;  
    que_front.push(front);  
    que_back.push(back);  
    visit[sx][sy] = 0;  
    visit[ex][ey] = 1;  
    color[sx][sy] = 1;  
    color[ex][ey] = 2;  
    int ans1 = 0, ans2 = 0;  
    while(!que_front.empty() || !que_back.empty())  
    {  
        if(!que_front.empty())  
        {  
            front = que_front.front();  
            que_front.pop();  
            for(int i = 0; i < 8; i++)  
            {  
                int dx = front.x + dir[i][0];  
                int dy = front.y + dir[i][1];  
                if(dx >= 0 && dx < 8 && dy >= 0 && dy < 8 && color[dx][dy] != 1)  
                {  
                    if(color[dx][dy] == 0)  
                    {  
                        knight tmp;  
                        tmp.x = dx; tmp.y = dy;  
                        tmp.step = front.step + 1;  
                        visit[dx][dy] = tmp.step;  
                        color[dx][dy] = 1;  
                        que_front.push(tmp);  
                    }  
                    else  
                        return front.step + visit[dx][dy];  
                }  
            }  
        }  
        if(!que_back.empty())  
        {  
            back = que_back.front();  
            que_back.pop();  
            for(int i = 0; i < 8; i++)  
            {  
                int dx = back.x + dir[i][0];  
                int dy = back.y + dir[i][1];  
                if(dx >= 0 && dx < 8 && dy >= 0 && dy < 8 && color[dx][dy] != 2)  
                {  
                    if(color[dx][dy] == 0)  
                    {  
                        knight tmp;  
                        tmp.x = dx; tmp.y = dy;  
                        tmp.step = back.step + 1;  
                        visit[dx][dy] = tmp.step;  
                        color[dx][dy] = 2;  
                        que_back.push(tmp);  
                    }  
                    else  
                        return back.step + visit[dx][dy];  
                }  
                  
            }  
        }  
    }  
    return -1;  
}