bfs学习

今天做到了bfs的练习，顺便写下心得。。。

bfs能解决搜索和最短路径的问题。

下面是学习心得：

typedef struct point                //定义点
{
    int x;
    int y;
}P;
bfs()
{
    int level[N];                   //记录队列中元素层数，即从起点到该点最短路径距离
    P father[NX][NY],queue[N];      //father用来记录父节点，queue用来记录队列
    int top=0;                      //top用来记录队列长度，并指向最后节点
    for(int i=0;i<nx;i++)
        for(int j=0;j<ny;j++)
        {
            father[i][j].x=-1;      //初始化father
            father[i][j].y=-1;
        }
    queue[top].x=x0;                //对起点进行赋值
    queue[top].y=y0;
    father[x0][y0].x=x0;
    father[x0][y0].y=y0;
    top++;
    for(int i=0;i<top;i++)
    {
        for(遍历第i个元素的邻接点)
        {
            if(邻接点没标记)          //若邻接点father为-1
            {
                父节点记为第i个元素;
                queue[top]赋值为该邻接点;      //把元素接到队列上
                level[top]=level[i]+1;          //元素层次为父节点层次+1
                top++;                          //队列长度+1
            }
            if(邻接点满足条件)
            {
                return ...;
            }
        }
    }
}

上一道训练题：

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.

我的题解（很水，有改进空间）：

#include<stdio.h>
#include<string.h>
int dx[8]={-1,-1,-2,-2,1,1,2,2};        //对应的xy变化表
int dy[8]={2,-2,1,-1,2,-2,1,-1};
typedef struct point
{
    int x;
    int y;
}P;
int bfs(char* s1,char* s2)
{
    int x1=(int)(s1[0]-'a'+1);          //转化为数字
    int x2=(int)(s2[0]-'a'+1);
    int y1=(int)(s1[1]-'0');
    int y2=(int)(s2[1]-'0');
    int level[80],top=0,flag;           //top用来记录队列长度，并可用于指向队尾
    P father[10][10],queue[80];         //father记录父节点，queue记录队列
    for(int i=0;i<10;i++)
        for(int j=0;j<10;j++)
        {
            father[i][j].x=-1;          //父节点初始化为-1
            father[i][j].y=-1;
        }
    level[top]=0;
    father[x1][y1].x=x1;
    father[x1][y1].y=y1;
    queue[top].x=x1;
    queue[top++].y=y1;
    if(x1==x2&&y1==y2)
        return 0;
    for(int i=0;i<top;i++)
    {
        flag=0;
        for(int j=0;j<8;j++)            //标记所有邻接点
        {
            int tempx,tempy;
            tempx=queue[i].x+dx[j];
            tempy=queue[i].y+dy[j];
            if(tempx<=8&&tempx>0&&tempy<=8&&tempy>0&&father[tempx][tempy].x==-1)
            {
                father[tempx][tempy].x=queue[i].x;
                father[tempx][tempy].y=queue[i].y;
                queue[top].x=tempx;
                queue[top].y=tempy;
                level[top++]=level[i]+1;      //level用来记录当前元素层次
            }
            if(tempx==x2&&tempy==y2)
            {
                flag=1;
                break;
            }
        }
        if(flag==1)
            return level[top-1];
    }
}
int main()
{
    char s1[3],s2[3];
    int num;
    while(scanf("%s%s",s1,s2)!=EOF)
    {
        num=bfs(s1,s2);         //输入
        printf("To get from %s to %s takes %d knight moves.\n",s1,s2,num);
    }
    return 0;
}