[HDU] 1372 Knight Moves 每步走马步（日字）的广搜

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=1372

方法：建模后的广搜是最基本的单位权值边单源最短路径。所以不需要使用优先队列，直接用普通队列来维护广搜即可。

感想：基础题。

代码：

#include <iostream>
#include <queue>
using namespace std;
bool visited[9][9];
char chessboard [9][9];
int derection[8][2];
struct cell
{
    int x,y;
    int currentSteps;
};
bool canGo(int x,int y)
{
   if(x<1||x>8 || y<1|| y>8)
       return false;
   return !visited[x][y];
}
int search(int startX, int startY,int endX, int endY)
{
    cell* cur = (cell*)malloc(sizeof(cell));
    cur->x = startX;
    cur->y= startY;
    cur->currentSteps =0;
    visited[cur->x][cur->y] = true;
    queue<cell*> Q;
    Q.push(cur);
    int result=-1;
    while(!Q.empty())
    {
        cur = Q.front();
        Q.pop();
        if(cur->x == endX && cur->y == endY)
        {
            result = cur->currentSteps;
            break;
        }
        else
        {
            for(int i = 0;i<8;i++)
            {
                int x = cur->x+derection[i][0];
                int y = cur->y+derection[i][1];
                if(canGo(x,y))
                {
                    cell* newCell = (cell*) malloc(sizeof(cell));
                    visited[x][y] = true;
                    newCell->x=x;
                    newCell->y=y;
                    newCell->currentSteps = cur->currentSteps+1;
                    Q.push(newCell);
                }
            }
        }
        delete[] cur;
    }
    while(!Q.empty())
    {
        delete[] Q.front();
        Q.pop();
    }
    return result;
}

void main()
{
    derection[0][0] = -2;derection[0][1] = 1;
    derection[1][0] = -1;derection[1][1] = 2;
    derection[2][0] =  1;derection[2][1] = 2;
    derection[3][0] =  2;derection[3][1] = 1;

    derection[4][0] =  2;derection[4][1] =-1;
    derection[5][0] =  1;derection[5][1] =-2;
    derection[6][0] = -1;derection[6][1] =-2;
    derection[7][0] = -2;derection[7][1] =-1;
    int s=0,t=0;
    char* start=new char[2];
    char* end=new char[2];
    while( scanf("%s %s",start,end)!= EOF )
    {
        int startX=start[1]-48;
        int startY=start[0]-96;
        int endX=end[1]-48;
        int endY=end[0]-96;
        for(int i =1;i<=8;i++)
            for(int j =1;j<=8;j++)
                visited[i][j]=false;
        int result = search(startX,startY,endX,endY);
        printf("To get from %s to %s takes %d knight moves.\n",start,end,result);
    }
}