• HDU1372,BFS象棋马走日


    简单的对于bfs的运用,但是还是写的太慢了写了TMD的1H,主要是不熟悉,以后慢慢熟悉就好了,模型基本已经能建立了,主要出现bug是在方向数组的运用上面,一定要记得是从0开始的,而不是从1开始的,导致错误。

    #include<cstdio>
    #include<cstdlib>
    #include<iostream>
    
    using namespace std;
    /*bfs*/
    
    //方向数组
    int nextx[8][2]={
        {-1,-2},
        {-2,-1},
        {1,2},
        {2,1},
        {1,-2},
        {2,-1},
        {-1,2},
        {-2,1}
    };
    
    /*
    1 0 3 0 5 6 7 8 9
    0 2 3 4 0 6 7 8 9 
    1 2 0 4 5 6 7 8 9
    0 2 3 4 0 6 7 8 9
    1 0 3 0 5 6 7 8 9 
    1 2 3 4 5 6 7 8 9
    
    */
    
    struct que
    {
        int x;//当前的A;
        int y;//当前的B;
        int step;//走过的步数
    };
    
    int main()
    {
        int i;//循环变量
        int y1,y2;
        int x1,x2;
        int nowX,nowY;
        char ch[10];
        
        int head=1;
        int tail=1;
    
        struct que ques[100];
    
        while (gets(ch) != NULL)
        {
            //将输入的数转换成对应数字
            x1 = ch[0]-'a'+1;
            y1 = ch[1]-'0';
            x2 = ch[3]-'a'+1;
            y2 = ch[4]-'0';
    
            //初始化地图,用于记录走过的点
            int maps[10][10]={0};
    
            head=1;
            tail=1;
    
            maps[x1][y1] = 1;//记录初始点
            ques[head].x = x1;
            ques[head].y = y1;
            ques[head].step = 0;
    
            tail++;
            nowX=x1;
            nowY=y1;
            if(nowX == x2 && nowY == y2)
            {
                goto f1;
            }
            
            while (head<tail)
            {
                for (i = 0; i < 8; i++)
                {
                    nowX = ques[head].x + nextx[i][0];
                    nowY = ques[head].y + nextx[i][1];
    
                    if(nowX<=0 || nowX>8 || nowY<=0 || nowY>8)
                        continue;
                    if(maps[nowX][nowY] == 0)
                    {
                        maps[nowX][nowY] = 1;
                        ques[tail].x = nowX;
                        ques[tail].y = nowY;
                        ques[tail].step = ques[head].step + 1;
                        tail++;
                    }
    
                    if(nowX == x2 && nowY == y2)
                    {
                        goto f1;
                    }
                }
                head++;
            }
    
            f1:printf("To get from %c%d to %c%d takes %d knight moves.
    ",ch[0],y1,ch[3],y2,ques[tail-1].step);
        }
        return 0;
    }
    
    /*
    
    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.
    
    
    */
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  • 原文地址:https://www.cnblogs.com/linkstar/p/5335545.html
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