zoukankan      html  css  js  c++  java
  • Chess

    There is a mobile piece and a stationary piece on the N×M chessboard. The available moves of the mobile piece are the same as set out in the image below. You need to capture the stationary piece by moving the mobile piece with the minimum amount of moves.



    Write a program to find out the minimum number moves to catch a piece.

    Time limit:1 second (java: 2 seconds)

    [Input]
    Several test cases can be included in the inputs. T, the number of cases is given in the first row of the inputs. After that, the test cases as many as T (T ≤ 20) are given in a row.
    N, the numbers of the rows and M, the number of columns of the chessboard are given in the first row of each test case.
    R & C is the location information of the attacking piece and S & K is the location of the defending pieces and are given in the row at the second line. However, the location of the uppermost end of the left end is (1, 1)

    [Output]
    For each test case, you should print "Case #T" in the first line where T means the case number.

    For each test case, you should output the minimum number of movements to catch a defending piece at the first line of each test case. If not moveable, output equals ‘-1’.

    [I/O Example]

    Input
    2
    9 9
    3 5 2 8
    20 20
    2 3 7 9

    Output
    Case #1

    2

    Case #2

    代码:

    #include <iostream>
    #include <stdio.h>
    #include <queue>
    #include <string.h>
    
    using namespace std;
    typedef struct
    {
        int x;
        int y;
        int level;
    }data;
    int mv[8][2] = {{-2,1},{-1,2},{1,2},{2,1},{2,-1},{1,-2},{-1,-2},{-2,-1}};
    
    int main()
    {
        //freopen("test.txt","r",stdin);
        int T;
        cin>>T;
        for(int t=1; t<=T; t++)
        {
            int n,m,r1,c1,r2,c2;
            cin>>n>>m;
            int a[n+1][m+1];
            memset(a,0,sizeof(int)*(n+1)*(m+1));
            cin>>r1>>c1>>r2>>c2;
            data d,d1,d2;
            queue<data> qt;
            d.x = r1;
            d.y = c1;
            d.level = 0;
            qt.push(d);
            a[d.x][d.y] = 2;
            int tmx,tmy,tml;
            int steps = 0;
            bool f = false;
            while(!qt.empty())
            {
                if(f)
                {
                    break;
                }
                d1 = qt.front();
                qt.pop();
                for(int k=0; k<8; k++)
                {
                    tmx = d1.x + mv[k][0];
                    tmy = d1.y + mv[k][1];
                    tml = d1.level + 1;
    
                    if(tmx>=1 && tmx<=n && tmy>=1 && tmy<=m && a[tmx][tmy] == 0)
                    {
                        if(tmx == r2 && tmy == c2)
                        {
                            steps = tml;
                            f = true;
                            break;
                        }
                        d2.x = tmx;
                        d2.y = tmy;
                        d2.level = tml;
                        qt.push(d2);
                        a[d2.x][d2.y] = 2;
                    }
                }
            }
            if(!f)
            {
                steps = -1;
            }
            cout<<"Case #"<<t<<endl;
            cout<<steps<<endl;
        }
        //cout << "Hello world!" << endl;
        return 0;
    }
    View Code
  • 相关阅读:
    (LeetCode 72)Edit Distance
    (LeetCode 53)Maximum Subarray
    (LeetCode 64)Minimum Path Sum
    (算法)关于随机数的生成
    (笔试题)数组A中任意两个相邻元素大小相差1,在其中查找某个数。
    (笔试题)分椰子
    (笔试题)和0交换的排序
    (笔试题)合法字符串
    (笔试题)被3和5整除的数的和
    (笔试题)程序运行时间
  • 原文地址:https://www.cnblogs.com/kingshow123/p/practicec2.html
Copyright © 2011-2022 走看看