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  • hdu 5012(bfs)

    题意:给你2个 骰子,让你通过翻转使第一个变成第二个,求最少翻转数

    思路:bfs


    #include<cstdio>
    #include<iostream>
    #include<cstring>
    #include<cstdlib>
    #include<cmath>
    #include<algorithm>
    #include<string>
    #include<map>
    #include<set>
    #include<vector>
    #include<queue>
    #include<stack>
    using namespace std;
    typedef long long ll;
    
    struct node
    {
        int arr[6];
        int ci;
    };
    
    int vis[1000000];
    int len;
    bool ju(node a, node b)
    {
        for (int i = 0; i < 6; i++)
            if (a.arr[i] != b.arr[i])
                return false;
        return true;
    }
    
    bool sol(node c)
    {
        int num = 0;
        for (int i = 0; i < 6; i++)
        {
            num = num * 10 + c.arr[i];
        }
        if(vis[num])
            return false;
        else{
            vis[num] = 1;
            return true;
        }
    }
    
    node turn(node a, int i)
    {
        node c;
        if (i == 1)
        {
            c.arr[0] = a.arr[3];
            c.arr[1] = a.arr[2];
            c.arr[2] = a.arr[0];
            c.arr[3] = a.arr[1];
            c.arr[4] = a.arr[4];
            c.arr[5] = a.arr[5];
        }
        if (i == 2)
        {
            c.arr[0] = a.arr[2];
            c.arr[1] = a.arr[3];
            c.arr[2] = a.arr[1];
            c.arr[3] = a.arr[0];
            c.arr[4] = a.arr[4];
            c.arr[5] = a.arr[5];
        }
        if (i == 3)
        {
            c.arr[0] = a.arr[5];
            c.arr[1] = a.arr[4];
            c.arr[2] = a.arr[2];
            c.arr[3] = a.arr[3];
            c.arr[4] = a.arr[0];
            c.arr[5] = a.arr[1];
        }
        if (i == 4)
        {
            c.arr[0] = a.arr[4];
            c.arr[1] = a.arr[5];
            c.arr[2] = a.arr[2];
            c.arr[3] = a.arr[3];
            c.arr[4] = a.arr[1];
            c.arr[5] = a.arr[0];
        }
        return c;
    }
    
    int main()
    {
        node p1,p2;
        while(scanf("%d%d%d%d%d%d",&p1.arr[0],&p1.arr[1],&p1.arr[2],&p1.arr[3],&p1.arr[4],&p1.arr[5]) != EOF)
        {
            scanf("%d%d%d%d%d%d",&p2.arr[0],&p2.arr[1],&p2.arr[2],&p2.arr[3],&p2.arr[4],&p2.arr[5]);
            memset(vis,0,sizeof(vis));
            queue<node> q;
            len = p1.ci = p2.ci = 0;
            q.push(p1);
            int flag = 0;
            while(!q.empty())
            {
                node tmp = q.front();
                q.pop();
                if(ju(tmp,p2))
                {
                    printf("%d
    ",tmp.ci);
                    flag = 1;
                    break;
                }
                for(int i = 1; i <= 4; i++)
                {
                    node c = turn(tmp,i);
                    if(sol(c))
                    {
                        c.ci = tmp.ci + 1;
                        q.push(c);
                    }
                }
            }
            if(!flag)
                printf("-1
    ");
        }
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/Przz/p/5409762.html
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