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  • 倒油瓶问题(平分水问题)(bfs)

    #include<iostream>
    #include<stdio.h>
    #include<string.h>
    #include<queue>
    using namespace std;
    int vist[105][105][105],a,b,c;
    struct node
    {
        int a,b,c;
        int step;
    }s[105];
    int sum=0;
    void bfs()
    {
        queue<node>q;
        memset(vist,0,sizeof(vist));
        node p1;
        p1.a=a;
        p1.b=0;
        p1.c=0;
        p1.step=0;
        q.push(p1);
        vist[p1.a][0][0]=1;
        while(!q.empty())
        {
            p1=q.front();
            q.pop();
            if((p1.a==a/2&&p1.b==a/2)||(p1.a==a/2&&p1.c==a/2)||(p1.b==a/2&&p1.c==a/2))
            {
                printf("%d ",p1.step);
                return;
            }
            node p2;   
            if(p1.a!=0)
            {
                if(p1.a>b-p1.b)
                {
                    p2.a=p1.a-(b-p1.b);
                    p2.b=b;
                    p2.c=p1.c;
                    p2.step=p1.step+1;
                }
                else
                {
                    p2.a=0;
                    p2.b=p1.b+p1.a;
                    p2.c=p1.c;
                    p2.step=p1.step+1;
                }
                if(!vist[p2.a][p2.b][p2.c])
                {
                    vist[p2.a][p2.b][p2.c]=1;
                    q.push(p2);
                }
            }
             
            if(p1.a!=0)
            {
                if(p1.a>c-p1.c)
                {
                    p2.a=p1.a-(c-p1.c);
                    p2.b=p1.b;
                    p2.c=c;
                    p2.step=p1.step+1;
                }
                else
                {
                    p2.a=0;
                    p2.b=p1.b;
                    p2.c=p1.c+p1.a;
                    p2.step=p1.step+1;
                }
                if(!vist[p2.a][p2.b][p2.c])
                {
                    vist[p2.a][p2.b][p2.c]=1;
                    q.push(p2);
                }
            }
             
            if(p1.b!=0)
            {
                if(p1.b>a-p1.a)
                {
                    p2.b=p1.b-(a-p1.a);
                    p2.a=a;
                    p2.c=p1.c;
                    p2.step=p1.step+1;
                }
                else
                {
                    p2.b=0;
                    p2.a=p1.a+p1.b;
                    p2.c=p1.c;
                    p2.step=p1.step+1;
                }
                if(!vist[p2.a][p2.b][p2.c])
                {
                    vist[p2.a][p2.b][p2.c]=1;
                    q.push(p2);
                }
            }
             
            if(p1.b!=0)
            {
                if(p1.b>c-p1.c)
                {
                    p2.b=p1.b-(c-p1.c);
                    p2.a=p1.a;
                    p2.c=c;
                    p2.step=p1.step+1;
                }
                else
                {
                    p2.b=0;
                    p2.a=p1.a;
                    p2.c=p1.c+p1.b;
                    p2.step=p1.step+1;
                }
                if(!vist[p2.a][p2.b][p2.c])
                {
                    vist[p2.a][p2.b][p2.c]=1;
                    q.push(p2);
                }
            }
             
            if(p1.c!=0)
            {
                if(p1.c>a-p1.a)
                {
                    p2.c=p1.c-(a-p1.a);
                    p2.a=a;
                    p2.b=p1.b;
                    p2.step=p1.step+1;
                }
                else
                {
                    p2.c=0;
                    p2.a=p1.a+p1.c;
                    p2.b=p1.b;
                    p2.step=p1.step+1;
                }
                if(!vist[p2.a][p2.b][p2.c])
                {
                    vist[p2.a][p2.b][p2.c]=1;
                    q.push(p2);
                }
            }
             
            if(p1.c!=0)
            {
                if(p1.c>b-p1.b)
                {
                    p2.c=p1.c-(b-p1.b);
                    p2.a=p1.a;
                    p2.b=b;
                    p2.step=p1.step+1;
                }
                else
                {
                    p2.c=0;
                    p2.a=p1.a;
                    p2.b=p1.b+p1.c;
                    p2.step=p1.step+1;
                }
                if(!vist[p2.a][p2.b][p2.c])
                {
                    vist[p2.a][p2.b][p2.c]=1;
                    q.push(p2);
                }
            }
        }
        printf("NO ");
    }
    int main()
    {
        while(scanf("%d%d%d",&a,&b,&c)>0&&(a+b+c))
        {
            if(a%2==1)
            {
                printf("NO ");
                continue;
            }
            bfs();
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/lengxia/p/4469726.html
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