poj 1184

经典的宽搜题目,感觉最好的办法应该是双向广搜。

不过用简单的启发式搜索可以飘过。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
int a,b;
char ans[1111111][7];
int inf[7]={1,1,10,100,1000,10000,100000};
struct D
{
    int key;
    char x,sum,now;
    bool operator <(const struct D & xx) const
    {
        if(sum==xx.sum)
        return now<xx.now;
        return sum>xx.sum;
    }
};
priority_queue <D> q;

int cal(int key,int x,int b)
{
    int from[7],to[7];
    for(int i=1;i<=6;i++)
    {
        from[i]=key%10;
        key/=10;
        to[i]=b%10;
        b/=10;
    }
    int ret=0;
    for(int i=1;i<=6;i++)
    if(i!=x)
    {
        ret+=from[i]!=to[i];
    }
    sort(from+1,from+1+6);
    sort(to+1,to+1+6);
    for(int i=1;i<=6;i++)
    {
        if(from[i]-to[i]>0)
        ret+=from[i]-to[i];
        else
        ret+=to[i]-from[i];
    }
    return ret;
}

int bfs(int a,int b)
{
    memset(ans,100,sizeof(ans));
    ans[a][1]=0;
    struct D xx;
    xx.key=a;
    xx.sum=0+cal(a,1,b);
    xx.x=1;
    xx.now=0;
    q.push(xx);
    while(1)
    {
        int key=q.top().key;
        int x=q.top().x;
        int tmp=ans[key][x];
        q.pop();
        if(key==b)
        return tmp;
        if(x<6&&tmp+1<ans[key][x+1])
        {
            xx.now=ans[key][x+1]=tmp+1;
            xx.key=key;
            xx.x=x+1;
            xx.sum=tmp+cal(key,x+1,b);
            q.push(xx);
        }
        int txt=inf[6-x+1];
        int keyx=key/txt%10;
        if(keyx>0&&tmp+1<ans[key-1*txt][x])
        {
            xx.now=ans[key-1*txt][x]=tmp+1;
            xx.key=key-1*txt;
            xx.x=x;
            xx.sum=tmp+cal(key-1*txt,x,b);
            q.push(xx);
        }
        if(keyx<9&&tmp+1<ans[key+1*txt][x])
        {
            xx.now=ans[key+1*txt][x]=tmp+1;
            xx.key=key+1*txt;
            xx.x=x;
            xx.sum=tmp+cal(key+1*txt,x,b);
            q.push(xx);
        }
        int tt=key/inf[6];
        int to=key%inf[6]+keyx*inf[6]-keyx*txt+tt*txt;
        if(tmp+1<ans[to][x])
        {
            xx.now=ans[to][x]=tmp+1;
            xx.key=to;
            xx.x=x;
            xx.sum=tmp+cal(to,x,b);
            q.push(xx);
        }
        tt=key%10;
        to=key/10*10+keyx-keyx*txt+tt*txt;
        if(tmp+1<ans[to][x])
        {
            xx.now=ans[to][x]=tmp+1;
            xx.key=to;
            xx.x=x;
            xx.sum=tmp+cal(to,x,b);
            q.push(xx);
        }
    }
}

int main()
{
//    freopen("in.txt","r",stdin);
    scanf("%d %d",&a,&b);
    int ans=bfs(a,b);
    cout<<ans<<endl;
    return 0;
}


 


 

 

posted @ 2013-08-09 23:43  javawebsoa  Views(268)  Comments(0Edit  收藏  举报