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;}