HDU-1043-八數位-代碼

解題分析過程見hdu1043八數位問題解題報告

 

//1043雙向搜尋-八數位#include <iostream>#include <string>#include <queue>using namespace std;char input[100];int state[10],s_num,e10[10],fac_n[10];char hash_T[400000],step[10000],hash_T2[400000];struct inf{    int pos;    char mode;};queue<inf> SQ;queue<inf> SQ2;inline int move(int num,char op);bool be_solved();inline int hash(int num);void BFS2();int main(){    int i,j;    for(e10[0]=1,i=1;i<=9;i++)        e10[i]=e10[i-1]*10;    for(fac_n[0]=0,fac_n[1]=1,i=2;i<=9;i++)        fac_n[i]=fac_n[i-1]*i;    while(gets(input))    {        for(i=strlen(input)-1,j=8;i>=0;i--)        {            if(input[i]!=' ')            {                if(input[i]=='x')                    state[9]=j+1;                else state[j--]=input[i]-'0';            }        }        for(s_num=0,i=9,j=1;i>0;i--,j*=10)            s_num+=state[i]*j;        if(!be_solved())            printf("unsolvable\n");        else         {            BFS2();            printf("\n");        }    }    return 0;}bool be_solved(){    int i,j,anti=0;    for(i=1;i<=8;i++)        for(j=1;j<i;j++)            if(state[i]<state[j])                anti++;    if(anti%2)return 0;    else return 1;}void BFS2(){    int to_num,to_hash,i;    char *phash,*phash2;    char op;    inf k,t;    memset(hash_T,0,sizeof(hash_T));    memset(hash_T2,0,sizeof(hash_T2));    while(!SQ2.empty())        SQ2.pop();    k.pos=s_num;k.mode=1;    SQ2.push(k);    k.pos=123456789;k.mode=2;    SQ2.push(k);    hash_T[hash(s_num)]='s';    hash_T2[hash(123456789)]='e';    while(!SQ2.empty())    {        k=SQ2.front();        SQ2.pop();        to_hash=hash(k.pos);        if(k.mode==1)            if(hash_T2[to_hash]!=0)break;            else phash=hash_T,phash2=hash_T2;        if(k.mode==2)            if(hash_T[to_hash]!=0)break;            else phash=hash_T2,phash2=hash_T;        t=k;        t.pos=move(k.pos,'r');        to_hash=hash(t.pos);        if(to_hash>=0&&phash[to_hash]==0)            phash[to_hash]='r',SQ2.push(t);        t.pos=move(k.pos,'l');        to_hash=hash(t.pos);        if(to_hash>=0&&phash[to_hash]==0)            phash[to_hash]='l',SQ2.push(t);        t.pos=move(k.pos,'u');        to_hash=hash(t.pos);        if(to_hash>=0&&phash[to_hash]==0)            phash[to_hash]='u',SQ2.push(t);        t.pos=move(k.pos,'d');        to_hash=hash(t.pos);        if(to_hash>=0&&phash[to_hash]==0)            phash[to_hash]='d',SQ2.push(t);    }    i=0;    to_hash=hash(k.pos);    to_num=k.pos;    while(hash_T[to_hash]!='s')    {        switch(step[i++]=hash_T[to_hash])        {        case 'r': op='l';break;        case 'l': op='r';break;        case 'u': op='d';break;        case 'd': op='u';break;        }        to_num=move(to_num,op);        to_hash=hash(to_num);    }    while(i>0)printf("%c",step[--i]);    to_hash=hash(k.pos);    to_num=k.pos;    while(hash_T2[to_hash]!='e')    {        switch(hash_T2[to_hash])        {        case 'r': op='l';break;        case 'l': op='r';break;        case 'u': op='d';break;        case 'd': op='u';break;        }        printf("%c",op);        to_num=move(to_num,op);        to_hash=hash(to_num);    }}inline int hash(int num){    int dig[10],i=9,j,sum,anti;    if(num==0)return -1;    while(num) dig[i]=num%10,num/=10,i--;    sum=(9-dig[9])*fac_n[8];    for(i=1;i<9;i++)    {        for(anti=0,j=1;j<i;j++)            if(dig[i]<dig[j])                anti++;        sum+=anti*fac_n[i-1];    }    return sum;}inline int move(int num,char op){    int t0,t1,t2;    switch(op)    {    case 'r':        if(num%10%3==0)return 0;        return num+1;    case 'l':        if((num-1)%10%3==0)return 0;        return num-1;    case 'u':        if(num%10-3<=0)return 0;        t0=9-num%10+1;        t1=num/e10[t0];        t2=t1%1000;        t1=t1-t2+(t2%100)*10+t2/100;        t1*=e10[t0];        return (t1+((num%e10[t0])-3));    case 'd':        if(num%10+3>9)return 0;        t0=9-num%10+1-3;        t1=num/e10[t0];        t2=t1%1000;        t1=t1-t2+(t2%10)*100+t2/10;        t1*=e10[t0];        return (t1+((num%e10[t0])+3));    }}