【hdu 1043】Eight

【题目链接】:http://acm.hdu.edu.cn/showproblem.php?pid=1043

【题意】



会给你很多组数据;

让你输出这组数据到目标状态的具体步骤;

【题解】



从12345678x这个目标状态开始bfs然后获取它能够到达的所有状态;

然后在队列中记录下执行这一步用的是那个操作;

递归输出就好;

map来判定这状态存不存在;

在队列里面存这个状态

就按照所给的存成1维的数组就好;

x用9代替;(用0也可以哦);

模拟八数码的操作就好;

因为是逆序的,所以记录的操作是相反的;

最后递归写输出的时候搞晕了；

又WA了好多次.



【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define ps push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%lld",&x)
#define ref(x) scanf("%lf",&x)

typedef pair<int, int> pii;
typedef pair<LL, LL> pll;

const int dx[9] = { 0,1,-1,0,0,-1,-1,1,1 };
const int dy[9] = { 0,0,0,-1,1,-1,1,-1,1 };
const double pi = acos(-1.0);
const int N = 200e4;
const int chushi[10] = { 0,1,2,3,4,5,6,7,8,9 };

struct abc
{
    int s[15];
    int pos,ope,pre;
};

map<int, int> dic;
abc dl[N];
int ts[15];
string t;

int zt(int s[15])
{
    int r = 0;
    rep1(i, 1, 9)
        r = r * 10 + s[i];
    return r;
}

void bfs()
{
    abc temp,temp1;
    rep1(i, 0, 9) temp.s[i] = chushi[i];
    temp.pos = 9;
    dic[zt(temp.s)] = 1;
    int head = 1, tail = 1;
    dl[1] = temp;
    while (head <= tail)
    {
        abc now = dl[head++];
        int ss[15];
        rep1(i, 1, 9) ss[i] = now.s[i];
        int t = now.pos;
        //up
        if (t > 3)
        {
            int tt = t - 3;
            swap(ss[tt], ss[t]);
            int pc = zt(ss);
            if (!dic[pc])
            {
                rep1(i, 1, 9) temp1.s[i] = ss[i];
                temp1.pos = tt, temp1.pre = head - 1, temp1.ope = 2;
                dl[++tail] = temp1;
                dic[pc] = tail;
            }
            swap(ss[tt], ss[t]);
        }
        //down
        if (t<7)
        {
            int tt = t + 3;
            swap(ss[tt], ss[t]);
            int pc = zt(ss);
            if (!dic[pc])
            {
                rep1(i, 1, 9) temp1.s[i] = ss[i];
                temp1.pos = tt, temp1.pre = head - 1, temp1.ope = 1;
                dic[pc] = tail+1;
                dl[++tail] = temp1;
            }
            swap(ss[tt], ss[t]);
        }
        //left
        if ((t % 3) != 1)
        {
            int tt = t - 1;
            swap(ss[tt], ss[t]);
            int pc = zt(ss);
            if (!dic[pc])
            {
                rep1(i, 1, 9) temp1.s[i] = ss[i];
                temp1.pos = tt, temp1.pre = head - 1, temp1.ope = 4;
                dic[pc] = tail+1;
                dl[++tail] = temp1;
            }
            swap(ss[tt], ss[t]);
        }
        //right
        if ((t % 3) != 0)
        {
            int tt = t + 1;
            swap(ss[tt], ss[t]);
            int pc = zt(ss);
            if (!dic[pc])
            {
                rep1(i, 1, 9) temp1.s[i] = ss[i];
                temp1.pos = tt, temp1.pre = head - 1, temp1.ope = 3;
                dic[pc] = tail+1;
                dl[++tail] = temp1;
            }
            swap(ss[tt], ss[t]);
        }
    }
}

void pri(int now){
    if (now == 1) return;
    int ju = dl[now].ope;
    if (ju == 1) printf("u");
    if (ju == 2) printf("d");
    if (ju == 3) printf("l");
    if (ju == 4) printf("r");
    pri(dl[now].pre);
}

int main()
{
    //freopen("F:\\rush.txt", "r", stdin);
    bfs();
    while (getline(cin,t)){
        int num = 0;
        int len = t.size();
        rep1(i,0,len-1)
        {
            if (t[i]=='x') t[i] = '9';
            if (t[i]>='0' && t[i]<='9') ts[++num] = t[i]-'0';
            if (num==9) break;
        }
        int pc=dic[zt(ts)];
        if (pc){
            pri(pc);
            puts("");
        }
        else
            puts("unsolvable");
    }
    //printf("\n%.2lf sec \n", (double)clock() / CLOCKS_PER_SEC);
    return 0;
}