【hihocoder 1312】搜索三·启发式搜索(普通广搜做法)

【题目链接】:http://hihocoder.com/problemset/problem/1312?sid=1092352

【题意】

【题解】

从末状态的123456780开始逆向搜;
看它能到达哪些状态;
到时候O(1)输出就可以了;
用map< int,int> dic来判重;
对于状态;
用数组表示;
然后把它转化成一个对应的十进制数;

【Number Of WA】

0

【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 pb push_back
#define fi first
#define se second
#define ms(x,y) memset(x,y,sizeof 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 int cs[9] = {1,2,3,4,5,6,7,8,0};
const double pi = acos(-1.0);
const int N = 110;

struct node
{
    int a[9],p,step;
};

node init;
queue <node> dl;
map <int,int> dic;
int a[9];

int has(int a[9])
{
    int x = 0;
    rep1(i,0,8)
        x = x*10 + a[i];
    return x;
}

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    ios::sync_with_stdio(false),cin.tie(0);//scanf,puts,printf not use

    rep1(i,0,8) init.a[i] = cs[i];
    init.p = 8,init.step = 0;
    dl.push(init);
    dic[has(init.a)] = 0;

    while (!dl.empty())
    {
        int p = dl.front().p;
        int now = dl.front().step;
        node temp;
        rep1(i,0,8) temp.a[i] = dl.front().a[i];
        dl.pop();

        //上
        if (p>2)
        {
            int tp = p-3;
            swap(temp.a[tp],temp.a[p]);
            int xzt = has(temp.a);
            if (dic.find(xzt)==dic.end())
            {
                dic[xzt] = now+1;
                temp.step = now+1;
                temp.p = tp;
                dl.push(temp);
            }
            swap(temp.a[tp],temp.a[p]);
        }
        //下
        if (p<6)
        {
            int tp = p+3;
            swap(temp.a[tp],temp.a[p]);
            int xzt = has(temp.a);
            if (dic.find(xzt)==dic.end())
            {
                dic[xzt] = now+1;
                temp.step = now+1;
                temp.p = tp;
                dl.push(temp);
            }
            swap(temp.a[tp],temp.a[p]);
        }

        //左
        if (p%3!=0)
        {
            int tp = p-1;
            swap(temp.a[tp],temp.a[p]);
            int xzt = has(temp.a);
            if (dic.find(xzt)==dic.end())
            {
                dic[xzt] = now+1;
                temp.step = now+1;
                temp.p = tp;
                dl.push(temp);
            }
            swap(temp.a[tp],temp.a[p]);
        }

        //右
        if (p%3!=2)
        {
            int tp = p+1;
            swap(temp.a[tp],temp.a[p]);
            int xzt = has(temp.a);
            if (dic.find(xzt)==dic.end())
            {
                dic[xzt] = now+1;
                temp.step = now+1;
                temp.p = tp;
                dl.push(temp);
            }
            swap(temp.a[tp],temp.a[p]);
        }
    }
    int t;
    cin >> t;
    while (t--)
    {
        rep1(i,0,8)
            cin >> a[i];
        int zt = has(a);
        if (dic.find(zt)!=dic.end())
            cout << dic[zt] << endl;
        else
            cout << "No Solution!" << endl;
    }
    return 0;
}