【hihocoder 1312】搜索三·启发式搜索(启发式搜索写法)

【题目链接】:http://hihocoder.com/problemset/problem/1312?sid=1092363

【题意】

【题解】



定义一个A*函数

f = step+val

这里的val是当前这个状态;每个点到目标状态的点的曼哈顿距离的绝对值;

(这个值肯定比真正需要花费的路程短)

step就为当前状态花费的步数;

把普通队列改成优先队列;

优先处理f值小的状态;

f值相同的,优先处理step值小的;

(也就是说f值大的不是不处理了,而是放到后面再处理)

这样就能较快地逼近目标状态了;

效果异常地棒

2s变成0.2s了!



【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 pre[9][2] =
{
    {2,2},
    {0,0},{0,1},{0,2},
    {1,0},{1,1},{1,2},
    {2,0},{2,1}
};
const double pi = acos(-1.0);
const int N = 110;

struct node
{
    int a[9],p,step,f;
    friend bool operator < (node x,node y)
    {
        if (x.f==y.f)
        {
            if (x.step==y.step)
                return true;
            else
                return x.step>y.step;
        }
        else
            return x.f>y.f;
    }
};

node init;
priority_queue <node> dl;
map <int,int> dic;
int a[9],goal;
int cs[9] = {1,2,3,4,5,6,7,8,0};

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

int val(int *a)
{
    int ret = 0;
    rep1(i,0,8)
    {
        int x = i/3,y = i%3;
        if (a[i]==0) continue;
        ret+=abs(pre[a[i]][0]-x)+abs(pre[a[i]][1]-y);
    }
    return ret;
}

int bfs()
{
    while (!dl.empty())
    {
        int p = dl.top().p;
        int now = dl.top().step;
        node temp;
        rep1(i,0,8) temp.a[i] = dl.top().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 (xzt==goal) return now+1;
            if (dic.find(xzt)==dic.end())
            {
                dic[xzt] = now+1;
                temp.step = now+1;
                temp.p = tp;
                temp.f = temp.step+val(temp.a);
                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 (xzt==goal) return now+1;
            if (dic.find(xzt)==dic.end())
            {
                dic[xzt] = now+1;
                temp.step = now+1;
                temp.p = tp;
                temp.f = temp.step+val(temp.a);
                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 (xzt==goal) return now+1;
            if (dic.find(xzt)==dic.end())
            {
                dic[xzt] = now+1;
                temp.step = now+1;
                temp.p = tp;
                temp.f = temp.step+val(temp.a);
                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 (xzt==goal) return now+1;
            if (dic.find(xzt)==dic.end())
            {
                dic[xzt] = now+1;
                temp.step = now+1;
                temp.p = tp;
                temp.f = temp.step+val(temp.a);
                dl.push(temp);
            }
            swap(temp.a[tp],temp.a[p]);
        }
    }
    return -1;
}

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

    int t;
    cin >> t;
    while (t--)
    {
        dic.clear();
        rep1(i,0,8)
        {
            cin >> a[i];
            if (a[i]==0) init.p = i;
        }

        rep1(i,0,8) init.a[i] = a[i];
        init.step = 0,init.f = init.step+val(init.a);
        dic[has(init.a)] = 0;
        if (has(init.a)==goal)
        {
            cout << 0 << endl;
            continue;
        }
        while (!dl.empty()) dl.pop();
        dl.push(init);
        int ans = bfs();
        if (ans==-1)
            cout <<"No Solution!"<<endl;
        else
            cout << ans << endl;

    }

    return 0;
}