【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;
}