题目描述

Scenario

The 15-puzzle has been around for over 100 years; even if you don’t know it by that name, you’ve seen it. It is constructed with 15 sliding tiles, each with a number from 1 to 15 on it, and all packed into a 4 by 4 frame with one tile missing. Let’s call the missing tile ‘x’; the object of the puzzle is to arrange the tiles so that they are ordered as:

1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 x

where the only legal operation is to exchange ‘x’ with one of the tiles with which it shares an edge. As an example, the following sequence of moves solves a slightly scrambled puzzle:

1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
5 6 7 8 5 6 7 8 5 6 7 8 5 6 7 8
9 x 10 12 9 10 x 12 9 10 11 12 9 10 11 12
13 14 11 15 13 14 11 15 13 14 x 15 13 14 15 x
r-> d-> r->

The letters in the previous row indicate which neighbor of the ‘x’ tile is swapped with the ‘x’ tile at each step; legal values are ‘r’,’l’,’u’ and ‘d’, for right, left, up, and down, respectively.

Not all puzzles can be solved; in 1870, a man named Sam Loyd was famous for distributing an unsolvable version of the puzzle, and
frustrating many people. In fact, all you have to do to make a regular puzzle into an unsolvable one is to swap two tiles (not counting the missing ‘x’ tile, of course).

In this problem, you will write a program for solving the less well-known 8-puzzle, composed of tiles on a three by three
arrangement.

题目分析

就是根据每两个位置之间的曼哈顿距离作为估值函数。。。然后用康拓展开判重。

代码

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <vector>
#include <queue>
#include <stack>
#include <iostream>
#include <string>
using namespace std;
const int MAXVIS = 363900;
struct State{unsigned short s[9];int h, g, mh;bool operator < (const State& s) const {return (g+h)>(s.g+s.h);}bool operator > (const State& s) const {if((g+h) == (s.g+s.h))return g < (s.g+s.h);return (g+h)<(s.g+s.h);}
}st, ed;
priority_queue<State> que;
char now[MAXVIS+5]; int pre[MAXVIS+5];
int _pow[9]; pair<int, int> show[9];
int fx[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
int ppos[9][2] = {{0, 0}, {0, 1}, {0, 2}, {1, 0}, {1, 1}, {1, 2}, {2, 0}, {2, 1}, {2, 2}};
int fppos[3][3] = {{0, 1, 2}, {3, 4, 5}, {6, 7, 8}};
char ch[4] = {'d', 'r', 'u', 'l'};
int vis[MAXVIS+5];
int _hash(unsigned short s[]){int ret = 0, count =0 ;for(int i=0;i<9;i++){count = 0;for(int j=i+1;j<9;j++)count += int(s[j]<s[i]);ret += count * _pow[8-i];}return ret;
}
int _abs(int u){return u>0?u:-u;}
int _h(unsigned short s[]){int ret = 0;for(int i=0;i<9;i++) if(s[i])ret += _abs(ppos[i][0]-show[s[i]].first) + _abs(ppos[i][1]-show[s[i]].second);return ret;
}
bool check_pos(const int& a, const int& b){if(a > 2 || b > 2 || a < 0 || b < 0) return false;return true;
}
stack<char> sta;
void Print(){while(!sta.empty()) sta.pop();int nowm = ed.mh;while(nowm != st.mh) {sta.push(now[nowm]);nowm = pre[nowm];}while(!sta.empty()){putchar(sta.top());sta.pop();}puts("");
}
void solve(){int counter = 0;for(int i=0;i<9;i++) if(st.s[i]){for(int j=i+1;j<9;j++) if(st.s[j]){if(st.s[i] > st.s[j]) counter++;}}if(counter & 1){cout<<"unsolvable"<<endl;return ;}memset(vis, 0x3f, sizeof vis);while(!que.empty()) que.pop();State tmp=st; State t2;now[tmp.mh] = 0;int pos;vis[tmp.mh] = tmp.h = _h(st.s); tmp.g = 0;que.push(tmp);while(!que.empty()){tmp = que.top(); que.pop();pos=0;if(tmp.mh == ed.mh){Print();return ;}if(tmp.g + tmp.h > vis[tmp.mh]) continue;while(tmp.s[pos]) pos++;for(int i=0;i<4;i++) {if(check_pos(ppos[pos][0] + fx[i][0], ppos[pos][1] + fx[i][1])){t2 = tmp;swap(t2.s[pos], t2.s[fppos[ppos[pos][0] + fx[i][0]][ppos[pos][1] + fx[i][1]]]);t2.mh = _hash(t2.s); t2.h = _h(t2.s); (++t2.g);if(t2.g + t2.h > vis[t2.mh]) continue;vis[t2.mh] = t2.g + t2.h;now[t2.mh] = ch[i]; pre[t2.mh] = tmp.mh;que.push(t2);}}}cout<<"unsolvable"<<endl;
}
int main(){char tstr[5];_pow[0] = 1;for(int i=1;i<=8;i++) _pow[i] = _pow[i-1] * i;ed.s[0] = 1; ed.s[1] = 2; ed.s[2] = 3; ed.s[3] = 4;ed.s[4] = 5; ed.s[5] = 6; ed.s[6] = 7; ed.s[7] = 8;ed.s[8] = 0;ed.mh = _hash(ed.s);for(int i=0;i<9;i++){show[ed.s[i]] = make_pair(i/3, i%3);}while(true){for(int i=0;i<9;i++){if(scanf("%s", tstr) == EOF) return 0;if(tstr[0] != 'x')st.s[i]=tstr[0]-'0';else st.s[i] = 0;}st.mh = _hash(st.s);if(st.mh == ed.mh) puts("");else solve();}return 0;
}