八数码总结

题意：

八数码问题也称为九宫问题。编号为1~8的8个正方形滑块被摆成3行3列，棋盘上还有一个空格，每次可以把与空格相邻的滑块移到空格中，而它原来的位置就成了新的空格。给定初始局面和目标局面，计算出最少的移动步数。

参考来源：八数码的八境界

方法1：

暴搜+queue，如果空格用’0’表示的话可以直接将状态压缩成一个9位整数，也可以用字符串表示状态，用结构体表示状态+步数+空格的位置。

代码：

/*
264137058
815736402
*/
#include<iostream>
#include<map>
#include<cstring>
#include<queue>
#include<cstdio>
using namespace std;
map<string, int>vis;
int m, n;
struct state{string str; int pos; int dist;};
state s;
string goal;
int dx[4] = {-1, 1, 0, 0};
int dy[4] = {0, 0, -1, 1};
int bfs()
{
queue<state>q;
q.push(s);
vis[s.str] = 1;
while(!q.empty()){
state t = q.front();q.pop();
if(t.str.compare(goal) == 0) return t.dist;
int p= t.pos;
int x= p / 3, y = p % 3;
for(int i = 0; i < 4; i++){
int newx = x + dx[i];
int newy = y + dy[i];
int newpos = newx * 3 + newy;
string a = t.str;
if(newx >= 0 && newx < 3 && newy >= 0 && newy < 3){
swap(a[p], a[newpos]);
if(vis[a]) continue;
vis[a] = 1;
q.push((state){a, newpos, t.dist + 1});
}
}
}
return -1;
}
int main (void)
{
string st;
int pos;
cin>>st;
cin>>goal;
for(int i = 0; i < 9; i++){
if(st[i] == '0')  pos = i;
}
s = (state){st, pos, 0};
int res = bfs();
if(res == -1) cout<<"No solution"<<endl;
else cout<<res<<endl;
}

方法2：

模拟队列+康拓展开/哈希/STL set判重，学习紫书上的代码~

注意其中康拓展开

代码：

康拓展开

#include<iostream>
#include<cstring>
#include<set>
#include<cstdio>
using namespace std;
typedef int state[9];//表示state 代表 int[9]
const int maxn = 1e7 + 5;
state st[maxn], sta, goal;
int dist[maxn];
int dx[4] = {-1, 1, 0, 0};
int dy[4] = {0, 0, -1, 1};
/*
2 6 4 1 3 7 0 5 8
8 1 5 7 3 6 4 0 2
*/
/******把0~8的全排列和0~362879的整数一一对应起来****/
int vis[400000], fact[10];
void init()
{
memset(vis, 0, sizeof(vis));
fact[0] = 1;
for(int i = 1; i < 9; i++) fact[i] = fact[i - 1] * i;
}
int Insert(int s)
{
int code = 0;
for(int i = 0; i < 9; i++){
int cnt = 0;
for(int j = i + 1; j < 9; j++)
if(st[s][j] < st[s][i]) cnt++;
code += fact[8 - i] * cnt;
}
if(vis[code]) return 0;
return vis[code] = 1;
}
int bfs()
{
init();
int fro = 0, rear = 1;
dist[fro] = 0;
Insert(fro);
while(fro < rear){
state& s = st[fro];
if (memcmp(goal, s, sizeof(s)) == 0) return dist[fro];
int pos;
for(pos = 0; pos < 9; pos++) if(!s[pos]) break;
int x = pos / 3, y = pos % 3;
for (int i = 0; i < 4; i++){
int newx = x + dx[i];
int newy = y + dy[i];
int newpos = newx * 3 + newy;
if (newx >= 0 && newx < 3 && newy >= 0 && newy < 3){
state& t = st[rear];
memcpy(&t, &s, sizeof(s));
t[newpos] = 0;
t[pos] = s[newpos];
dist[rear] = dist[fro] + 1;
if(Insert(rear)) rear++;
}
}
fro++;
}
return -1;
}
int main (void)
{
for(int i = 0; i < 9; i++) cin>>st[0][i];
for(int i = 0; i < 9; i++) cin>>goal[i];
int res = bfs();
if(res != -1) cout<<res<<endl;
else cout<<"No solution"<<endl;
return 0;
}

哈希：

const int hashsize = 1000003;//大质数
int head[hashsize], next[hashsize];
void init()
{
memset(head, 0, sizeof(head));
}
int Hash(int s)
{
int tmp = 0;
for(int i = 0; i < 9; i++)
tmp = tmp * 10 + st[s][i];
return tmp % hashsize;
}
int Insert(int s)
{
int h = Hash(s);
int u = head[h];
while(u){
if(memcpy(st[u], st[s], sizeof(st[s])) == 0)  return 0;
u = next[u];
}
next[s] = head[h];
head[h] = s;
return 1;
}

set判重：

set<int>vis;
void init()
{
vis.clear();
}
int Insert(int s)
{
int v  = 0;
for(int i = 0; i < 9; i++) v = v* 10 + st[s][i];
if(vis.count(v)) return 0;
vis.insert(v);
return 1;
}

方法3：

模拟队列+双向bfs~~这里判重随便选一个就好啦~~节约一半的时间和一半的空间，真心快多了~

代码：

#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
typedef int state[9];
const int maxn = 1e7 + 5;
state st[maxn];
int dx[4] = {-1, 1, 0, 0};
int dy[4] = {0, 0, -1, 1};
int fact[maxn], vis[maxn], dist[3][maxn];
/*
2 6 4 1 3 7 0 5 8
8 1 5 7 3 6 4 0 2
*/
void init()
{
memset(vis, 0, sizeof(vis));
fact[0] = 1;
for(int i = 1; i < 9; i++)
fact[i] = fact[i - 1] * i;
}
int canto(state s)
{
int code = 0;
for(int i = 0; i < 9; i++){
int cnt = 0;
for(int j = i + 1; j < 9; j++){
if(s[j] < s[i]) cnt++;
}
code += fact[8 - i] * cnt;
}
return code;
}
int bfs()
{
init();
int cs = canto(st[0]);
int ct = canto(st[1]);
dist[1][cs] = 0;
dist[2][ct] = 0;
vis[cs] = 1;
vis[ct] = 2;
int fro = 0, rear  = 2;
while(fro < rear){
state& s = st[fro];
cs = canto(s);
int pos;
for(pos = 0; pos < 9; pos++) if(!s[pos]) break;
int x = pos / 3, y = pos % 3;
for (int i = 0; i < 4; i++){
int newx = x + dx[i];
int newy = y + dy[i];
int newpos = newx * 3 + newy;
if (newx >= 0 && newx < 3 && newy >= 0 && newy < 3){
state& t = st[rear];
memcpy(&t, &s, sizeof(s));
t[pos] = s[newpos];
t[newpos] = s[pos];
ct = canto(t);
if(!vis[ct]){
vis[ct] = vis[cs];
dist[vis[cs]][ct] = dist[vis[cs]][cs] + 1;
rear++;
}else if(vis[ct] != vis[cs]){
return dist[vis[ct]][ct] + dist[vis[cs]][cs] + 1;
}
}
}
fro++;
}
return -1;
}
int main (void)
{
for(int i = 0; i < 9; i++) cin>>st[0][i];
for(int i = 0; i < 9; i++) cin>>st[1][i];
int res = bfs();
if(res != -1) cout<<res<<endl;
else cout<<"No solution"<<endl;
return 0;
}

方法4：

A* + 优先级队列（小顶堆）+ 曼哈顿距离

A*搜索算法

代码：

#include<iostream>
#include<cstring>
#include<cmath>
#include<queue>
/*
1 2 3 4 5 6 7 8 0
1 2 3 4 5 6 0 7 8

2 6 4 1 3 7 0 5 8
8 1 5 7 3 6 4 0 2
*/
using namespace std;
struct state
{
int a[9];
int pos;
int f,g;
int can;
bool operator < ( const state s2)const {
return f > s2.f;
}
};
const int maxn = 1e7 + 5;
state sta, goal;
int dx[4] = {-1, 1, 0, 0};
int dy[4] = {0, 0, -1, 1};
int fact[maxn], vis[maxn];
void init()
{
fact[0] = 1;
for(int i = 1; i < 9; i++)
fact[i] = fact[i - 1] * i;
}
int canto(state s)
{
int code = 0;
for(int i = 0; i < 9; i++){
int cnt = 0;
for(int j = i + 1; j < 9; j++){
if(s.a[j] < s.a[i]) cnt++;
}
code += fact[8 - i] * cnt;
}
return code;
}
int geth(state be)
{
int res = 0;
for(int i = 0; i < 9; i++){
int tmp = goal.a[i];
for(int j = 0; j < 9; j++){
if(be.a[j] == tmp)
res += abs(i/3 - j/3) + abs(i % 3 - j % 3);
break;
}
}
return res;
}
int astar()
{
priority_queue<state> q;
q.push(sta);
vis[sta.can] = 1;
while(!q.empty()){
state s = q.top();
q.pop();
if(s.can == goal.can) return s.g;
int x = s.pos / 3, y = s.pos % 3;
for(int i = 0; i < 4; i++){
int newx = x + dx[i];
int newy = y + dy[i];
int newpos = newx * 3 + newy;
if (newx >= 0 && newx < 3 && newy >= 0 && newy < 3){
state t = s;
t.a[t.pos] = s.a[newpos];
t.a[newpos] = s.a[t.pos];
t.pos = newpos;
t.can = canto(t);
if(vis[t.can]) continue;
t.g = s.g + 1;
t.f= geth(t) + t.g;
q.push(t);
vis[t.can] = 1;
}
}
}
return -1;
}
void print (state t)
{
for(int i = 0; i < 9; i++)
cout<<t.a[i]<<' ';
cout<<endl;
cout<<"pos"<<t.pos<<endl;
cout<<"can"<<t.can<<endl;
cout<<"f"<<t.f<<endl;
}
int main (void)
{
init();
for(int i = 0; i < 9; i++){
cin>>sta.a[i];
if(sta.a[i] == 0){
sta.pos = i;
}
}
sta.g = 0;
sta.can = canto(sta);
//print(sta);
for(int i = 0; i < 9; i++) cin>>goal.a[i];
goal.can = canto(goal);
//print(goal);

int res = astar();
if(res != -1) cout<<res<<endl;
else cout<<"No solution"<<endl;
return 0;
}

方法5：

IDA* + 曼哈顿距离

代码：

#include<iostream>
#include<queue>
#include<cstring>
#include<cmath>
using namespace std;
const int maxn = 1e7 + 5;
int maxdepth;
int dx[4] = {1, 0 , 0, -1};
int dy[4] = {0, 1, -1, 0};
/*
1 2 3 4 5 6 7 8 0
1 2 3 4 5 6 0 7 8

2 6 4 1 3 7 0 5 8
8 1 5 7 3 6 4 0 2
*/
struct state
{
int a[9];
int pos;
};
state st[maxn];
state goal, sta;
int geth(state be)
{
int res = 0;
for(int i = 0; i < 9; i++){
int tmp = goal.a[i];
if(tmp == 0) continue;
for(int j = 0; j < 9; j++){
if(be.a[j] == tmp){
res += abs(i/3 - j/3) + abs(i % 3 - j % 3);
break;
}
}
}
return res;
}
void print(state s)
{
for(int i = 0; i < 9; i++)
cout<<s.a[i]<<' ';
cout<<endl;
}
int dfs(state s, int h, int depth, int prei)
{
if(h == 0) return 1;
// print(s);
if(depth + h > maxdepth) return 0;
int x = s.pos / 3, y = s.pos % 3;
for (int i = 0; i < 4; i++){
if(prei + i == 3) continue;
int newx = x + dx[i];
int newy = y + dy[i];
int newpos = newx * 3 + newy;
if (newx >= 0 && newx < 3 && newy >= 0 && newy < 3){
state t = s;
t.a[s.pos]  = s.a[newpos];
t.a[newpos] = s.a[s.pos];
t.pos = newpos;

if(dfs(t, geth(t), depth + 1, i)) return 1;
}
}
return 0;
}
int idastar()
{
int h = geth(sta);
maxdepth = h;
while(!dfs(sta, h, 0, -1) && maxdepth < 100) {
maxdepth++;
}
return maxdepth;
}
int main (void)
{
for(int i = 0; i < 9; i++){
cin>>sta.a[i];
if(sta.a[i] == 0){
sta.pos = i;
}
}
for(int i = 0; i < 9; i++){
cin>>goal.a[i];
}
int res = idastar();
if(res < 100) cout<<res<<endl;
else cout<<"No solution"<<endl;
return 0;
}

方法6：

有关逆序数

空格在同一行上进行移动，1~8的排列逆序数不变，空格在同一列上进行移动，1~8的排列逆序数发生变化的对数为偶数。所以初始状态的逆序数与目标态同奇偶则有解，反之无解；

这一优化可以用在上面的所有方法中，也是一个很重要的剪枝。

代码：

int Reverse(state s)
{
int cnt = 0;
for(int i = 0; i < 9; i++){
int tmp = s.a[i];
if(tmp == 0) continue;
for(int j = i +1; j < 9; j++){
if(s.a[j] && s.a[i] > s.a[j]) cnt++;
}
}
return cnt % 2;
}

真的是细节决定一切。。。竟然写了一天。。。一半时间还都用来调试。。。写的时候就应该心思细腻，检查的时候模拟一遍过程，任何细节都不能放过！