Hoj 1868 八数码问题
题目链接：http://acm.hit.edu.cn/hoj/problem/view?id=1868
超时的写法：容易想到的是一遍搜索，把所有的状态都保留下来，我们只要建立0-8的全排列和0-362879的对应起来，就可以了。其中fact数组保留了以当前标号开始的全排列的个数。但是很遗憾，这样是会TLE的。
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
typedef int State[9];
#define Maxn 400000
#define Maxm 362900
State st[Maxn],goal;
int dist[Maxn];
int vis[Maxm];
int fact[9];
int total = 0;
int dx[] = {-1,1,0,0};
int dy[] = {0,0,-1,1};
void init_lookup_table()
{
fact[0] = 1;
for(int i=1;i<9;i++) fact[i] = fact[i-1] * i;
}
int try_to_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()
{
int front = 1,rear = 2;
while(front<rear)
{
State & s = st[front];
if(memcmp(goal,s,sizeof(s)) == 0) return front;
int z;
for(z = 0;z<9;z++) if(!s[z]) break;
int x = z/3,y = z%3;
for(int d = 0;d<4;d++)
{
int newx = x + dx[d];
int newy = y + dy[d];
int newz = newx*3 + newy;
if(newx>=0 && newx <3 && newy>=0 && newy<3)
{
State &t = st[rear];
memcpy(&t,&s,sizeof(s));
t[newz] = s[z];
t[z] = s[newz];
dist[rear] = dist[front] + 1;
if(try_to_insert(rear)) rear++;
}
}
front++;
total = rear;
}
return -1;
}
void init()
{
memset(vis,0,sizeof(vis));
memset(dist,0,sizeof(dist));
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
#endif
int t;
int ans;
scanf(" %d",&t);
init_lookup_table();
while(t--)
{
init();
for(int i=0; i<9; i++) scanf(" %d",&st[1][i]);
for(int i=0;i<9;i++) scanf(" %d",&goal[i]);
ans = bfs();
if(ans == -1) puts("-1");
else printf("%d\n",dist[ans]);
}
return 0;
}
另一种超时的方法是用哈希来建立映射。
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
typedef int State[9];
#define Maxn 400000
#define Hash_size 1000003
State st[Maxn],goal;
int dist[Maxn];
int head[Hash_size],next[Maxn];
int total = 0;
int dx[] = {-1,1,0,0};
int dy[] = {0,0,-1,1};
int hash_handle(State &s)
{
int v = 0;
for(int i=0;i<9;i++) v = v*10 + s[i];
return v % Hash_size;
}
void init_lookup_table()
{
memset(head,0,sizeof(head));
}
int try_to_insert(int s)
{
int h = hash_handle(st[s]);
int u = head[h];
while(u)
{
if(memcmp(st[u],st[s],sizeof(st[s])) == 0) return 0;
u = next[u];
}
next[s] = head[h];
head[h] = s;
return 1;
}
int bfs()
{
int front = 1,rear = 2;
while(front<rear)
{
State & s = st[front];
if(memcmp(goal,s,sizeof(s)) == 0) return front;
int z;
for(z = 0;z<9;z++) if(!s[z]) break;
int x = z/3,y = z%3;
//printf("front = %d,rear = %d\n",front,rear);
//printf("x = %d,y = %d\n",x,y);
for(int d = 0;d<4;d++)
{
int newx = x + dx[d];
int newy = y + dy[d];
int newz = newx*3 + newy;
if(newx>=0 && newx <3 && newy>=0 && newy<3)
{
State &t = st[rear];
memcpy(&t,&s,sizeof(s));
t[newz] = s[z];
t[z] = s[newz];
dist[rear] = dist[front] + 1;
if(try_to_insert(rear)) rear++;
}
}
front++;
total = rear;
}
return -1;
}
void init()
{
//memset(vis,0,sizeof(vis));
memset(dist,0,sizeof(dist));
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
#endif
int t;
int ans;
scanf(" %d",&t);
while(t--)
{
init_lookup_table();
init();
for(int i=0; i<9; i++) scanf(" %d",&st[1][i]);
for(int i=0;i<9;i++) scanf(" %d",&goal[i]);
ans = bfs();
if(ans == -1) puts("-1");
else printf("%d\n",dist[ans]);
}
return 0;
}
以上两种超时的原因是只有一个方向的搜索，所以我改成了双广BFS+Hash。
但是依然超时，我不知道为什么，求告知。
代码:
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <queue>
#include <algorithm>
using namespace std;
#define Hash_size 455809
struct State
{
int num;
int zero;
int g;
State()
{
}
};
State first,last;
struct Hash_node
{
State st;
Hash_node * next;
Hash_node()
{
next = NULL;
}
};
Hash_node hash_node[2][Hash_size];
int dx[] = {-1,1,0,0};
int dy[] = {0,0,-1,1};
int a[9],b[9];
queue<State> q1,q2;
bool same(State x,State y)
{
if(x.num == y.num) return true;
return false;
}
int getHash(State s)
{
return s.num % Hash_size ;
}
int try_to_insert(State x,int kind)
{
int h = getHash(x);
Hash_node * u = &hash_node[kind][h];
while(u->next!=NULL)
{
if(same(x,u->next->st)) return 0;
u = u->next;
}
u->next = new Hash_node;
u->next->st = x;
return 1;
}
bool exist(State x,int kind)
{
int h = getHash(x) ;
Hash_node * u= &hash_node[kind][h];
while(u->next!=NULL)
{
if(same(x,u->next->st)) return true;
u = u->next;
}
return false;
}
int handle(int num,int zero1,int zero2)
{
int st[9];
for(int i=8;i>=0;i--)
{
st[i] = num%10;
num/=10;
}
int x = st[zero1];
st[zero1] = st[zero2];
st[zero2] = x;
num = 0;
for(int i=0;i<=8;i++)
{
num = num*10 + st[i];
}
return num;
}
int bfs()
{
int code;
int x,y,newx,newy,z,newz,d;
while(!q1.empty()) q1.pop();
while(!q2.empty()) q2.pop();
q1.push(first);
q2.push(last);
try_to_insert(first,0);
try_to_insert(last,1);
while(!q1.empty() || !q2.empty())
{
int n1 = q1.size();
int n2 = q2.size();
//处理环A
if(n1 < n2 || n2 == 0)
{
//一层扩展
for(int i=0; i<n1; i++)
{
State s = q1.front();
q1.pop();
//在对方这个状态是否存在
if(exist(s,1)) return s.g + q2.front().g;
z = s.zero;
x = z/3;
y = z%3;
for(d=0; d<4; d++)
{
newx = x + dx[d];
newy = y + dy[d];
newz = newx*3 + newy;
if(newx>=0 && newx <3 && newy>=0 && newy<3)
{
State t = s;
t.num = handle(s.num,z,newz);
t.g = s.g + 1;
t.zero = newz;
if(try_to_insert(t,0)) q1.push(t);
}
}
}
}
//处理环B
else
{
//一层扩展
for(int i=0; i<n2; i++)
{
State s = q2.front();
q2.pop();
if(exist(s,0)) return s.g + q1.front().g;
z = s.zero;
x = z/3;
y = z%3;
for(d=0; d<4; d++)
{
newx = x + dx[d];
newy = y + dy[d];
newz = newx*3 + newy;
if(newx>=0 && newx <3 && newy>=0 && newy<3)
{
State t = s;
t.num = handle(s.num,z,newz);
t.g = s.g + 1;
t.zero = newz;
if(try_to_insert(t,1)) q2.push(t);
}
}
}
}
}
return 0;
}
//求逆序数对，要排除第二个数字是0的那种情况
bool reversePair()
{
int ansA = 0,ansB = 0;
for(int i=0; i<9; i++)
{
for(int j=i+1; j<9; j++)
{
if(a[i]>a[j] && a[j]!=0) ansA++;
if(b[i]>b[j] && b[j]!=0) ansB++;
}
}
if((ansA + ansB)%2) return true;
return false;
}
bool preJudge()
{
if(same(first,last))
{
puts("0");
return false;
}
else if(reversePair())
{
puts("-1");
return false;
}
return true;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif
int t;
int ans;
int zero;
int v;
scanf(" %d",&t);
while(t--)
{
v = 0;
for(int i=0; i<9; i++)
{
scanf(" %d",&a[i]);
if(a[i] == 0) zero = i;
v = v*10 + a[i];
}
first.g = 0;
first.num = v;
first.zero = zero;
v = 0;
for(int i=0; i<9; i++)
{
scanf(" %d",&b[i]);
if(b[i] == 0) zero = i;
v = v*10 + b[i];
}
last.g = 0;
last.num = v;
last.zero = zero;
if(preJudge())
{
ans = bfs();
printf("%d\n",ans);
}
}
return 0;
}
于是，我又写了一发双广BFS+全排列映射。A掉了。
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <queue>
#include <algorithm>
using namespace std;
#define Maxn 400000
#define Maxm 400000
struct State
{
int a[9];
int zero;
int g;
State()
{
}
State(int b[9])
{
for(int i=0; i<9; i++)
{
a[i]=b[i];
if(b[i]==0) zero=i;
}
g=0;
}
};
int dist[2][Maxn];
int fact[9];
int vis[2][Maxm];
State first,last;
int dx[] = {-1,1,0,0};
int dy[] = {0,0,-1,1};
queue<State> q1,q2;
void init_lookup_table()
{
fact[0] = 1;
for(int i=1; i<9; i++) fact[i] = fact[i-1] * i;
}
bool same(State x,State y)
{
for(int i=0; i<9; i++)
{
if(x.a[i]!= y.a[i]) return false;
}
return true;
}
int getcode(State x)
{
int code=0;
for(int i=0; i<9; i++)
{
int cnt=0;
for(int j=i+1; j<9; j++)
{
if(x.a[i]>x.a[j]) cnt++;
}
code+=fact[8-i]*cnt;
}
return code;
}
int try_to_insert(State x,int kind)
{
int code = getcode(x);
if(vis[kind][code]) return 0;
dist[kind][code] = x.g;
return vis[kind][code] = 1;
}
int bfs()
{
int code;
int x,y,newx,newy,z,newz,d;
memset(dist,0,sizeof(dist));
memset(vis,0,sizeof(vis));
while(!q1.empty()) q1.pop();
while(!q2.empty()) q2.pop();
q1.push(first);
q2.push(last);
try_to_insert(first,0);
try_to_insert(last,1);
while(!q1.empty() || !q2.empty())
{
int n1 = q1.size();
int n2 = q2.size();
//处理环A
if(n1 < n2 || n2 == 0)
{
//一层扩展
for(int i=0; i<n1; i++)
{
State s = q1.front();
q1.pop();
code = getcode(s);
//在对方这个状态是否存在
if(vis[1][code]) return dist[0][code] + dist[1][code];
z = s.zero;
x = z/3;
y = z%3;
for(d=0; d<4; d++)
{
newx = x + dx[d];
newy = y + dy[d];
newz = newx*3 + newy;
if(newx>=0 && newx <3 && newy>=0 && newy<3)
{
State t = s;
t.a[newz] = s.a[z];
t.a[z] = s.a[newz];
t.g = s.g + 1;
t.zero = newz;
if(try_to_insert(t,0)) q1.push(t);
}
}
}
}
//处理环B
else
{
//一层扩展
for(int i=0; i<n2; i++)
{
State s = q2.front();
q2.pop();
code = getcode(s);
//在对方这个状态是否存在
if(vis[0][code]) return dist[0][code] + dist[1][code];
z = s.zero;
x = z/3;
y = z%3;
for(d=0; d<4; d++)
{
newx = x + dx[d];
newy = y + dy[d];
newz = newx*3 + newy;
if(newx>=0 && newx <3 && newy>=0 && newy<3)
{
State t = s;
t.a[newz] = s.a[z];
t.a[z] = s.a[newz];
t.g = s.g + 1;
t.zero = newz;
if(try_to_insert(t,1)) q2.push(t);
}
}
}
}
}
return 0;
}
//求逆序数对，要排除第二个数字是0的那种情况
int reversePair(State s)
{
int ans = 0;
for(int i=0; i<9; i++)
{
for(int j=i+1; j<9; j++) if(s.a[i]>s.a[j] && s.a[j]!=0) ans++;
}
return ans;
}
bool preJudge()
{
if(same(first,last))
{
puts("0");
return false;
}
else if((reversePair(first) + reversePair(last))%2)
{
puts("-1");
return false;
}
return true;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif
int t;
int ans;
int begin[9],end[9];
scanf(" %d",&t);
init_lookup_table();
while(t--)
{
for(int i=0; i<9; i++) scanf(" %d",&begin[i]);
State a(begin);
first = a;
for(int i=0; i<9; i++) scanf(" %d",&end[i]);
State b(end);
last = b;
if(preJudge())
{
ans = bfs();
printf("%d\n",ans);
}
}
return 0;
}