此题我用了2种方法去做，bfs和双向 bfs 现在还在学A*，准备学会了再用A*去试试，单向bfs只过了poj，双向bfs全部都过了，具体思想就是搜索加判重，有用hash，有用康托展开，不过康托展开比较方便，因为毕竟可以一一对应，就不用判重的时候还要比较9个数了，康托展开的计算方法是：从最高位开始，在它位数前面比它小的所有数的数量再乘以它（位数-1）的阶乘，再把每一位的这个值给加起来，就是要求的。讲起来很晦涩，其实就是代表数字排列顺序的先后，比如321的康托展开 = 2×2！+ 1×1！=5是第5位，因为是从0开始数的，所以就是3！-1就是5了，312=2×2！=4 还有123 = 0，这样就可以让0-8总共9个数一一和9！一一对应起来了。
HOJ10466代码：这只用求步数就可以了不用给出路径，但是数据挺强大的，我双向BFS还用了828ms。
#include<cstdio>
#include<ctype.h>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
using namespace std;
#define MAX 500000
#define NMAX 1000007
int cantor[9],vis[MAX],dist[MAX];
int ans;
bool found;
short dx[4]= {-1,0,1,0},dy[4]= {0,1,0,-1};
typedef struct status
{
int state[9];
}status;
queue<status>q1;
queue<status>q2;
status st,go;
void initcantor()
{
cantor[0] = 1;
for(int i = 1; i < 9; i++)
cantor[i] = cantor[i-1]*i;
}
void init()
{
ans=0;
memset(dist,0,sizeof(dist));
memset(vis,0,sizeof(vis));
while(!q1.empty()) q1.pop();
while(!q2.empty()) q2.pop();
}
int gethash(status &s)
{
int code = 0;
for(int i = 0; i < 9; i++)
{
int cnt = 0;
for(int j = i+1; j < 9; j++)
if(s.state[i] > s.state[j]) cnt++;
code += cantor[8-i]*cnt;
}
return code;
}
void bfs_expend(queue<status>&q,int flag)
{
int ff;
if(flag == 1)
ff = 2;
else
ff = 1;
status temp = q.front();
q.pop();
int z;
for(z = 0; temp.state[z] != 0; z++);
int x = z/3,y = z%3;
int h = gethash(temp);
for(int i = 0; i < 4; i++)
{
status t1=temp;
int newx = x+dx[i],newy = y+dy[i];
if(newx>=0 && newx < 3 && newy>=0 && newy < 3)
{
int newz = 3*newx+newy;
int k = t1.state[newz];
t1.state[z] = k;
t1.state[newz] = 0;
int newh = gethash(t1);
if(vis[newh] == ff)
{
ans = dist[newh] + dist[h]+1;
found = true;
return;
}
if(vis[newh] != flag)
{
q.push(t1);
vis[newh] = flag;
dist[newh] = dist[h] + 1;
}
}
}
}
void TBFS()
{
found = false;
memset(vis,0,sizeof(vis));
vis[gethash(st)] = 1;
vis[gethash(go)] = 2;
q1.push(st);
q2.push(go);
while(!q1.empty() || !q2.empty())
{
if(!q1.empty())
bfs_expend(q1,1);
if(found)
return;
if(!q2.empty())
bfs_expend(q2,2);
if(found)
return;
}
}
int main()
{
initcantor();
init();
int i,j;
for(i = 0; i < 9; i++) scanf("%d",&st.state[i]);
for(i = 0; i < 9; i++) scanf("%d",&go.state[i]);
if(memcmp(st.state,go.state,sizeof(st.state)) == 0)
{
printf("0\n");
return 0;
}
TBFS();
if(!ans)
printf("-1\n");
else
printf("%d\n",ans);
return 0;
}
poj和hdu的代码
#include<cstdio>
#include<ctype.h>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
using namespace std;
#define MAX 365000
int cantor[9],vis[MAX];
int move1[MAX],move2[MAX];
int father1[MAX],father2[MAX];
int path[1000],record;
bool found;
short dx[4]= {-1,0,1,0},dy[4]= {0,1,0,-1};
char pos1[5] = "urdl";
char pos2[5] = "dlur";
typedef struct status
{
int state[9];
} status;
queue<status>q1;
queue<status>q2;
stack<char>ss;
status st,go;
void initcantor()
{
cantor[0] = 1;
for(int i = 1; i < 9; i++)
cantor[i] = cantor[i-1]*i;
}
void init()
{
record=0;
found = false;
memset(vis,0,sizeof(vis));
while(!q1.empty()) q1.pop();
while(!q2.empty()) q2.pop();
}
int gethash(status &s)
{
int code = 0;
for(int i = 0; i < 9; i++)
{
int cnt = 0;
for(int j = i+1; j < 9; j++)
if(s.state[i] > s.state[j]) cnt++;
code += cantor[8-i]*cnt;
}
return code;
}
void bfs_expend(queue<status>&q,int flag)
{
int ff;
if(flag == 1)
ff = 2;
else
ff = 1;
status temp = q.front();
q.pop();
int z;
for(z = 0; temp.state[z] != 0; z++);
int x = z/3,y = z%3;
int h = gethash(temp);
for(int i = 0; i < 4; i++)
{
status t1=temp;
int newx = x+dx[i],newy = y+dy[i];
if(newx>=0 && newx < 3 && newy>=0 && newy < 3)
{
int newz = 3*newx+newy;
int k = t1.state[newz];
t1.state[z] = k;
t1.state[newz] = 0;
int newh = gethash(t1);
if(vis[newh] == ff)
{
record = newh;
if(ff == 2)
{
move1[newh] = i;
father1[newh] = h;
}
else
{
move2[newh] = i;
father2[newh] = h;
}
found = true;
return;
}
if(vis[newh] != flag)
{
if(flag == 1)
{
move1[newh] = i;
father1[newh] = h;
}
else
{
move2[newh] = i;
father2[newh] = h;
}
q.push(t1);
vis[newh] = flag;
}
}
}
}
void TBFS()
{
vis[gethash(st)] = 1;
vis[gethash(go)] = 2;
q1.push(st);
q2.push(go);
while(!q1.empty() || !q2.empty())
{
if(!q1.empty())
bfs_expend(q1,1);
if(found)
return;
if(!q2.empty())
bfs_expend(q2,2);
if(found)
return;
}
}
void find_path(int start,int end)
{
int temp = record;
ss.push(pos1[ move1[temp] ]);
while(father1[temp] != start)
{
temp = father1[temp];
ss.push(pos1[ move1[temp] ]);
}
while(!ss.empty())
{
printf("%c",ss.top());
ss.pop();
}
temp = record;
printf("%c",pos2[ move2[temp] ]);
while(father2[temp] != end)
{
temp = father2[temp];
printf("%c",pos2[ move2[temp] ]);
}
printf("\n");
}
int main()
{
initcantor();
int i,j;
for(i = 0; i < 8; i++) go.state[i] = i+1;
go.state[8] = 0;
int coo=0;
char temp;
while(cin>>temp)
{
init();
if(temp == ‘x‘) st.state[0] = 0;
else st.state[0] = temp-‘0‘;
for(i = 1; i < 9; i++)
{
cin>>temp;
if(temp == ‘x‘) st.state[i] = 0;
else st.state[i] = temp-‘0‘;
}
coo=0;
for(i = 1; i < 9; i++)
for(j = 0; j < i; j++)
{
if(!st.state[i] || !st.state[j]) continue;
if(st.state[i] < st.state[j])
coo++;
}
if(coo%2 == 1) printf("unsolvable\n");
else
{
TBFS();
find_path(gethash(st),gethash(go));
}
}
return 0;
}
poj 1077 + HOJ 10466 + hdu 1043 eight,布布扣,bubuko.com
poj 1077 + HOJ 10466 + hdu 1043 eight
原文地址：http://blog.csdn.net/glqac/article/details/24909997
