UESTC 485 Game（康托展开，bfs打表）
Game Time Limit: 4000/2000MS (Java/Others) Memory Limit: 65535/65535KB (Java/Others)
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Today I want to introduce an interesting game to you. Like eight puzzle, it is a square board with 99 positions, but it filled by 99 numbered tiles. There is only one type of valid move, which is to rotate one row or column. That is, three tiles in a row or column are moved towards the head by one tile and the head tile is moved to the end of the row or column. So it has 1212 different moves just as the picture left. The objective in the game is to begin with an arbitrary configuration of tiles, and move them so as to get the numbered tiles arranged as the target configuration.
title
Now the question is to calculate the minimum steps required from the initial configuration to the final configuration. Note that the initial configuration is filled with a permutation of 11 to 99, but the final configuration is filled with numbers and * (which can be any number).
Input The first line of input contains an integer TT (T≤1000T≤1000), which is the number of data sets that follow.
There are 66 lines in each data set. The first three lines give the initial configuration and the next three lines give the final configuration.
Output For every test case, you should output Case #k: first, where kk indicates the case number and starts at 11. Then the fewest steps needed. If he can’t move to the end, just output No Solution! (without quotes).
Sample input and output Sample Input Sample Output 2 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 9 8 1 2 3 4 5 6 7 8 9 8 * 9 5 3 7 2 * * Case #1: No Solution! Case #2: 7
利用康托展开进行bfs预处理。题目给的一个起始的九宫格，和一个目标的九宫格。 不能直接用目标的九宫格去找起始的九宫格，会超时，应该根据把起始九宫格当作 1 2 3 4 5 6 7 8 9 然后确定目标九宫格是怎么样的，这样就可以直接用之前打的表了。预处理就是处理1 2 3 4 5 6 7 8 9到每种九宫格的步数
关于康托展开，给出一篇博文吧 http://blog.csdn.net/dacc123/article/details/50952079
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>
#include <queue>
using namespace std;
struct Node
{
int a[5][5];
int sta;
};
queue<Node> q;
int b[10];
int fac[10];
int vis[400000];
int pre[400000];
int ans;
int f1[10];
int f2[10];
int tran[10];
char ch[10];
bool used[10];
Node cyk;
void facfun()
{
fac[0]=1;
for(int i=1;i<=9;i++)
{
fac[i]=i*fac[i-1];
}
}
int kt(Node q)
{
int cnt=0;
for(int i=1;i<=3;i++)
for(int j=1;j<=3;j++)
b[++cnt]=q.a[i][j];
int sum=0,num=0;
for(int i=1;i<=9;i++)
{
num=0;
for(int j=i+1;j<=9;j++)
{
if(b[i]>b[j])
num++;
}
sum+=num*fac[9-i];
}
return sum;
}
void bfs(Node t)
{
q.push(t);
vis[t.sta]=1;
pre[t.sta]=0;
while(!q.empty())
{
Node term=q.front();
q.pop();
for(int i=1;i<=12;i++)
{
Node temp=term;
if(i<=3)
{
temp.a[i][1]=term.a[i][3];
temp.a[i][2]=term.a[i][1];
temp.a[i][3]=term.a[i][2];
}
else if(i>3&&i<=6)
{
temp.a[i-3][1]=term.a[i-3][2];
temp.a[i-3][2]=term.a[i-3][3];
temp.a[i-3][3]=term.a[i-3][1];
}
else if(i>6&&i<=9)
{
temp.a[1][i-6]=term.a[3][i-6];
temp.a[2][i-6]=term.a[1][i-6];
temp.a[3][i-6]=term.a[2][i-6];
}
else if(i>9&&i<=12)
{
temp.a[1][i-9]=term.a[2][i-9];
temp.a[2][i-9]=term.a[3][i-9];
temp.a[3][i-9]=term.a[1][i-9];
}
int state=kt(temp);
if(vis[state])
continue;
temp.sta=state;
vis[state]=1;
pre[state]=pre[term.sta]+1;
q.push(temp);
}
}
}
void init()
{
memset(vis,0,sizeof(vis));
memset(pre,-1,sizeof(pre));
facfun();
Node st;int cnt=0;
for(int i=1;i<=3;i++)
for(int j=1;j<=3;j++)
st.a[i][j]=++cnt;
st.sta=0;
bfs(st);
}
int anspos;
void dfs(int i)
{
if(i==10)
{
/*for(int p=1;p<=3;p++)
{
for(int k=1;k<=3;k++)
{
cout<<cyk.a[p][k]<<" ";
}
cout<<endl;
}*/
int c=pre[kt(cyk)];
if(c==-1) return;
ans=min(ans,c);return;
}
if(f2[i]==0)
{
for(int j=1;j<=9;j++)
{
if(!used[j])
{
used[j]=true;
int y=i%3,x;
if(y==0){x=i/3;y=3;}
else {x=i/3+1;}
cyk.a[x][y]=j;
dfs(i+1);
used[j]=false;
}
}
}
else
{
int y=i%3,x;
if(y==0){x=i/3;y=3;}
else {x=i/3+1;}
cyk.a[x][y]=f2[i];
dfs(i+1);
}
}
int main()
{
int t;
scanf("%d",&t);
init();
int cas=0;
while(t--)
{
memset(used,0,sizeof(used));
for(int i=1;i<=9;i++)
{
scanf("%d",&f1[i]);
tran[f1[i]]=i;
}
for(int i=1;i<=9;i++)
{
scanf("%s",ch);
f2[i]=ch[0]-'0';
if(f2[i]>=1&&f2[i]<=9)
f2[i]=tran[f2[i]],used[f2[i]]=true;
else
f2[i]=0;
}
ans=1000000;
dfs(1);
if(ans>=1000000)
printf("Case #%d: No Solution!\n",++cas);
else
printf("Case #%d: %d\n",++cas,ans);
}
return 0;
}腾讯云开发者
