2286 The Rotation Game (1,状压 2,迭代加深搜索)

L - The Rotation Game

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

Submit Status

Description

The rotation game uses a # shaped board, which can hold 24 pieces of square blocks (see Fig.1). The blocks are marked with symbols 1, 2 and 3, with exactly 8 pieces of each kind.

Initially, the blocks are placed on the board randomly. Your task is to move the blocks so that the eight blocks placed in the center square have the same symbol marked. There is only one type of valid move, which is to rotate one of the four lines, each consisting of seven blocks. That is, six blocks in the line are moved towards the head by one block and the head block is moved to the end of the line. The eight possible moves are marked with capital letters A to H. Figure 1 illustrates two consecutive moves, move A and move C from some initial configuration.

Input

The input consists of no more than 30 test cases. Each test case has only one line that contains 24 numbers, which are the symbols of the blocks in the initial configuration. The rows of blocks are listed from top to bottom. For each row the blocks are listed from left to right. The numbers are separated by spaces. For example, the first test case in the sample input corresponds to the initial configuration in Fig.1. There are no blank lines between cases. There is a line containing a single `0' after the last test case that ends the input.

Output

For each test case, you must output two lines. The first line contains all the moves needed to reach the final configuration. Each move is a letter, ranging from `A' to `H', and there should not be any spaces between the letters in the line. If no moves are needed, output `No moves needed' instead. In the second line, you must output the symbol of the blocks in the center square after these moves. If there are several possible solutions, you must output the one that uses the least number of moves. If there is still more than one possible solution, you must output the solution that is smallest in dictionary order for the letters of the moves. There is no need to output blank lines between cases.

Sample Input

1 1 1 1 3 2 3 2 3 1 3 2 2 3 1 2 2 2 3 1 2 1 3 31 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 30

Sample Output

AC2DDHH2

题意：有八个方向可以让数列移动，求最小移动次数使中间八个数值相同（数列移动规则看图·）

分析：两种方法：1.状态压缩（小白上有介绍）。观察可知终止状态分三种情况，中间八个要么为1，要么为2，或者为3，则先预处理，中间为1时，四周16个都不为一，中间为2时，16个不为2，中间为3时，16个不为3，排列组合可知状态有C(8,24)种，状态有大约70多w，还是可求的，但无论我怎么写，在poj可以过，4.594s,在uva果断超时，有知道怎么实现的同学告知在下（uva上要求3s，poj上15s）

2。迭代加深。从小枚举深度，加剪掉 现有步数加预测最快到达步数还不能到达的

先粘状态压缩的代码：(4.594s)

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<queue>#include<cmath>#include<algorithm>#define LL long longusing namespace std;const int INF=0x3f3f3f3f;int forturn[8][7]={{0,2,6,11,15,20,22},{1,3,8,12,17,21,23},{10,9,8,7,6,5,4},{19,18,17,16,15,14,13},{23,21,17,12,8,3,1},{22,20,15,11,6,2,0},{13,14,15,16,17,18,19},{4,5,6,7,8,9,10}};int in[25],st[3],goal,symans;char pans[50];bool done[1<<26];struct node{int x,w,id;char p[50];node(int xx=0,int ww=0,int i=0,char* pp=""){x=xx;w=ww,strcpy(p,pp);id=i;}}qx[1200000];int solve(int x,int c){int i=0,xx=x;for(i=0;i<6;i++){    if(1&(x>>forturn[c][i]))xx^=(1<<forturn[c][i]);xx=xx|(((x>>forturn[c][i+1])&1)<<forturn[c][i]);}    if(1&(x>>forturn[c][6]))xx^=(1<<forturn[c][6]);    xx=xx|(((x>>forturn[c][0])&1)<<forturn[c][6]);return xx;}void init(){int i,j;st[0]=st[1]=st[2]=0;for(j=1;j<=3;j++)for(i=23;i>=0;i--)st[j-1]=(st[j-1]<<1)|(in[i]==j);}int bfs(){int i,j,last=INF;int _start=0,_end=0;memset(done,0,sizeof done);for(i=0;i<3;i++){qx[_end]=node(st[i],0,i,"");_end=(_end+1)%1200000;done[st[i]+i*(1<<24)]=1;}while(_start!=_end){node e=qx[_start];_start=(_start+1)%1200000;if(e.w==last)return 1;for(i=0;i<8;i++){int cur=solve(e.x,i);if(done[cur+e.id*(1<<24)])continue;char ssa[50];strcpy(ssa,e.p);j=strlen(e.p);ssa[j]='A'+i;if(j<29)ssa[j+1]='\0';if(cur==goal&&last==INF){last=e.w+1;strcpy(pans,ssa);symans=e.id;}if(cur==goal&&strcmp(ssa,pans)<0){strcpy(pans,ssa);symans=e.id;}qx[_end]=node(cur,e.w+1,e.id,ssa);_end=(_end+1)%1200000;done[cur+e.id*(1<<24)]=1;}}return -1;}int main(){int i;//("code.txt","r",stdin);while(~scanf("%d",&in[0])){if(!in[0])break;for(i=1;i<24;i++)scanf("%d",&in[i]);goal=235968;init();if(st[0]==goal||st[1]==goal||st[2]==goal){printf("No moves needed\n%d\n",in[6]);continue;}bfs();printf("%s\n%d\n",pans,symans+1);}return 0;}

迭代加深代码：

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;int in[24];int turn[8][7]={0,2,6,11,15,20,22,   1,3,8,12,17,21,23,   10,9,8,7,6,5,4,   19,18,17,16,15,14,13,   23,21,17,12,8,3,1,   22,20,15,11,6,2,0,   13,14,15,16,17,18,19,   4,5,6,7,8,9,10};int aim[8]={6,7,8,11,12,15,16,17};int opp[8]={5,4,7,6,1,0,3,2};int ans[50],D;int togoal(){    int i,cnt[3]={0};    for(i=0;i<8;i++)        cnt[in[aim[i]]-1]++;    return 8-max(cnt[0],max(cnt[1],cnt[2]));}void forturn(int cur){    int tmp=in[turn[cur][0]];    for(int i=0;i<6;i++)        in[turn[cur][i]]=in[turn[cur][i+1]];    in[turn[cur][6]]=tmp;}int dfs(int place){    if(place==D+1)return 0;    for(int i=0;i<8;i++)    {        forturn(i);        ans[place]=i;        int tmp;        if((tmp=togoal())==0)return 1;        if(place+tmp<=D)        {            if(dfs(place+1))return 1;        }        forturn(opp[i]);    }    return 0;}int main(){    int i;    //freopen("code.txt","r",stdin);    while(~scanf("%d",&in[0]))    {        if(!in[0])break;        for(i=1;i<24;i++)            scanf("%d",&in[i]);        if(togoal()==0)        {            printf("No moves needed\n%d\n",in[aim[0]]);continue;        }        for(D=1;!dfs(1);D++)            ;        for(i=1;i<=D;i++)            printf("%c",ans[i]+'A');        printf("\n%d\n",in[aim[0]]);    }    return 0;}