Eight II
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 130000/65536 K (Java/Others)Total Submission(s): 4103 Accepted Submission(s): 878
In this game, you are given a 3 by 3 board and 8 tiles. The tiles are numbered from 1 to 8 and each covers a grid. As you see, there is a blank grid which can be represented as an 'X'. Tiles in grids having a common edge with the blank grid can be moved into that blank grid. This operation leads to an exchange of 'X' with one tile.
We use the symbol 'r' to represent exchanging 'X' with the tile on its right side, and 'l' for the left side, 'u' for the one above it, 'd' for the one below it.
A state of the board can be represented by a string S using the rule showed below.
The problem is to operate an operation list of 'r', 'u', 'l', 'd' to turn the state of the board from state A to state B. You are required to find the result which meets the following constrains:
1. It is of minimum length among all possible solutions.
2. It is the lexicographically smallest one of all solutions of minimum length.
The input of each test case consists of two lines with state A occupying the first line and state B on the second line.
It is guaranteed that there is an available solution from state A to B.
The first line is in the format of "Case x: d", in which x is the case number counted from one, d is the minimum length of operation list you need to turn A to B.
S is the operation list meeting the constraints and it should be showed on the second line.
The title is easy to understand. It is a Huarong Road game. The only thing to pay attention to is to output the shortest path. When the length is the same, the dictionary order should be the smallest.
This question is the same as the hdu1430 method, except that 1430 is a bfs+cantor expansion to save all the states, and there are nine kinds of starting points, so only nine bfs are required, and the path cannot be directly saved by violence, I used ans[ x][e] to indicate that the starting point is x, go to the operation character whose Cantor value is e, and then use a pre[x][e] to indicate the previous Cantor value of this operation. In addition, this question card time card has a pruning time, that is, if the previous step is u, then this step does not need d, and the other directions are the same. In addition, I would like to praise this method of writing functions in the structure, really Very useful! !
By the way, because the starting point of each input is different, and the positions of other numbers are also different, so use a p to save the position of x, (in the process of bfs, p is also used to indicate where x is, And remember to update), and then use a re[] array to represent the correspondence of other numbers other than x, and then you can.
建议先做一下hdu1430，这是1430的题解。点击打开链接
上代码。
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<bitset>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(x,y) memset(x,y,sizeof(x))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=362880+10;
int fact[10]= {1,1,2,6,24,120,720,5040,40320,362880};
struct node {
char arr[10];
int val;
int p;
string step;
void contor() {
val=0;
for(int i=0; i<9; i++) {
int t=0;
for(int j=i+1; j<9; j++) {
if(arr[j]<arr[i])t++;
}
val+=t*fact[9-i-1];
}
val++;
}
void change_d(int x) {
char temp=arr[x];
arr[x]=arr[x+3];
arr[x+3]=temp;
p=x+3;
step+='d';
}
void change_l(int x) {
p=x-1;
char temp=arr[x];
arr[x]=arr[x-1];
arr[x-1]=temp;
step+='l';
}
void change_r(int x) {
p=x+1;
char temp=arr[x];
arr[x]=arr[x+1];
arr[x+1]=temp;
step+='r';
}
void change_u(int x) {
p=x-3;
char temp=arr[x];
arr[x]=arr[x-3];
arr[x-3]=temp;
step+='u';
}
node() {}
node(char *s) {
int i,j;
for(i = 0; i<strlen(s); i++) {
arr[i] = s[i];
}
}
};
int vis[N];
int pre[9][N];
char ans[9][N];
char re[10];
node s,t;
void bfs(int x) {
memset(pre[x],-1,sizeof(pre[x]));
memset(vis,0,sizeof(vis));
s.contor();
s.p=x;
queue<node>q;
vis[s.val]=1;
q.push(s);
node now,v;
while(!q.empty()) {
now=q.front();
q.pop();
v=now;
int e=v.val;
int len=v.step.length();
//d
if(v.p<6&&ans[x][e]!='u') {
v.change_d(v.p);
v.contor();
if(!vis[v.val]) {
vis[v.val]=1;
ans[x][v.val]='d';
pre[x][v.val]=e;
q.push(v);
}
}
//l
v=now;
if(v.p!=0&&v.p!=3&&v.p!=6&&ans[x][e]!='r') {
v.change_l(v.p);
v.contor();
if(!vis[v.val]) {
vis[v.val]=1;
ans[x][v.val]='l';
pre[x][v.val]=e;
q.push(v);
}
}
//r
v=now;
if(v.p!=2&&v.p!=5&&v.p!=8&&ans[x][e]!='l') {
v.change_r(v.p);
v.contor();
if(!vis[v.val]) {
vis[v.val]=1;
ans[x][v.val]='r';
pre[x][v.val]=e;
q.push(v);
}
}
//u
v=now;
if(v.p>2&&ans[x][e]!='d') {
v.change_u(v.p);
v.contor();
if(!vis[v.val]) {
vis[v.val]=1;
ans[x][v.val]='u';
pre[x][v.val]=e;
q.push(v);
}
}
}
}
int main() {
s = node("X12345678");
bfs(0);
s = node("1X2345678");
bfs(1);
s = node("12X345678");
bfs(2);
s = node("123X45678");
bfs(3);
s = node("1234X5678");
bfs(4);
s = node("12345X678");
bfs(5);
s = node("123456X78");
bfs(6);
s = node("1234567X8");
bfs(7);
s = node("12345678X");
bfs(8);
char ss[10];
int T;
cin>>T;
int cas=1;
while(T--) {
scanf("%s",ss);
int p;
for(int i=0,j=0; i<9; i++) {
if(ss[i]=='X'){
p=i;
}else{
re[ss[i]-'0']=++j; //记住 是j
}
}
scanf("%s",ss);
for(int i=0; i<9; i++) {
if(ss[i]!='X')
t.arr[i]=re[ss[i]-'0'];
else t.arr[i]='X';
}
t.contor();
int sum=t.val;
string aa;
aa="";
while(sum!=-1){
aa+=ans[p][sum];
sum=pre[p][sum];
}
printf("Case %d: %d\n",cas++,aa.size()-1);//为什么要-1呢？因为第一种状态是空字符，不是路径
for(int i=aa.size()-2;i>=0;i--)
printf("%c",aa[i]);
printf("\n");
}
}