HDU356 Eight II(康拓展开+预处理)

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双向BFS会TLE， T_T ，但是离线就不会了

虽然TLE了，还是贴出来留个纪念吧

#include <iostream>
#include <stdlib.h>
#include <queue>
#include <string.h>
#include <algorithm>
#include <stdio.h>

using namespace std;
const int MAXN=1000000;//最多是9!/2
struct node
{
bool type;
int num[10];
int sta;
int loc;
};
int move[4][2] = {{0,1},{0,-1},{1,0},{-1,0}};
char index[5] = "rldu";
char index2[5] = "lrud";
int fac[]={1,1,2,6,24,120,720,5040,40320,362880};//康拖展开判重
// 0!1!2!3! 4! 5! 6! 7! 8! 9!
bool vis1[MAXN], vis2[MAXN];//标记
string path1[MAXN];
string path2[MAXN];
int aim;
int cantor(int s[])//康拖展开求该序列的hash值
{
int sum=0;
for(int i=0;i<9;i++)
{
int num=0;
for(int j=i+1;j<9;j++)
if(s[j]<s[i])num++;
sum+=(num*fac[9-i-1]);
}
return sum+1;
}

bool check( string str1, string str2)
{
string a, b;
a = str1, b = str2;
if(a > b)
return 1;
return 0;
}
void bfs( node a, node b)
{
if(a.sta == b.sta)
{
cout<<path1[a.sta]<<endl;
return ;
}
queue<node> que;
while(!que.empty()) que.pop();

vis1[a.sta] = vis2[b.sta] = 1;
path1[a.sta] = path2[b.sta] = "";
que.push(a);
que.push(b);
node tem, now;

while(!que.empty())
{
now = que.front();
que.pop();

int x = now.loc/3;
int y = now.loc%3;

for( int i = 0; i < 4; i++ )
{
int tx = x + move[i][0];
int ty = y + move[i][1];
if(ty < 0 || ty > 2 || tx < 0 || tx > 2) continue;

tem = now;
tem.loc = tx*3 + ty;
tem.num[now.loc] = tem.num[tem.loc];
tem.num[tem.loc] = 0;
tem.sta = cantor(tem.num);

if(tem.type == 0 && !vis1[tem.sta])
{
vis1[tem.sta] = 1;
path1[tem.sta] = path1[now.sta] + index[i];
que.push(tem);
}
if(tem.type == 1)
{
if(!vis2[tem.sta])
{
vis2[tem.sta] = 1;
path2[tem.sta] = path2[now.sta] + index2[i];
for( int i = 0; i < 9; i++)
que.push(tem);
}
else if(check(path2[tem.sta], path2[now.sta] + index2[i]))
path2[tem.sta] = path2[now.sta] + index2[i];

}
if((tem.type == 0 && vis2[tem.sta]) || (tem.type == 1 && vis1[tem.sta]))
{
cout<<path1[tem.sta].size()+path2[tem.sta].size()<<endl;
reverse(path2[tem.sta].begin(),path2[tem.sta].end());
cout<<path1[tem.sta]<<path2[tem.sta]<<endl;
return ;
}
}
}
}
void input( node &a)
{
char s;
for(int i = 0; i < 9; i++)
{
cin>>s;
if(s == 'X')
{
a.num[i] = 0;
a.loc = i;
}
else
a.num[i] = s - '0';
}

}

int main()
{
node a,b;
int t,ca = 1;
cin>>t;
while(ca<=t)
{
memset(vis1, 0, sizeof(vis1));
memset(vis2, 0, sizeof(vis2));
input(a);
input(b);
a.sta = cantor(a.num);
b.sta = cantor(b.num);
a.type = 0;
b.type = 1;
printf("Case %d: ",ca++);
bfs(a, b);
}
}

正确答案

#include <iostream>

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <stack>
#include <string>
using namespace std;

struct node
{
int x,y;
char map[5][5];
node() {}
node(char *s)
{
int i,j;
int xx = 0,yy = 0;
for(i = 0; i<strlen(s); i++)
{
map[xx][yy] = s[i];
if(s[i] == 'X')
{
x = xx;
y = yy;
}
yy++;
if(yy == 3)
{
xx++;
yy = 0;
}
}
}
};

node s;
char str[20];
int num[20],hash[10];
bool vis[500000];
int pre[10][500000],ans[10][500000];
int to[4][2] = {1,0,0,-1,0,1,-1,0};
char way[10] = "dlru";

int solve(node a)//康拓
{
int i,j,k,cnt,ans = 0;
int b[20];
for(i = 0; i<3; i++)
{
for(j = 0; j<3; j++)
{
b[3*i+j] = a.map[i][j];
cnt = 0;
for(k = 3*i+j-1; k>=0; k--)
{
if(b[k]>b[3*i+j])
cnt++;
}
ans+=hash[3*i+j]*cnt;
}
}
return ans;
}

void bfs(int p)
{
memset(pre[p],-1,sizeof(pre[p]));
memset(vis,false,sizeof(vis));
node a,next;
queue<node> Q;
Q.push(s);
vis[solve(s)] = true;
while(!Q.empty())
{
a = Q.front();
Q.pop();
int sa = solve(a);
for(int i = 0; i<4; i++)
{
next = a;
next.x+=to[i][0];
next.y+=to[i][1];
if(next.x<0 || next.x>2 || next.y<0 || next.y>2)
continue;
next.map[a.x][a.y] = next.map[next.x][next.y];
next.map[next.x][next.y] = 'X';
int sb = solve(next);
if(vis[sb])
continue;
vis[sb] = true;
pre[p][sb] = sa;
ans[p][sb] = way[i];
Q.push(next);
}
}
}

int main()
{
int t,i,j,k,cas = 1;
hash[0] = 1;
for(i = 1; i<10; i++)
hash[i] = hash[i-1]*i;

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);

scanf("%d",&t);
while(t--)
{
scanf("%s",str);
int p;
for(i = 0,j = 0; i<9; i++)//保存位置，因为前面预处理的都是位置
{
if(str[i]=='X') p = i;
else
num[str[i]-'0'] = j++;
}
scanf("%s",str);
for(i = 0; i<9; i++)//求出目标状态每个数在原状态的位置
{
if(str[i]=='X')
continue;
str[i] = num[str[i]-'0']+'1';
}
s = node(str);//由目标态逆推到初始态
int sum = solve(s);
string ss="";
while(sum!=-1)
{
ss+=ans[p][sum];
sum = pre[p][sum];
}
printf("Case %d: %d\n",cas++,ss.size()-1);
for(i = ss.size()-2; i>=0; i--)//由于方案是逆推，输出也要逆推
printf("%c",ss[i]);
printf("\n");
}

return 0;
}

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