POJ 1184 聪明的打字员

简直难到没朋友。

双向bfs + 剪枝。

剪枝策略：

对于2--5位置上的数，只有当光标在对应位置时通过swap ，up，down来改变，那么当当前位置没有达到目标状态时，left和right无意义。

好了，只剪掉这里就过掉了。。。

还有比较炫酷的方法实现枚举720种排列。。。然后状压什么的。。。功力不够完全看不懂。。。。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <queue>
#include <cmath>
#include <stack>
#include <map>

#pragma comment(linker, "/STACK:1024000000");
#define EPS (1e-8)
#define LL long long
#define ULL unsigned long long
#define _LL __int64
#define _INF 0x3f3f3f3f
#define Mod 9999991

using namespace std;

short int mark[1000000][7][2];

struct Q
{
    int ans,sta,site,fl;
};

int s[10];

int pre[] = {1000000,100000,10000,1000,100,10,1};
            //     0      1     2    3   4  5 6
inline int Cal(int num,int site,int ch)
{
    int temp = (num/pre[site])%10 + ch;
    if(0 <= temp && temp <= 9)
        return num/pre[site-1]*pre[site-1] + temp*pre[site] + num%pre[site];
    return num;
}

inline int Swap(int num,int site,int m)
{
    int t1 = (num/pre[site])%10;
    int t2 = (num/pre[m])%10;

    if(t1 == t2)
        return num;

    num = num/pre[site-1]*pre[site-1] + t2*pre[site] + num%pre[site];
    num = num/pre[m-1]*pre[m-1] + t1*pre[m] + num%pre[m];
    return num;
}

bool Judge(int site,int a,int b)
{
    if((a/pre[site])%10 == (b/pre[site])%10)
        return true;
    return false;
}

int bfs(int a,int b)
{
    memset(mark,-1,sizeof(mark));

    queue<Q> q;
    Q s,t;

    s.sta = a;
    s.fl = 0;
    s.ans = 0;
    s.site = 1;

    mark[s.sta][s.site][s.fl] = 0;
    q.push(s);

    s.sta = b;
    s.ans = 0;
    s.fl = 1;

    for(int i = 1;i <= 6; ++i)
    {
        s.site = i;
        mark[s.sta][s.site][s.fl] = 0;
        q.push(s);
    }

    while(q.empty() == false)
    {
        t = q.front();
        q.pop();

       // printf("site = %2d sta = %6d fl = %2d ans = %2d\n",t.site,t.sta,t.fl,t.ans);

        if(mark[t.sta][t.site][t.fl^1] != -1)
        {
            return mark[t.sta][t.site][t.fl^1] + t.ans;
        }

        s.fl = t.fl;
        s.ans = t.ans+1;

        //left
        s.sta = t.sta;

        if(t.site > 1 && (t.site < 2 || t.site > 5 || Judge(t.site,t.sta,t.fl == 0 ? b : a)))
        {
            s.site = t.site-1;
            if(mark[s.sta][s.site][s.fl] == -1)
            {
                mark[s.sta][s.site][s.fl] = s.ans;
                q.push(s);
            }
        }

        //right
        if(t.site < 6 && (t.site < 2 || t.site > 5 || Judge(t.site,t.sta,t.fl == 0 ? b : a)))
        {
            s.site = t.site+1;
            if(mark[s.sta][s.site][s.fl] == -1)
            {
                mark[s.sta][s.site][s.fl] = s.ans;
                q.push(s);
            }
        }

        //up
        s.sta = Cal(t.sta,t.site,1);
        s.site = t.site;
        if(mark[s.sta][s.site][s.fl] == -1)
        {
            mark[s.sta][s.site][s.fl] = s.ans;
            q.push(s);
        }

        //down
        s.sta = Cal(t.sta,t.site,-1);

        if(mark[s.sta][s.site][s.fl] == -1)
        {
            mark[s.sta][s.site][s.fl] = s.ans;
            q.push(s);
        }

        //swap0
        if(t.site != 1)
        {
            s.sta = Swap(t.sta,t.site,1);

            if(mark[s.sta][s.site][s.fl] == -1)
            {
                mark[s.sta][s.site][s.fl] = s.ans;
                q.push(s);
            }
        }

        //swap1
        if(t.site != 6)
        {
            s.sta = Swap(t.sta,t.site,6);
            if(mark[s.sta][s.site][s.fl] == -1)
            {
                mark[s.sta][s.site][s.fl] = s.ans;
                q.push(s);
            }
        }
    }
    return 0;
}

int main()
{
    //freopen("data.txt","r",stdin);
    int a,b;
    scanf("%d %d",&a,&b);
    printf("%d\n",bfs(a,b));

    return 0;
}

POJ 1184 聪明的打字员