POJ 3131 Cubic Eight-Puzzle 双向BFS + HASH

标签：style   os   for   io   amp   ios   

超赞的一道搜索题，麻烦到没朋友，不过思路还是很清楚的。

双向搜索的时候需要注意一个地方就是起始状态只有一个，目标状态最多的时候感觉会有512个，所以要控制两个方向搜索的层数。第一次知道双向搜索还能控制层数，简直点赞。

#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 6000007

using namespace std;

char Map[5][5];

int Judge(char a,char b,char c)
{
    if(a == 'W')
    {
        if(b == 'R')
            return 0;
        return 5;
    }
    else if(a == 'R')
    {
        if(b == 'W')
            return 1;
        return 3;
    }
    else if(a == 'B')
    {
        if(b == 'R')
            return 2;
        return 4;
    }
    else return 6;
}

struct Q
{
    int sta,x,y,fl,ans;
};

int mark[3000010][2];

int jx[] = { 0, 0, 1,-1};
int jy[] = {-1, 1, 0, 0};

int re[10][10];

int bit[10];

struct H
{
    int site,data,next;
}st[Mod];

int head[Mod];

int Top_H;

int Query(int data,int Top)
{
    int temp = data%Mod;

    int p = head[temp];
    for(;p != -1; p = st[p].next)
    {
        if(st[p].data == data)
            return st[p].site;
    }

    p = Top_H++;
    st[p].data = data;
    st[p].site = Top;
    st[p].next = head[temp];
    head[temp] = p;
    return -1;
}

int bfs(int x,int y)
{
    Top_H = 0;
    memset(mark,-1,sizeof(mark));

    int Top = 0;

    int i,j;

    for(i = 0;i < Mod; ++i)
        head[i] = -1;

    int sta = 0;

    for(i = 0; i < 3; ++i)
    {
        for(j = 0; j < 3; ++j)
            if(i == x && j == y)
                sta += Judge('E','E','E')*bit[i*3+j+1];
            else
                sta += Judge('W','B','R')*bit[i*3+j+1];
    }

    Q f,t;
    queue<Q> q;

    f.sta = sta;
    f.ans = 0,f.x = x,f.y = y,f.fl = 0;
    q.push(f);

    Query(f.sta,Top);

    mark[Top][0] = 0;
    Top++;

    f.ans = 0;
    f.fl = 1;

    for(i = 0; i < 512; ++i)
    {
        sta = 0,x = 0,y = 0;
        for(j = 0; j < 9; ++j)
        {
            if(Map[x][y] == 'E')
            {
                f.x = x;
                f.y = y;
                sta += Judge('E','E','E')*bit[x*3+y+1];
            }
            else if(i&(1<<j))
            {
                if(Map[x][y] == 'W')
                    sta += Judge('W','R','B')*bit[x*3+y+1];
                else if(Map[x][y] == 'R')
                    sta += Judge('R','B','W')*bit[x*3+y+1];
                else
                    sta += Judge('B','R','W')*bit[x*3+y+1];
            }
            else
            {
                if(Map[x][y] == 'W')
                    sta += Judge('W','B','R')*bit[x*3+y+1];
                else if(Map[x][y] == 'R')
                    sta += Judge('R','W','B')*bit[x*3+y+1];
                else
                    sta += Judge('B','W','R')*bit[x*3+y+1];
            }
            y++;
            if(y == 3)
                y = 0,x++;
        }
        f.sta = sta;
        q.push(f);
        if(Query(f.sta,Top) == -1)
        {
            mark[Top++][1] = 0;
        }
        else
        {
            mark[Query(f.sta,Top)][1] = 0;
        }
    }

    int site,tsite,tdata;

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

        if((f.fl == 1 && f.ans > 9) || (f.fl == 0 && f.ans > 20))
            continue;

        if(mark[Query(f.sta,Top)][f.fl^1] != -1)
        {
            return mark[Query(f.sta,Top)][f.fl^1] + f.ans;
        }
        site = f.x*3 + f.y;

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

        for(i = 0; i < 4; ++i)
        {
            t.x = f.x + jx[i];
            t.y = f.y + jy[i];

            if(0 <= t.x && t.x <= 2 && 0 <= t.y && t.y <= 2)
            {
                tsite = t.x*3+t.y;
                tdata = f.sta/bit[tsite+1]%7;

                t.sta = f.sta/bit[tsite]*bit[tsite] + 6*bit[tsite+1] + f.sta%bit[tsite+1];
                t.sta = t.sta/bit[site]*bit[site] + re[tdata][i]*bit[site+1] + t.sta%bit[site+1];

                if(Query(t.sta,Top) == -1)
                {
                    mark[Top++][t.fl] = t.ans;
                    q.push(t);
                }
                else if(mark[Query(t.sta,Top)][t.fl] == -1)
                {
                    mark[Query(t.sta,Top)][t.fl] = t.ans;
                    q.push(t);
                }
            }
        }
    }
    return -1;
}

int main()
{
    re[0][0] = 1;
    re[0][1] = 1;
    re[0][2] = 2;
    re[0][3] = 2;

    re[1][0] = 0;
    re[1][1] = 0;
    re[1][2] = 4;
    re[1][3] = 4;

    re[2][0] = 3;
    re[2][1] = 3;
    re[2][2] = 0;
    re[2][3] = 0;

    re[3][0] = 2;
    re[3][1] = 2;
    re[3][2] = 5;
    re[3][3] = 5;

    re[4][0] = 5;
    re[4][1] = 5;
    re[4][2] = 1;
    re[4][3] = 1;

    re[5][0] = 4;
    re[5][1] = 4;
    re[5][2] = 3;
    re[5][3] = 3;

    int i,j,x,y;

    bit[9] = 1;

    for(i = 8; i >= 0; --i)
        bit[i] = bit[i+1]*7;

    while(scanf("%d %d",&y,&x) && (x||y))
    {

        for(i = 0; i < 3; ++i)
        {
            for(j = 0; j < 3; ++j)
            {
                scanf("%*c%c",&Map[i][j]);
            }
        }

        printf("%d\n",bfs(x-1,y-1));
    }

    return 0;
}

POJ 3131 Cubic Eight-Puzzle 双向BFS + HASH,布布扣,bubuko.com

POJ 3131 Cubic Eight-Puzzle 双向BFS + HASH

标签：style   os   for   io   amp   ios   

原文地址：http://blog.csdn.net/zmx354/article/details/36187339