【BFS】uva10047The Monocycle

/*
本题的特殊之处，到达一个格子时，因为朝向不同，以及接触地面的颜色不同，
会处于不同的状态；；；；；；；；；
把(x, y, d, c)作为一个结点，表示所在位置(x, y)，方向为d,颜色为c;;;;;
------------------------------------------------------------------------
在方向上我们把前，左，右编号为0,1,2；；；；
颜色，从蓝色开始编号为0,1,2,3；；；；；；；；；；
--------------------------------------------------------------------------
vis[x1][y1][0][0] = 1;起点S进入队列
node a, b;
a.x = x1;
a.y = y1;
a.d = 0;
a.c = 0;
a.t = 0;
q.push(a);
-----------------------------------------------------------------------
b.t = a.t + 1;前进方向；；；时间加1
b.c = (a.c + 1)%5;扩展后的颜色
b.x = a.x + d[a.d][0];扩展后的位置
b.y = a.y + d[a.d][1];
b.d = a.d;方向不变
if(judge(b))
{
    if(b.x == x2 && b.y == y2 && b.c == 0)是否到达终点
        return b.t;
    vis[b.x][b.y][b.d][b.c] = 1;
    q.push(b);
}
----------------------------------------------------------------------------------
b = a;向右转
b.t ++;
b.d = (b.d + 1)%4;扩展后的方向
if(judge(b))
{
    vis[b.x][b.y][b.d][b.c] = 1;
    q.push(b);
}
b = a;向左转
b.t ++;
b.d = (b.d - 1 + 4)%4;
if(judge(b))
{
    vis[b.x][b.y][b.d][b.c] = 1;
    q.push(b);
}
-------------------------------------------------------------------
for(int i=0; i<m; i++)寻找始末点S和T点
{
    for(int j=0; j<n; j++)
    {
        if(g[i][j] == 'S')
        {
            x1 = i;
            y1 = j;
            break;
        }
    }
}
for(int i=0; i<m; i++)
{
    for(int j=0; j<n; j++)
    {
        if(g[i][j] == 'T')
        {
            x2 = i;
            y2 = j;
            break;
        }
    }
}
------------------------------------------------------------------------------

*/
#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
#include <algorithm>
#define INF 0x3f3f3f3f

using namespace std;

const int MAXN = 30;

struct node
{
    int x, y, d, c, t;
};

int m, n;
char g[MAXN][MAXN];
int vis[MAXN][MAXN][6][6];
int x1, x2, y1, y2;
int d[4][2] = {{-1,0},{0,1},{1,0},{0,-1}};

int judge(node b)
{
    if(b.x < 0 || b.x >= m || b.y < 0 || b.y >= n)
        return false;
    if(g[b.x][b.y] == '#')
        return false;
    if(vis[b.x][b.y][b.d][b.c])
        return false;
    return true;
}

int bfs()
{
    queue<node> q;
    vis[x1][y1][0][0] = 1;
    node a, b;
    a.x = x1;
    a.y = y1;
    a.d = 0;
    a.c = 0;
    a.t = 0;
    q.push(a);
    while(!q.empty())
    {
        a = q.front();
        q.pop();
        b.t = a.t + 1;
        b.c = (a.c + 1)%5;
        b.x = a.x + d[a.d][0];
        b.y = a.y + d[a.d][1];
        b.d = a.d;
        if(judge(b))
        {
            if(b.x == x2 && b.y == y2 && b.c == 0)
                return b.t;
            vis[b.x][b.y][b.d][b.c] = 1;
            q.push(b);
        }
        b = a;
        b.t ++;
        b.d = (b.d + 1)%4;
        if(judge(b))
        {
            vis[b.x][b.y][b.d][b.c] = 1;
            q.push(b);
        }
        b = a;
        b.t ++;
        b.d = (b.d - 1 + 4)%4;
        if(judge(b))
        {
            vis[b.x][b.y][b.d][b.c] = 1;
            q.push(b);
        }
    }
    return -1;
}

int main()
{
    //freopen("input.txt","r",stdin);
    int kase = 1;
    while(scanf("%d%d",&m,&n) != EOF && (n+m))
    {
        if(kase != 1)
            printf("\n");
        printf("Case #%d\n",kase++);
        getchar();
        for(int i=0; i<m; i++)
            gets(g[i]);
        for(int i=0; i<m; i++)
        {
            for(int j=0; j<n; j++)
            {
                if(g[i][j] == 'S')
                {
                    x1 = i;
                    y1 = j;
                    break;
                }
            }
        }
        for(int i=0; i<m; i++)
        {
            for(int j=0; j<n; j++)
            {
                if(g[i][j] == 'T')
                {
                    x2 = i;
                    y2 = j;
                    break;
                }
            }
        }
        memset(vis, 0, sizeof(vis));
        int time = bfs();
        if(time == -1)
            printf("destination not reachable\n");
        else
            printf("minimum time = %d sec\n",time);
    }
    return 0;
}

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