inpu add align needed 方向 square nod character req

Swipe Bo

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1549 Accepted Submission(s): 315


Problem Description “Swipe Bo” is a puzzle game that requires foresight and skill.
The main character of this game is a square blue tofu called Bo. We can swipe up / down / left / right to move Bo up / down / left / right. Bo always moves in a straight line and nothing can stop it except a wall. You need to help Bo find the way out.
The picture A shows that we needs three steps to swipe Bo to the exit (swipe up, swipe left, swipe down). In a similar way, we need only two steps to make Bo disappear from the world (swipe left, swipe up)!

Look at the picture B. The exit is locked, so we have to swipe Bo to get all the keys to unlock the exit. When Bo get all the keys, the exit will unlock automatically .The exit is considered inexistent if locked. And you may notice that there are some turning signs, Bo will make a turn as soon as it meets a

turning signs. For example, if we swipe Bo up, it will go along the purple line.
Now, your task is to write a program to calculate the minimum number of moves needed for us to swipe Bo to the exit.
Input The input contains multiple cases, no more than 40.
The first line of each test case contains two integers N and M (1≤N, M≤200), which denote the sizes of the map. The next N lines give the map’s layout, with each line containing M characters. A character is one of the following: ‘#‘: represents the wall; ‘S‘ represents the start point of the Bo; ‘E‘ represents the exit; ‘.‘ represents an empty block; ‘K’ represents the key, and there are no more than 7 keys in the map; ‘L‘,‘U‘,‘D‘,‘R‘ represents the turning sign with the direction of left, up, down, right.
Output For each test case of the input you have to calculate the minimal amount of moves which are necessary to make Bo move from the starting point to the exit. If Bo cannot reach the exit, output -1. The answer must be written on a single line.
Sample Input

5 6
######
#....#
.E...#
..S.##
.#####
5 6
######
#....#
.....#
SEK.##
.#####
5 6
######
#....#
....K#
SEK.##
.#####
5 6
######
#....#
D...E#
S...L#
.#####

Sample Output

3
2
7
-1

Source 2013 Multi-University Training Contest 4

題意：有一個迷宮，包括墻、空白格子、起點S、終點E、方向格子（LRUD）和鑰匙K。要求例如以下：

（1）每次轉彎僅僅能在碰到墻壁時（每次轉彎的選擇和初始時從S出發的方向選擇均稱為一次操作）；

（2）對於方向格子。若到達該格子，無論周圍是不是墻，必須轉向該格子指示的方向（這個不算一次操作）。

（3）若迷宮中沒有鑰匙存在，則求出S到E的最少操作次數；若有鑰匙。則必須先遍歷到每一個鑰匙之後才幹去E（在這個過程中能夠經過E也就是E不算做障礙）。

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

const int N  = 225;
const int inf = 1<<29;
struct node{
    int x,y,sta,stp;
};
int n,m,k;
char mapt[N][N];
int K[N][N];
bool vist[N][N][1<<7];
int dir[4][2]={0,-1,0,1,-1,0,1,0};

int judge1(node& now,int &e){
    int flag=0;
        if(now.x<0||now.x>=n||now.y<0||now.y>=m)
            return 0;
        if(mapt[now.x][now.y]!=‘#‘){

            if(mapt[now.x][now.y]==‘L‘)
                    e=0,flag=1;
            else if(mapt[now.x][now.y]==‘R‘)
                        e=1,flag=1;
            else if(mapt[now.x][now.y]==‘U‘)
                        e=2,flag=1;
            else if(mapt[now.x][now.y]==‘D‘)
                        e=3,flag=1;
             else if(mapt[now.x][now.y]==‘K‘)
                        now.sta|=(1<<K[now.x][now.y]);
            if(flag&&vist[now.x][now.y][now.sta])
                return 0;
            else if(flag) vist[now.x][now.y][now.sta]=1;  //固定方向的位置。能夠直接標記
            return 1;
        }
        else    //遇到墻，退一格，在當前位置停止
        {
            now.x-=dir[e][0];
            now.y-=dir[e][1];
            return 2;
        }

}
int bfs(int sx,int sy){
    queue<node>q;
    node now,pre;

    for(int i=0; i<n; i++)
        for(int j=0; j<m; j++)
         for(int sta=0; sta<(1<<k); sta++)
          vist[i][j][sta]=0;

        now.x=sx,now.y=sy,now.sta=0,now.stp=0;
        q.push(now);
        vist[now.x][now.y][now.sta]=1;

    while(!q.empty()){
        pre=q.front(); q.pop();
        pre.stp++;

        for(int te=0; te<4; te++){
            int e=te;
            now=pre;
            while(1){   //找到一個停止點。或不能走。或走過了。則跳出
                now.x+=dir[e][0];
                now.y+=dir[e][1];

                int flag=judge1(now,e);

                if(flag==0)break;
                if(flag==1&&mapt[now.x][now.y]==‘E‘&&now.sta==(1<<k)-1){
                    return now.stp;
                }
                if(flag==2){
                    if(vist[now.x][now.y][now.sta])break;
                    vist[now.x][now.y][now.sta]=1;
                    q.push(now);
                    break;
                }
            }
        }
    }

    return -1;
}
int main()
{
    int sx,sy;
    while(scanf("%d%d",&n,&m)>0){
            k=0;
        for(int i=0; i<n; i++){
            scanf("%s",mapt[i]);
            for(int j=0; j<m; j++)
            if(mapt[i][j]==‘S‘)
             sx=i,sy=j;
            else if(mapt[i][j]==‘K‘)
                K[i][j]=k++;
        }
        printf("%d\n",bfs(sx,sy));
    }
}

hdu Swipe Bo(bfs＋狀態壓縮)錯了多次的題