HDU4634:Swipe Bo(BFS+狀壓)

標籤：hdu

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. 
InputThe 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. 
OutputFor 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

327-1

 
題意：給出一個迷宮，沒有遇到特殊的點，就一直沿著同一方向一直往前走，一旦遇到牆了就停下，然後可以自由選擇方向前進，URLD幾個是強制轉換方向，E是終點，如果圖中有K，那麼要經過所有的K獲得鑰匙之後，門才會開，才能離開
思路：狀態壓縮記錄所有訪問的狀態

#include <iostream>#include <stdio.h>#include <string.h>#include <stack>#include <queue>#include <map>#include <set>#include <vector>#include <math.h>#include <bitset>#include <algorithm>#include <climits>using namespace std;#define ls 2*i#define rs 2*i+1#define UP(i,x,y) for(i=x;i<=y;i++)#define DOWN(i,x,y) for(i=x;i>=y;i--)#define MEM(a,x) memset(a,x,sizeof(a))#define W(a) while(a)#define gcd(a,b) __gcd(a,b)#define LL long long#define N 205#define MOD 1000000007#define INF 0x3f3f3f3f#define EXP 1e-8#define rank rank1const int mod = 10007;struct node{    int x,y,step,key;};int n,m,sx,sy,knum,kk[N][N];char g[N][N];bool vis[N][N][4][1<<7];bool used[N][N][1<<7];int to[4][2]= {1,0,0,1,-1,0,0,-1};int check(int x,int y){    if(x<0||y<0||x>=n||y>=m)        return 1;    return 0;}int bfs(){    int i,j;    MEM(vis,0);    MEM(used,0);    queue<node> Q;    node a,next;    a.x = sx;    a.y = sy;    a.step = 0;    a.key = 0;    Q.push(a);    used[a.x][a.y][0] = 1;    while(!Q.empty())    {        a = Q.front();        Q.pop();        for(i = 0; i<4; i++)        {            int mx=to[i][0],my=to[i][1];            int x,y,k,d;            x = a.x,y=a.y,k=a.key;            while(1)            {                if(g[x][y]=='D')                    mx=1,my=0;                if(g[x][y]=='R')                    mx=-0,my=1;                if(g[x][y]=='U')                    mx=-1,my=0;                if(g[x][y]=='L')                    mx=0,my=-1;                if(mx==1&&my==0)                    d=0;                if(mx==0&&my==1)                    d=1;                if(mx==-1&&my==0)                    d=2;                if(mx==0&&my==-1)                    d=3;                if(vis[x][y][d][k])                    break;                vis[x][y][d][k] = 1;                x+=mx,y+=my;                if(check(x,y)) break;                if(g[x][y]=='#') break;                if(g[x][y]=='E'&&k==(1<<knum)-1)                    return a.step+1;                if(g[x][y]=='D')                    mx=1,my=0;                if(g[x][y]=='R')                    mx=-0,my=1;                if(g[x][y]=='U')                    mx=-1,my=0;                if(g[x][y]=='L')                    mx=0,my=-1;                if(g[x][y]=='K')                    k|=kk[x][y];                if(!check(x+mx,y+my)&&g[x+mx][y+my]=='#')                {                    if(used[x][y][k])                        break;                    used[x][y][k] = 1;                    next.x = x;                    next.y = y;                    next.key = k;                    next.step=a.step+1;                    Q.push(next);                    break;                }            }        }    }    return -1;}int main(){    int i,j;    while(~scanf("%d%d",&n,&m))    {        knum = 0;        for(i = 0; i<n; i++)        {            scanf("%s",g[i]);            for(j = 0; j<m; j++)            {                if(g[i][j]=='S')                {                    sx=i,sy=j;                }                if(g[i][j]=='K')                {                    kk[i][j]=1<<knum;                    knum++;                }            }        }        printf("%d\n",bfs());    }    return 0;}

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HDU4634:Swipe Bo(BFS+狀壓)