典型的bfs模拟 (广度优先搜索) ，不过有好多细节要注意，比如图中如果是 R# 走到这个R的话就无限往右走了，这样就挂了~肯定到不了出口。还有一种容易造成死循环的，比如
#E##
DLLL
D. .U
D.SU
RRRU
这样的话就必须要标记下当前位置某个方向获得钥匙的状态是否被访问过了，获得钥匙的状态可以状态压缩来表示。
墙角如果遇到转弯了是不会加step的！
#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std; struct Point {
int x, y, step, dir, now;
Point() {}
Point(int x, int y, int step, int dir, int now) : x(x), y(y), step(step), dir(dir), now(now) {}
}cur; queue<Point> q;
bool vis[4][1<<7][202][202];
int n, m, key, id[202][202], mp[333];
int dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, 1, -1};
char s[202][202];
int full ; int bfs() {
full = (1<<key)-1;
while(!q.empty()) {
cur = q.front();
q.pop();
int x = cur.x, y = cur.y, dir = cur.dir, step = cur.step, now = cur.now;
int xx = x + dx[dir], yy = y + dy[dir];
if(xx < 1 || yy < 1 || xx > n || yy > m) continue;
// 模拟走的路程，细节要注意
while(s[xx][yy] != '#') {
x = xx; y = yy;
if(s[xx][yy] == 'K') {
now |= id[xx][yy];
}
if(s[xx][yy] == 'E') {
if(now == full) return step;
}
if(mp[s[xx][yy]] >= 0) {
dir = mp[s[xx][yy]];
if(vis[dir][now][xx][yy]) break;
vis[dir][now][xx][yy] = 1;
}
xx += dx[dir]; yy += dy[dir];
if(xx < 1 || yy < 1 || xx > n || yy > m) break;
}
if(xx < 1 || yy < 1 || xx > n || yy > m) continue;
if(s[xx][yy] == '#' && mp[s[x][y]] == -1) {
for(int i = 0;i < 4; i++) if(i != dir) {
if(vis[i][now][x][y]) continue;
vis[i][now][x][y] = 1;
xx = x+dx[i];yy = y + dy[i];
if(s[xx][yy] == '#') continue;
cur = Point(x, y, step+1, i, now);
q.push(cur);
}
}
}
return -1;
} int main(){
memset(mp, -1, sizeof(mp));
mp['D'] = 0; mp['U'] = 1; mp['R'] = 2 ; mp['L'] = 3;
int i, j, x, y, k, l;
while(scanf("%d%d", &n, &m) != -1) {
for(i = 1;i <= n; i++)
scanf("%s", s[i]+1);
key = 0;
for(i = 1;i <= n; i++) {
for(j = 1;j <= m; j++)
if(s[i][j] == 'K') {
id[i][j] = 1<<key;
key++;
}
else if(s[i][j] == 'S')
x = i, y = j;
}
memset(vis ,0, sizeof(vis));
while(!q.empty()) q.pop();
for(i = 0;i < 4; i++) {
cur = Point(x, y, 1, i, 0);
vis[i][0][x][y] = 1;
q.push(cur);
}
printf("%d\n", bfs());
}
return 0;
}