[uva816]AbbottsRevenge Abbott的复仇(经典迷宫BFS)

这题思路就普通的BFS加上一个维度朝向，主要是要注意输入，输出，以及细节的处理

 

#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
using namespace std;
const int MAX = 9+1;
const int DIR = 4;
const int TURN = 3;
bool have_edge[MAX][MAX][DIR][TURN];//所在位置 ， 朝向 ， 要走的方向
int d[MAX][MAX][DIR];//最短路径 同时可以判断结点是否访问过
#define Node(x,n) x[n.r][n.c][n.dir]
struct node
{
int r , c, dir;
node(){}
node(int r, int c, int dir):r(r),c(c),dir(dir){}
};
struct node p[MAX][MAX][DIR];//父节点 打印 路径
const char *dirs = "NESW";//clockwise
const char *turns = "LFR";
inline int dir_id(char c) {return strchr(dirs,c)-dirs;}
inline int turn_id(char c) {return strchr(turns,c)-turns;}
const int MAX_NAME_LEN = 42;
char MazeName[MAX_NAME_LEN];//20个字符，还有空格。。。
int r0,c0;
int r1,c1,dir1;
int r2,c2;
int dr[] = {-1, 0, 1, 0};//clockwise
int dc[] = { 0, 1, 0,-1};
bool read()
{
gets(MazeName);//lineread
//
if(!strcmp(MazeName,"END")) return false;
char s[10]; int r, c;
if(!~scanf("%d%d%s",&r0,&c0,s)) return false;
printf("%sn",MazeName);
memset(have_edge,false,sizeof(have_edge));
dir1 = dir_id(s[0]);
r1 = r0 + dr[dir1]; c1 = c0 + dc[dir1];
scanf("%d%d",&r2,&c2);
while(scanf("%d",&r),r) {
scanf("%d",&c);
while(scanf("%s",s),*s != '*'){
char *p = s; int dir = dir_id(*s);//!!
while(*(++p)) { have_edge[r][c][dir][turn_id(*p)] = true;}
}
}
getchar();//吸收换行符
return true;
}
void print_ans(node u)
{
vector<node> ans;
for(;;){
ans.push_back(u);
if(Node(d,u) == 0) break;
u = Node(p,u);
}
ans.push_back(node(r0,c0,dir1));
int cnt = 0;
for(int i = ans.size() - 1; i >= 0; i--){
if(cnt%10 == 0) printf(" ");
printf(" (%d,%d)", ans[i].r, ans[i].c);
if(++cnt%10 == 0) puts("");
}
if(ans.size()%10) puts("");//注意
}
node walk(node &u,int turn){
int dir = u.dir;
if(turn == 0) dir = (dir + 3)%4;//rev
else if(turn == 2) dir = (dir + 1)%4;//clockwise
return node(u.r + dr[dir], u.c + dc[dir], dir );
}
inline bool inside(int r,int c) { return r > 0&&r <= 9 && c > 0 && c <= 9; }
void solve()
{
queue<node> q;
node u(r1,c1,dir1);//c1 - >c2
memset(d, -1, sizeof(d));
d[u.r][u.c][u.dir] = 0;
q.push(u);
while(!q.empty()){
u = q.front();q.pop();
if(u.r == r2 && u.c == c2) { print_ans(u); return ;}
for(int i = 0; i < 3; i++) {
if(!have_edge[u.r][u.c][u.dir][i]) continue;
node v = walk(u,i);
if(inside(v.r, v.c) && d[v.r][v.c][v.dir] < 0){
d[v.r][v.c][v.dir] = Node(d,u) + 1;
Node(p,v) = u;
q.push(v);//push(u) ...
}
}
}
printf("
No Solution Possiblen");
}
int main()
{
#ifdef local
freopen("in2.txt","r",stdin);
// freopen("myout.txt","w",stdout);
#endif // local
while(read()){
solve();
}
return 0;
}