UVA - 816 Abbott's Revenge（bfs）

题意：迷宫从起点走到终点，进入某点的朝向不同，可以出去的方向也不同，输出最短路。

分析：因为朝向决定接下来在该点可以往哪里走，所以每个点需要有三个信息：x,y,d(坐标和进入该点的朝向)，所以将起点的下一个点当做初始状态

注意理解题意:进入交叉点的朝向与从哪个方向进交叉点正好相反

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 1, 0, -1};
const int dc[] = {1, 0, -1, 0};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int MAXN = 10000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
char s[30];
bool edge[15][15][5][5];
int dis[15][15][5];
const char* a = "ESWN";//与数组dr,dc方向对应
const char* b = "LFR";
struct Node{
int x, y, d;//当前点的坐标，及进入该点的朝向

Node(){}
Node(int xx, int yy, int dd):x(xx), y(yy), d(dd){}
}num[15][15][5];//记录该点的上一个点
int dir_id(char c){
return strchr(a, c) - a;
}
int turn_id(char c){
return strchr(b, c) - b;
}
bool judge(int x, int y){
return x >= 1 && x <= 9 && y >= 1 && y <= 9;
}
Node next(const Node&t, int turn){//turn与字符数组b下标对应
int dir = t.d;//向前走朝向不变
if(!turn){//向左走
dir = (dir + 3) % 4;//减1等价于加3

}
else if(turn == 2){
dir = (dir + 1) % 4;
}
return Node(t.x + dr[dir], t.y + dc[dir], dir);
}
void print_ans(Node t, int tx, int ty, int dir){
stack<Node> st;
while(1){
st.push(t);
if(dis[t.x][t.y][t.d] == 0) break;
t = num[t.x][t.y][t.d];
}
st.push(Node(tx, ty, dir));
int cnt = 0;
while(!st.empty()){
Node tmp = st.top();
st.pop();
++cnt;
if(cnt % 10 == 1) printf(" ");
printf(" (%d,%d)", tmp.x, tmp.y);
if(cnt % 10 == 0) printf("n");
}
if(cnt % 10 != 0) printf("n");
}
bool bfs(int sx, int sy, int dir, int ex, int ey){
memset(dis, -1, sizeof dis);
queue<Node> q;
q.push(Node(sx, sy, dir));
dis[sx][sy][dir] = 0;
int tx = sx - dr[dir];//迷宫起点
int ty = sy - dc[dir];
while(!q.empty()){
Node t = q.front();
q.pop();
if(t.x == ex && t.y == ey){
print_ans(t, tx, ty, dir);
return true;
}
for(int i = 0; i < 3; ++i){
Node v = next(t, i);
if(edge[t.x][t.y][t.d][i] && judge(v.x, v.y) && dis[v.x][v.y][v.d] < 0){
dis[v.x][v.y][v.d] = dis[t.x][t.y][t.d] + 1;
num[v.x][v.y][v.d] = t;
q.push(v);
}
}
}
return false;
}
int main(){
while(scanf("%s", s) == 1){
if(strcmp(s, "END") == 0) return 0;
printf("%sn", s);
memset(edge, 0, sizeof edge);
int sx, sy, ex, ey;
char dir;
scanf("%d%d %c%d%d", &sx, &sy, &dir, &ex, &ey);
int d = dir_id(dir);//进起点的下一个点的朝向
int nx = sx + dr[d];//起点的下一个位置
int ny = sy + dc[d];
int x;
while(scanf("%d", &x) == 1){
if(x == 0) break;
int y;
scanf("%d", &y);
while(scanf("%s", s) == 1){
if(s[0] == '*') break;
int tmp_dir = dir_id(s[0]);
int len = strlen(s);
for(int i = 1; i < len; ++i){
int tmp_turn = turn_id(s[i]);
edge[x][y][tmp_dir][tmp_turn] = 1;
}
}
}
bool ok = bfs(nx, ny, d, ex, ey);//起点的下一个点因为已知，所以将其当为开始的点
if(!ok){
printf("
No Solution Possiblen");
}
}
return 0;
}