例题6-14　Abbott的复仇（Abbott's Revenge, ACM/ICPC World Finals 2000, UVa 816）

#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <algorithm>
#include <functional>
#include <utility>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <ctime>
#include <cmath>
#include <cctype>
#define CLEAR(a, b) memset(a, b, sizeof(a))
#define IN() freopen("in.txt", "r", stdin)
#define OUT() freopen("out.txt", "w", stdout)
#define LL long long
#define maxn 15
#define maxm 6000005
#define mod  10007
#define INF 1000000007
#define EPS 1e-7
#define PI 3.1415926535898
#define N 4294967296
using namespace std;
//-------------------------CHC------------------------------//
struct Node {
int r, c, dir;
Node(int r = 0, int c = 0, int dir = 0) : r(r), c(c), dir(dir) { }
};
bool a[maxn][maxn][4][3];
const char dirs[] = "NESW";
const char turns[] = "LRF";
const int dr[] = { -1, 0, 1, 0 };
const int dc[] = { 0, 1, 0, -1 };
int d[maxn][maxn][4];
Node p[maxn][maxn][4];

int Dir(int ch) { return strchr(dirs, ch) - dirs; }
int Turn(int ch) { return strchr(turns, ch) - turns; }

Node walk(Node u, int turn) {
int dir = u.dir, r, c;
if (turn == 0) dir = (u.dir + 3) % 4;
if (turn == 1) dir = (u.dir + 1) % 4;
r = u.r + dr[dir];
c = u.c + dc[dir];
return Node(r, c, dir);
}

void print(Node u, int sr, int sc) {
stack<Node> sta;
printf("  (%d,%d)", sr, sc);
while (d[u.r][u.c][u.dir]) {
sta.push(u);
u = p[u.r][u.c][u.dir];
}
sta.push(u);
int cnt = 1;
while (sta.size()) {
Node cur = sta.top(); sta.pop();
if (cnt % 10 == 0) putchar(' ');
printf(" (%d,%d)", cur.r, cur.c);
if (cnt % 10 == 9) putchar('\n');
cnt++;
}
if (cnt % 10) putchar('\n');	//细节
}

void BFS(Node s, int tr, int tc, int sr, int sc) {
CLEAR(d, -1);
queue<Node> q;
q.push(s);
d[s.r][s.c][s.dir] = 0;
while (q.size()) {
Node u = q.front(); q.pop();
//printf("r = %d, c = %d, dir = %d\n", u.r, u.c, u.dir);
if (u.r == tr && u.c == tc) { print(u, sr, sc); return; }
for (int i = 0; i < 3; ++i) {
Node v = walk(u, i);
if (a[u.r][u.c][u.dir][i] && d[v.r][v.c][v.dir] == -1) {
d[v.r][v.c][v.dir] = d[u.r][u.c][u.dir] + 1;
p[v.r][v.c][v.dir] = u;
q.push(v);
}
}
}
puts("  No Solution Possible");
}

int main() {
//IN(); OUT();
char s[25];
while (scanf("%s", s) && strcmp(s, "END")) {
CLEAR(a, 0);
puts(s);
int sr, sc, tr, tc, r0, c0, dir0, r, c;
scanf("%d%d%s%d%d", &sr, &sc, s, &tr, &tc);
dir0 = Dir(s[0]);
r0 = sr + dr[dir0];
c0 = sc + dc[dir0];
while (scanf("%d", &r) && r) {
scanf("%d", &c);
while (scanf("%s", s) && strcmp(s, "*")) {
for (int i = 1; i < strlen(s); ++i)
a[r][c][Dir(s[0])][Turn(s[i])] = true;
}
}
BFS(Node(r0, c0, dir0), tr, tc, sr, sc);
}
return 0;
}