习题6-12 筛子难题（A Dicey Problem, ACM/ICPC World Finals 1999, UVa810）

0. bfs迷宫求解的题。只不过这题状态多了两种，依旧是水题。
1. 记录下由色子的上前到右的映射，且对面的点数和为7。
2. 起点相同，入队列的判定要做出一点改变 (dis <= 0)。
3. 为了减少首坐标的输出特判，可以在输出name时先不输出换行。

#include <iostream>#include <string>#include <vector>#include <stack>#include <queue>#include <deque>#include <set>#include <map>#include <algorithm>#include <sstream>#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 100005#define mod 1000000007#define INF 1000000007#define eps 1e-5#define PI 3.1415926535898#define N 26using namespace std;//-------------------------CHC------------------------------//int n, m;struct Node {int r, c;int t, f;Node(int r = 0, int c = 0, int t = 0, int f = 0) : r(r), c(c), t(t), f(f) { }};Node s;int G[maxn][maxn];int r[7][7];int d[maxn][maxn][7][7];Node p[maxn][maxn][7][7];const int dx[] = { -1, 1, 0, 0 };const int dy[] = { 0, 0, -1, 1 };bool inside(Node u) { return u.r >= 1 && u.c >= 1 && u.r <= n && u.c <= m; }bool same(Node u) { return u.r == s.r && u.c == s.c; }Node walk(Node u, int i) {Node v(u.r + dx[i], u.c + dy[i]);if (i == 0) v.t = u.f, v.f = 7 - u.t;else if (i == 1) v.t = 7 - u.f, v.f = u.t;else if (i == 2) v.t = r[u.t][u.f], v.f = u.f;else v.t = 7 - r[u.t][u.f], v.f = u.f;return v;}void print(Node u) {vector<pair<int, int>> v;bool first = true;for (;;) {v.push_back(make_pair(u.r, u.c));if (u.r == s.r && u.c == s.c) {if (first) first = false;else break;}u = p[u.r][u.c][u.t][u.f];}for (int i = v.size() - 1, cnt = 0; i >= 0; --i, ++cnt) {if (cnt % 9 == 0) printf("\n  ");printf("(%d,%d)", v[i].first, v[i].second);if (i) putchar(',');}puts("");}void bfs() {CLEAR(d, -1);queue<Node> q;q.push(s);d[s.r][s.c][s.t][s.f] = 0;bool first = true;while (q.size()) {Node u = q.front(); q.pop();if (first) first = false;else if (same(u)) {print(u);return;}for (int i = 0; i < 4; ++i) {Node v = walk(u, i);if (inside(v) && d[v.r][v.c][v.t][v.f] <= 0 && (u.t == G[v.r][v.c] || G[v.r][v.c] == -1)) {d[v.r][v.c][v.t][v.f] = d[u.r][u.c][u.t][u.f] + 1;p[v.r][v.c][v.t][v.f] = u;q.push(v);}}}puts("\n  No Solution Possible");}int main() {r[1][2] = r[2][6] = r[6][5] = r[5][1] = 3;r[1][3] = r[3][6] = r[6][4] = r[4][1] = 5;r[2][3] = r[3][5] = r[5][4] = r[4][2] = 1;r[2][1] = r[6][2] = r[5][6] = r[1][5] = 4;r[3][1] = r[6][3] = r[4][6] = r[1][4] = 2;r[3][2] = r[5][3] = r[4][5] = r[2][4] = 6;char name[25];while (~scanf("%s", name) && strcmp(name, "END")) {CLEAR(G, 0);scanf("%d%d%d%d%d%d", &n, &m, &s.r, &s.c, &s.t, &s.f);for (int i = 1; i <= n; ++i)for (int j = 1; j <= m; ++j)scanf("%d", &G[i][j]);printf("%s", name);bfs();}return 0;}