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

#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 26
using 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;
}