华容道

华容道最优解，请将代码末的棋盘输入，程序会打印每一步走法。

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <map>#define Rep(i, x, y) for (int i = x; i <= y; i ++)#define Dwn(i, x, y) for (int i = x; i >= y; i --)#define RepE(i, x) for (int i = pos[x]; i; i = g[i].nex)#define v g[i].y#define col(x, y) map[x][y]#define Nexts sv = encode(map); if (find(sv, d + 1)) { ans[d] = s; return 1; }using namespace std;typedef long long ll;typedef double db;const int n = 5, m = 4;int px[14], py[14], l, lim = 1;int mx[4] = { 0, 1, 0, -1 }, my[4] = { 1, 0, -1, 0 };ll ans[600];map<ll, int> vis, step;ll encode(int map[n + 2][m + 2]) {int t1 = 3, t2 = 7;bool vd[14];memset(vd, 0, sizeof(vd)); vd[0] = 1;Rep(i, 1, n) {Rep(j, 1, m) {int c;if (!vd[ c = map[i][j] ]) {if (c >= 7) px[t2] = i, py[t2] = j, t2 ++;else px[c] = i, py[c] = j;vd[c] = 1;}}}ll s = 0;Dwn(i, 10, 1) s = (s * 4 + py[i] - 1) * 5 + px[i] - 1;return s;}void decode(ll s, int map[n + 2][m + 2]) {Rep(i, 1, 10) px[i] = s % 5 + 1, s /= 5, py[i] = s % 4 + 1, s /= 4;Rep(i, 0, n + 1) Rep(j, 0, m + 1) map[i][j] = 0;Rep(i, 0, 1) Rep(j, 0, 1) map[ px[1] + i ][ py[1] + j ] = 1;Rep(i, 0, 1) map[ px[2] ][ py[2] + i ] = 2;Rep(k, 3, 6) Rep(i, 0, 1) map[ px[k] + i ][ py[k] ] = k;Rep(k, 7, 10) map[ px[k] ][ py[k] ] = k;}bool find(ll s, int d) {if (d > lim || step[s] && step[s] < d) return 0;if (vis[s] == lim) return 0;vis[s] = lim; step[s] = d;int map[n + 2][m + 2];decode(s, map);int _x[3], _y[3], t = 0;Rep(i, 1, n) Rep(j, 1, m) if (!map[i][j]) _x[++ t] = i, _y[t] = j;if (map[5][2] == 1 && map[5][3] == 1) { printf("%d\n", d); ans[l = d] = s; return 1; }ll sv;Rep(i, 1, 2) {int x = _x[i], y = _y[i], c, x1, y1, x2, y2;Rep(j, 0, 3) {x1 = x + mx[j], y1 = y + my[j];c = map[x1][y1];x2 = x + mx[j] * 2, y2 = y + my[j] * 2;if (c >= 7) {map[x][y] = c, map[x1][y1] = 0;Nexts;map[x1][y1] = c;} else if (c > 1 && c == map[x2][y2]) {map[x][y] = c, map[x2][y2] = 0;Nexts;map[x2][y2] = c;}}map[x][y] = 0;}Rep(j, 0, 3) {int c1, c2, r;c1 = map[ _x[1] + mx[j] ][ _y[1] + my[j] ];if (!c1) continue ;c2 = map[ _x[2] + mx[j] ][ _y[2] + my[j] ];if (c1 == c2) {r = c1 == 1 ? 2 : 1;Rep(i, 1, 2) {map[ _x[i] ][ _y[i] ] = c1;map[ _x[i] + mx[j] * r ][ _y[i] + my[j] * r ] = 0;}Nexts;Rep(i, 1, 2) {map[ _x[i] ][ _y[i] ] = 0;map[ _x[i] + mx[j] * r ][ _y[i] + my[j] * r ] = c1;}}}return 0;}int main(){int t1 = 7, map[n + 2][m + 2];Rep(i, 0, n + 1) Rep(j, 0, m + 1) map[i][j] = 0;char s[8];Rep(i, 1, n) {gets (s + 1);Rep(j, 1, m) {if (s[j] == ' ') map[i][j] = 0;else map[i][j] = s[j] == '*' ? t1 ++ : s[j] - '0';}}while (!find( encode(map), 1 )) {lim ++;if (lim > 200) { puts("no solution"); return 0 ; }}cout << l << endl;Rep(t, 1, l) {system("pause");decode(ans[t], map);Rep(i, 1, n) {Rep(j, 1, m) {if (map[i][j] >= 7) printf("*");else if (!map[i][j]) printf(" ");else printf("%d", map[i][j]);}puts("");}}return 0;}/*3114311452265**6*  **/