UVa 10181

題目：15數碼求解。

分析：搜索，A*。利用A*算法求解15數碼。利用逆序數判斷是否可解。

            可解：左右移動時、逆序數不變，上下移動時、逆序數變化奇偶性、加上行變換次數，奇偶性不變；

                       （這裡計算逆序數是，不計入0，這個是空位）

            估值：哈密爾頓距離 + 當前步數（這兩項可以有個比例權值）；

                       （哈密爾頓距離為每個數字的當前位置和目標位置的行數差 + 列數差，的總和）；

            搜索：這裡直接利用bfs的框架，使用map判重，每次想四個方向搜索即可；

說明：跑了9秒多，( ⊙ o ⊙ )！

#include <iostream>#include <cstdlib>#include <string>#include <queue>#include <map>using namespace std;typedef struct _qnode{string state;string path;int    step;int    h;bool operator < (const _qnode &n) const    {        return h + step > n.h + n.step;    }    _qnode(string State, string Path, int Step, int H) {state = State;path  = Path;step  = Step;h     = H;}}qnode;string array_to_string(int data[]){string str = "123456789abcdef0";for (int i = 0; i < 16; ++ i) {str[i] = data[i] + '0';}return str;}int target[16] = {15, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14};int get_h(string state){int count = 0, place[16];for (int i = 0; i < 16; ++ i) {place[state[i]-'0'] = i;}for (int i = 0; i < 16; ++ i) {count += abs(place[i]%4 - target[i]%4) + abs(place[i]/4 - target[i]/4);}return count;}int get_zero_index(string state){for (int i = 0; i < 16; ++ i) {if (state[i] == '0') {return i;}}return -1;}char move_step[4] = {'D', 'R', 'U', 'L'};int  dxy[4][2] = {{1,0}, {0,1}, {-1,0}, {0,-1}};int  A_star(int start[], int over[]){string s_state = array_to_string(start);string e_state = array_to_string(over);map <string, int> M;M[s_state] = 1;priority_queue <qnode> Q;Q.push(qnode(s_state, "", 0, get_h(s_state))); while (!Q.empty()) {qnode now = Q.top(); Q.pop();if (now.step > 50) {continue;}if (now.state == e_state) {cout << now.path << endl;return 0;}int index = get_zero_index(now.state);int x = index/4, y = index%4;for (int k = 0; k < 4; ++ k) {int xx = x + dxy[k][0];int yy = y + dxy[k][1];if (xx >= 0 && xx < 4 && yy >= 0 && yy < 4) {string new_state = now.state;swap(new_state[index], new_state[xx*4+yy]);if (!M[new_state]) {M[new_state] = 1;Q.push(qnode(new_state, now.path + move_step[k], now.step + 1, get_h(new_state)));}}}}return -1;}int inversion(int data[]){int count = 0;for (int i = 0; i < 16; ++ i) {if (data[i]) {for (int j = 0; j < i; ++ j) {count += data[j] > data[i];}}else {count += i/4;}}return count%2;}int input[16];int ans[16] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 0};int main(){int n;while (cin >> n)while (n --) {for (int i = 0; i < 16; ++ i) {cin >> input[i];}if (!inversion(input) || A_star(input, ans) == -1) {cout << "This puzzle is not solvable." << endl;}}return 0;}