SRM 588 D2 L3：GameInDarknessDiv2，DFS

题目来源：http://community.topcoder.com/stat?c=problem_statement&pm=12710

采用DFS搜索，第一次写的时候忘了加访问标志，结果状态空间呈指数增长（主要是因为有大量重复的状态），根本算不出结果，后来加入访问标志数组 v 后，就能保证不访问重复的状态的了。这道题目的启示就是使用DFS一定要记住确保不访问重复的状态，有些时候很容易就忘了这一点，导致算法失败。

代码如下：

 

#include <algorithm>

#include <iostream>

#include <sstream>



#include <string>

#include <vector>

#include <stack>

#include <deque>

#include <queue>

#include <set>

#include <map>



#include <cstdio>

#include <cstdlib>

#include <cctype>

#include <cmath>

#include <cstring>



using namespace std;





/*************** Program Begin **********************/

vector <string> f;

string M;

bool v[2600][50][50];

bool BobWin = false;

void move(int Ax, int Ay, int Bx, int By, int steps)

{

	v[steps][Bx][By] = true;

	if (steps == M.size()) {

		BobWin = true;

		return;

	} else {

		int nAx = Ax;

		int nAy = Ay;

		switch (M[steps]) {

		case 'U':

			nAy = Ay - 1;

			break;

		case 'R':

			nAx = Ax + 1;

			break;

		case 'L':

			nAx = Ax - 1;

			break;

		case 'D':

			nAy = Ay + 1;

			break;

		}

		if (nAx == Bx && nAy == By) {

			return;

		}

		int nBx = Bx;

		int nBy = By;

		// 上

		nBx = Bx;

		nBy = By - 1;

		if ( nBy >= 0 && '.' == f[nBy][nBx] && !(nAx == nBx && nAy == nBy) && !v[steps+1][nBx][nBy] ) {

			move(nAx, nAy, nBx, nBy, steps+1);

		}

		// 下

		nBx = Bx;

		nBy = By + 1;

		if ( nBy <= f.size() - 1 && '.' == f[nBy][nBx] && !(nAx == nBx && nAy == nBy) && !v[steps+1][nBx][nBy] ) {

			move(nAx, nAy, nBx, nBy, steps+1);

		}

		// 左

		nBx = Bx - 1;

		nBy = By;

		if ( nBx >= 0 && '.' == f[nBy][nBx] && !(nAx == nBx && nAy == nBy) && !v[steps+1][nBx][nBy] ) {

			move(nAx, nAy, nBx, nBy, steps+1);

		}

		// 右

		nBx = Bx + 1;

		nBy = By;

		if ( nBx <= f[0].size() - 1 && '.' == f[nBy][nBx] && !(nAx == nBx && nAy == nBy) && !v[steps+1][nBx][nBy] ) {

			move(nAx, nAy, nBx, nBy, steps+1);

		}

	}

}



class GameInDarknessDiv2 {

public:

	string check(vector <string> field, vector <string> moves) {

		string res = "";

		f = field;

		int Ax = 0, Ay = 0, Bx = 0, By = 0;

		for (int i = 0; i < f.size(); i++) {

			for (int j = 0; j < f[0].size(); j++) {

				if ('A' == f[i][j]) {

					Ay = i;

					Ax = j;

					f[i][j] = '.';

				} else if ('B' == f[i][j]) {

					By = i;

					Bx = j;

					f[i][j] = '.';

				}

			}

		}

		M = "";

		for (int i = 0; i < moves.size(); i++) {

			M += moves[i];

		}

		BobWin = false;

		memset(v, 0, sizeof(v));

		move(Ax, Ay, Bx, By, 0);

		if (BobWin) {

			return "Bob wins";

		} else {

			return "Alice wins";

		}

	}

};

/************** Program End ************************/