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 ************************/