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