leetcode79——Word Search

/*
Word Search
Given a 2D board and a word, find if the word exists in the grid.

The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.

For example,
Given board =

[
["ABCE"],
["SFCS"],
["ADEE"]
]
word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.
*/
#include<iostream>
#include<vector>
using namespace std;

class Solution {
public:
bool exist(vector<vector<char> > &board, string word) {
const int m = board.size();
const int n = board[0].size();
vector<vector<bool> > visited(m, vector<bool>(n, false));
for (int i = 0; i < m; ++i)
for (int j = 0; j < n; ++j)
if (dfs(board, word, 0, i, j, visited))
return true;
return false;
}
private:
static bool dfs(const vector<vector<char> > &board, const string &word,
int index, int x, int y, vector<vector<bool> > &visited) {
if (index == word.size())
return true;														// 收敛条件
if (x < 0 || y < 0 || x >= board.size() || y >= board[0].size())
return false;														// 越界，终止条件
if (visited[x][y]) return false;										// 已经访问过，剪枝
if (board[x][y] != word[index]) return false;							// 不相等，剪枝
visited[x][y] = true;
bool ret = dfs(board, word, index + 1, x - 1, y, visited) ||			// 上
dfs(board, word, index + 1, x + 1, y, visited) ||						// 下
dfs(board, word, index + 1, x, y - 1, visited) ||						// 左
dfs(board, word, index + 1, x, y + 1, visited);							// 右
visited[x][y] = false;
return ret;
}
};

int main(){
vector<vector<char> > board;
string word="SEE";

char m;
for(int i=0;i<3;i++){
vector<char> b;
for(int j=0;j<4;j++){
//b.clear();
cin>>m;
b.push_back(m);
}
board.push_back(b);
}
Solution eg;
cout<<eg.exist(board,word)<<endl;
for(int i=0;i<3;i++){
for(int j=0;j<4;j++)
cout<<board[i][j];
cout<<endl;
}
cout<<endl;
return 0;
}