LC 980. Unique Paths III

On a 2-dimensional grid, there are 4 types of squares:

• 1 represents the starting square.  There is exactly one starting square.
• 2 represents the ending square.  There is exactly one ending square.
• 0 represents empty squares we can walk over.
• -1 represents obstacles that we cannot walk over.

Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once.

Example 1:

Input: [[1,0,0,0],[0,0,0,0],[0,0,2,-1]]
Output: 2
Explanation: We have the following two paths:
1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2)
2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2)

Example 2:

Input: [[1,0,0,0],[0,0,0,0],[0,0,0,2]]
Output: 4
Explanation: We have the following four paths:
1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3)
2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3)
3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3)
4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3)

Example 3:

Input: [[0,1],[2,0]]
Output: 0
Explanation:
There is no path that walks over every empty square exactly once.
Note that the starting and ending square can be anywhere in the grid.

Note:

1. 1 <= grid.length * grid[0].length <= 20

Runtime: 0 ms, faster than 100.00% of C++ online submissions for Unique Paths III.

//
// Created by yuxi on 2019/1/21.
//

#include <vector>
#include <iostream>
using namespace std;

class Solution {
public:
  int cntzero;
  int ret;
  vector<vector<int>> dirs = {{0,1},{0,-1},{-1,0},{1,0}};
  int uniquePathsIII(vector<vector<int>>& grid) {
    vector<vector<int>> records(2, vector<int>(2,0));
    ret = 0;
    cntzero = 0;
    for(int i=0; i<grid.size(); i++) {
      for(int j=0; j < grid[0].size(); j++) {
        if(grid[i][j] == 1) {
          records[0][0] = i;
          records[0][1] = j;
        } else if(grid[i][j] == 2){
          records[1][0] = i;
          records[1][1] = j;
        } else if(grid[i][j] == 0) cntzero++;
      }
    }
    int cnt = 0;
    vector<bool> used(grid.size()*grid[0].size(), false);
    vector<vector<int>> path;
    helper(grid, path, records[0], records[1], cnt, used);
    //cout << ret << endl;
    return ret;
  }
  void helper(vector<vector<int>>& grid, vector<vector<int>>& path, vector<int> s, vector<int>& e, int cnt, vector<bool>& used) {
//    for(int i=0; i<path.size(); i++) {
//      cout << "("<< path[i][0] << " " << path[i][1] << ")" << " ";
//    }
    //printgird(grid);
    int N = grid.size(), M = grid[0].size();
    if(s[0] == e[0] && s[1] == e[1]) {
//      cout << "(" << s[0] << " " << s[1] << ")" << " " << endl;
      if(cnt == cntzero) ret++;
      return;
    }
    // cout << endl;
//    used[s[0]*N+s[1]] = true;
    grid[s[0]][s[1]] = -2;
    path.push_back({s[0],s[1]});
    for(auto& dir : dirs) {
      int newx = dir[0] + s[0], newy = dir[1] + s[1];
      if(newx >= 0 && newx < N && newy >= 0 && newy < M && grid[newx][newy] != -2 && grid[newx][newy] != 1 && grid[newx][newy] != -1) {
        int newcnt = cnt;
        if(grid[newx][newy] == 0) newcnt++;
        helper(grid, path, {newx, newy}, e, newcnt, used);
      }
    }
    grid[s[0]][s[1]] = 0;
//    used[s[0]*N+s[1]] = false;
    path.pop_back();
  }

  void printgird(vector<vector<int>>& grid) {
    int N = grid.size(), M = grid[0].size();
    for(int i=0; i<N; i++) {
      for(int j=0; j<M; j++) {
        cout << grid[i][j] << " ";
      }
      cout << endl;
    }
  }
};

原文地址：https://www.cnblogs.com/ethanhong/p/10306766.html