LintCode 1044 · Largest Plus Sign (DP好题)

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本文介绍了一个编程问题，如何在一个2D网格中找出除给定列表中0以外，最大且轴对齐的加号形状（由4条等长臂构成），返回其大小。给出了使用动态规划和unordered_set优化的解决方案。

1044 · Largest Plus Sign
Algorithms
Medium

Description
In a 2D grid from (0, 0) to (N-1, N-1), every cell contains a 1, except those cells in the given list mines which are 0. What is the largest axis-aligned plus sign of 1s contained in the grid? Return the order of the plus sign. If there is none, return 0.

An “axis-aligned plus sign of 1s of order k” has some center grid[x][y] = 1 along with 4 arms of length k-1 going up, down, left, and right, and made of 1s. This is demonstrated in the diagrams below. Note that there could be 0s or 1s beyond the arms of the plus sign, only the relevant area of the plus sign is checked for 1s.

Examples of Axis-Aligned Plus Signs of Order k:

Order 1:
000
010
000

Order 2:
00000
00100
01110
00100
00000

Order 3:
0000000
0001000
0001000
0111110
0001000
0001000
0000000
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1.N will be an integer in the range [1, 500].
2.mines will have length at most 5000.
3.mines[i] will be length 2 and consist of integers in the range [0, N-1].

Example
Example 1:
Input: N = 5, mines = [[4, 2]]
Output: 2

Explanation:
11111
11111
11111
11111
11011
In the above grid, the largest plus sign can only be order 2.  There is more than one.

Example 2:
Input: N = 2, mines = []
Output: 1

Explanation:
11
11
So the max size of '+' is 1.

Example 3:
Input: N = 1, mines = [[0,0]]
Output: 0

Explanation:
There is no '+' in the matrix.

解法1：参考的网上的解法。
注意:

res = max(dp[row][col], res); 是在4个方向都走完之后再执行。
用set<vector>会超时，所以降维用set。
用unordered_set<vector> 编译会有问题，可能需要自定义bool operator<(vector &a, vector &b)。下次再验证。

class Solution {
public:
    /**
     * @param n: size of 2D grid
     * @param mines: in the given list
     * @return: the order of the plus sign
     */
    int orderOfLargestPlusSign(int n, vector<vector<int>> &mines) {
        unordered_set<int> mineSet;
        for (auto mine : mines) {
            mineSet.insert(mine[0] * n + mine[1]);
        }
        vector<vector<int>> dp(n, vector<int>(n, 0));
        int res = 0;
        for (int row = 0; row < n; row++) {
            int count = 0;
            for (int col = 0; col < n; col++) {
                if (mineSet.find(row * n + col) != mineSet.end()) {
                    count = 0;
                } else {
                    count++;
                }
                dp[row][col] = count;
                //res = max(dp[row][col], res);
            }
        }
        for (int row = 0; row < n; row++) {
            int count = 0;
            for (int col = n - 1; col >= 0; col--) {
                if (mineSet.find(row * n + col) != mineSet.end()) {
                    count = 0;
                } else {
                    count++;
                }
                dp[row][col] = min(dp[row][col], count);
                //res = max(dp[row][col], res);
            }
        }
        for (int col = 0; col < n; col++) {
            int count = 0;
            for (int row = 0; row < n; row++) { 
                if (mineSet.find(row * n + col) != mineSet.end()) {
                    count = 0;
                } else {
                    count++;
                }
                dp[row][col] = min(dp[row][col], count);
                //res = max(dp[row][col], res);
            }
        }
        for (int col = 0; col < n; col++) {
            int count = 0;
            for (int row = n - 1; row >= 0; row--) {
                if (mineSet.find(row * n + col) != mineSet.end()) {
                    count = 0;
                } else {
                    count++;
                }
                dp[row][col] = min(dp[row][col], count);
                res = max(dp[row][col], res);
            }
        }
        return res;
    }
};