【LeetCode】1001. Grid Illumination 解题报告（C++）
作者： 负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/
题目地址：https://leetcode.com/problems/available-captures-for-rook/
题目描述
On a N x N grid of cells, each cell (x, y) with 0 <= x < N and 0 <= y < N has a lamp.
Initially, some number of lamps are on. lamps[i] tells us the location of the i-th lamp that is on. Each lamp that is on illuminates every square on its x-axis, y-axis, and both diagonals (similar to a Queen in chess).
For the i-th query queries[i] = (x, y), the answer to the query is 1 if the cell (x, y) is illuminated, else 0.
After each query (x, y) [in the order given by queries], we turn off any lamps that are at cell (x, y) or are adjacent 8-directionally (ie., share a corner or edge with cell (x, y).)
Return an array of answers. Each value answer[i] should be equal to the answer of the i-th query queries[i].
Example 1:
Input: N = 5, lamps = [[0,0],[4,4]], queries = [[1,1],[1,0]]
Output: [1,0]
Explanation:
Before performing the first query we have both lamps [0,0] and [4,4] on.
The grid representing which cells are lit looks like this, where [0,0] is the top left corner, and [4,4] is the bottom right corner:
1 1 1 1 1
1 1 0 0 1
1 0 1 0 1
1 0 0 1 1
1 1 1 1 1
Then the query at [1, 1] returns 1 because the cell is lit. After this query, the lamp at [0, 0] turns off, and the grid now looks like this:
1 0 0 0 1
0 1 0 0 1
0 0 1 0 1
0 0 0 1 1
1 1 1 1 1
Before performing the second query we have only the lamp [4,4] on. Now the query at [1,0] returns 0, because the cell is no longer lit.
Note:
- 1 <= N <= 10^9
- 0 <= lamps.length <= 20000
- 0 <= queries.length <= 20000
- lamps[i].length == queries[i].length == 2
题目大意
给出了一个N*N的格子空间，在lams[i]位置上有灯，每个灯会照亮相同x方向、相同y方向、和两条对角线方向共四个方向。我们给出了一系列的queries，这个queries[i]代表查询该位置是否有亮光，同时每次查询的话会把该位置和该位置的8联通方向的亮灯全部关掉。如果queries[i]有光亮的话，那么返回1，否则返回0，问最后的查询结果是多少。
解题方法
哈希
这个题目其实已经告诉我们，类似于象棋的皇后问题。那么就联想起前面做过的51. N-Queens问题，在N皇后问题中，判断两个点是否相同的x和y坐标当然容易，判断两点是否在对角线上怎么做呢？
在同一条左斜线上的点，方程式都形如x+y = c，也就是他们的坐标之和相等
在同一条右斜线上的点，方程式都刑辱y = x+c，也就是他们的坐标之差相等
所以，如果知道了这个结论，我们只需要四个字典，分别保存每个点的横坐标、纵坐标、x + y、x - y，然后如果有两个点的满足其中任何一个相等就说明两者共线。
代码还是很简单的，只是别手误就行。
C++代码如下：
class Solution {
public:
vector<int> gridIllumination(int N, vector<vector<int>>& lamps, vector<vector<int>>& queries) {
unordered_map<int, int> xcount;
unordered_map<int, int> ycount;
unordered_map<int, int> l_diagcount;
unordered_map<int, int> r_diagcount;
set<pair<int, int>> lset;
for (auto l : lamps) {
++xcount[l[0]];
++ycount[l[1]];
++l_diagcount[l[0] + l[1]];
++r_diagcount[l[0] - l[1]];
lset.insert({l[0], l[1]});
}
vector<int> res;
for (auto q : queries) {
if (xcount[q[0]] || ycount[q[1]] || l_diagcount[q[0] + q[1]] || r_diagcount[q[0] - q[1]]) {
res.push_back(1);
} else {
res.push_back(0);
}
for (int i = -1; i <= 1; ++i) {
for (int j = -1; j <= 1; ++j) {
pair<int, int> xy = {q[0] + i, q[1] + j};
if (lset.count(xy)) {
--xcount[xy.first];
--ycount[xy.second];
--l_diagcount[xy.first + xy.second];
--r_diagcount[xy.first - xy.second];
lset.erase(xy);
}
}
}
}
return res;
}
};
日期
2019 年 2 月 24 日 —— 周末又结束了
