LintCode 434. 岛屿的个数II（并查集）

文章目录

- 1. 题目
- 2. 解题

1. 题目

给定 n, m, 分别代表一个二维矩阵的行数和列数, 并给定一个大小为 k 的二元数组A.
初始二维矩阵全0.
二元数组A内的k个元素代表k次操作, 设第 i 个元素为 (A[i].x, A[i].y), 表示把二维矩阵中下标为A[i].x行A[i].y列的元素由海洋变为岛屿.
问在每次操作之后, 二维矩阵中岛屿的数量. 你需要返回一个大小为k的数组.

样例 1:
输入: n = 4, m = 5, 
A = [[1,1],[0,1],[3,3],[3,4]]
输出: [1,1,2,2]
解释: 
0.  00000000000000000000
1.  00000010000000000000
2.  01000010000000000000
3.  01000010000000000010
4.  01000010000000000011样例 2:
输入: n = 3, m = 3, 
A = [[0,0],[0,1],[2,2],[2,1]]
输出: [1,1,2,2]
注意事项
设定0表示海洋, 1代表岛屿, 并且上下左右相邻的1为同一个岛屿.

https://www.lintcode.com/problem/number-of-islands-ii/description

2. 解题

并查集求解连通分量个数

/*** Definition for a point.* struct Point {*     int x;*     int y;*     Point() : x(0), y(0) {}*     Point(int a, int b) : x(a), y(b) {}* };*/class Solution {
public:/*** @param n: An integer* @param m: An integer* @param operators: an array of point* @return: an integer array*/vector<int> f;int island = 0;void merge(int a, int b){int fa = find(a), fb = find(b);if(fa != fb){island--;f[fa] = fb;}}int find(int a){if(a == f[a]) return a;return f[a] = find(f[a]);}vector<int> numIslands2(int n, int m, vector<Point> &operators) {// write your code heref.resize(n*m);for(int i = 0; i < m*n; ++i)f[i] = i;unordered_set<int> landmark;//保存陆地，压缩为一维vector<vector<int>> dir = {{1,0},{0,1},{-1,0},{0,-1}};vector<int> ans(operators.size());for(int i = 0; i < operators.size(); ++i){int x = operators[i].x;int y = operators[i].y;int idx = m*x + y;if(!landmark.count(idx)){	//新的陆地landmark.insert(idx);island++;for(int k = 0; k < 4; ++k){	//周围的地方int nx = x+dir[k][0];int ny = y+dir[k][1];int nidx = m*nx+ny;if(nx>=0 && nx <n && ny>=0 && ny < m && landmark.count(nidx)){	// 新陆地的四周在界内，且是陆地merge(idx, nidx);// 合并}}}ans[i] = island;}return ans;}
};