算法.图论-习题全集(Updating)

本节设置的意义

主要就是为了复习图论算法, 尝试从题目解析的角度,更深入的理解图论算法…

并查集篇

并查集简介以及常见技巧

并查集是一种用于大集团查找, 合并等操作的数据结构, 常见的方法有

• find: 用来查找元素在大集团中的代表元素(这里使用的是扁平化的处理)
• isSameSet: 用来查找两个元素是不是一个大集团的(其实就是find的应用)
• union: 用来合并两大集团的元素

关于并查集打标签的技巧, 其实我们之前的size数组也是一种打标签的逻辑, 其实打标签就是给每一个集团的代表节点打上标签即可, 还有我们在并查集的题目中通常会设置一个sets作为集合的总数目(每次合并–), 这是一个常见的技巧, 并查集的细节我们在这里不进行过多的介绍, 在之前的章节中有细致的描述…

并查集板子(洛谷)

这里我们的并查集的板子使用的是洛谷的板子(小挂大的优化都没必要其实)

// 并查集模版(洛谷)
// 本实现用递归函数实现路径压缩，而且省掉了小挂大的优化，一般情况下可以省略
// 测试链接 : https://www.luogu.com.cn/problem/P3367import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;public class Main{public static int MAXN = 10001;public static int[] father = new int[MAXN];public static int n;public static void build() {for (int i = 0; i <= n; i++) {father[i] = i;}}public static int find(int i) {if (i != father[i]) {father[i] = find(father[i]);}return father[i];}public static boolean isSameSet(int x, int y) {return find(x) == find(y);}public static void union(int x, int y) {father[find(x)] = find(y);}public static void main(String[] args) throws IOException {BufferedReader br = new BufferedReader(new InputStreamReader(System.in));StreamTokenizer in = new StreamTokenizer(br);PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));while (in.nextToken() != StreamTokenizer.TT_EOF) {n = (int) in.nval;build();in.nextToken();int m = (int) in.nval;for (int i = 0; i < m; i++) {in.nextToken();int z = (int) in.nval;in.nextToken();int x = (int) in.nval;in.nextToken();int y = (int) in.nval;if (z == 1) {union(x, y);} else {out.println(isSameSet(x, y) ? "Y" : "N");}}}out.flush();out.close();br.close();}}

情侣牵手问题

本题的突破点就是如果一个大集团里面有 n 对情侣, 那么我们至少要交换 n - 1次(通过把情侣进行编号)

// 这次我们尝试使用轻量版的并查集来解决这道题
class Solution {private static final int MAX_CP = 31;private static final int[] father = new int[MAX_CP];private static int sets = 0;private static int find(int i) {if (i != father[i]) {father[i] = find(father[i]);}return father[i];}private static boolean isSameSet(int a, int b) {return find(a) == find(b);}private static void union(int a, int b) {int fa = find(a);int fb = find(b);if (fa != fb) {father[fa] = fb;sets--;}}// 初始化并查集private static void build(int n) {for (int i = 0; i < n; i++) {father[i] = i;}sets = n;}public int minSwapsCouples(int[] row) {build(row.length / 2);for (int i = 0; i < row.length; i += 2) {int n1 = row[i] / 2;int n2 = row[i + 1] / 2;union(n1, n2);}return row.length / 2 - sets;}
}

相似的字符串组

其实就是枚举每一个位置, 然后判断是不是一组的就OK了

// 还是使用一下轻量级的并查集板子
class Solution {private static final int MAX_SZ = 301;private static final int[] father = new int[MAX_SZ];private static int sets = 0;private static int find(int i) {if (i != father[i]) {father[i] = find(father[i]);}return father[i];}private static boolean isSameSet(int a, int b) {return find(a) == find(b);}private static void union(int a, int b) {int fa = find(a);int fb = find(b);if (fa != fb) {father[fa] = fb;sets--;}}// 初始化并查集private static void build(int n) {for (int i = 0; i < n; i++) {father[i] = i;}sets = n;}public int numSimilarGroups(String[] strs) {build(strs.length);// 主流程的时间复杂度是 O(n ^ 2), 遍历strs的每一个位置int m = strs[0].length();for (int i = 0; i < strs.length; i++) {for (int j = i + 1; j < strs.length; j++) {// 获取到两个字符串, 然后计算两个字符串的不同字符数量String s1 = strs[i];String s2 = strs[j];int diff = 0;for (int k = 0; k < m && diff < 3; k++) {if (s1.charAt(k) != s2.charAt(k))diff++;}if (diff == 0 || diff == 2)union(i, j);}}return sets;}
}

岛屿数量(并查集做法)

这道题的解法非常多, 比如多源 BFS , 洪水填充(其实就是递归加回溯) , 还有今天介绍的并查集的方法(这个方法不是最好的)

// 这个题的并查集做法只要注意一点就可以了: 把一个二维下标转化为一维下标
class Solution {private static final int MAX_SZ = 301 * 301;private static final int[] father = new int[MAX_SZ];private static int sets = 0;private static int row = 0;private static int col = 0;// 模拟bfs的move数组private static final int[] move = { -1, 0, 1, 0, -1 };private static int find(int i) {if (i != father[i]) {father[i] = find(father[i]);}return father[i];}private static boolean isSameSet(int a, int b) {return find(a) == find(b);}private static void union(int a, int b) {int fa = find(a);int fb = find(b);if (fa != fb) {father[fa] = fb;sets--;}}private static void build(char[][] grid, int rl, int cl) {row = rl;col = cl;sets = 0;for (int i = 0; i < row; i++) {for (int j = 0; j < col; j++) {if (grid[i][j] == '1') {sets++;father[getIndex(i, j)] = getIndex(i, j);}}}}public int numIslands(char[][] grid) {// 初始化并查集并统计 '1' 的数量build(grid, grid.length, grid[0].length);// 遍历grid进行合并for (int i = 0; i < row; i++) {for (int j = 0; j < col; j++) {// 向四个方向扩展if (grid[i][j] == '1') {for (int k = 0; k < 4; k++) {int nx = i + move[k];int ny = j + move[k + 1];if (nx >= 0 && nx < row && ny >= 0 && ny < col && grid[nx][ny] == '1') {union(getIndex(i, j), getIndex(nx, ny));}}}}}return sets;}// 二维下标转一维下标private static int getIndex(int i, int j) {return i * col + j;}
}

省份数量

没什么可说的, 就是一个简单的并查集的思路

class Solution {// 这其实也是一个并查集的题private static final int MAXM = 201;private static final int[] father = new int[MAXM];private static final int[] size = new int[MAXM];private static int sets = 0;private static int find(int i){if(i != father[i]){father[i] = find(father[i]);}return father[i];}private static boolean isSameSet(int a, int b){return find(a) == find(b);}private static void union(int a, int b){if(!isSameSet(a, b)){int fa = find(a);int fb = find(b);if(size[fa] > size[fb]){father[fb] = fa;size[fa] += size[fb];}else{father[fa] = fb;size[fb] += size[fa];}sets--;}}private static void build(int n){for(int i = 0; i < n; i++){father[i] = i;size[i] = 1;}sets = n;}public int findCircleNum(int[][] isConnected) {// 初始化并查集build(isConnected.length);for(int i = 0; i < isConnected.length; i++){int[] info = isConnected[i];for(int j = 0; j < info.length; j++){if(info[j] == 1){union(i, j);}}}return sets;}
}

移除最多的同行或同列石头

其实就是每一个集团最后都会被消成一个元素, 我们中间用哈希表加了一些关于离散化的处理的技巧

// 使用一下轻量版本的并查集加上哈希表进行离散化的操作
class Solution {private static Map<Integer, Integer> rowFirst = new HashMap<>();private static Map<Integer, Integer> colFirst = new HashMap<>();private static final int MAXM = 1001;private static final int[] father = new int[MAXM];private static int sets = 0;private static int find(int i){if(i != father[i]){father[i] = find(father[i]);}return father[i];}private static boolean isSameSet(int a, int b){return find(a) == find(b);}private static void union(int a, int b){int fa = find(a);int fb = find(b);if(fa != fb){father[fa] = fb;sets--;}}// 初始化并查集private static void build(int n){for(int i = 0; i < n; i++){father[i] = i;}sets = n;}public int removeStones(int[][] stones) {// 清空哈希表rowFirst.clear();colFirst.clear();// 初始化并查集build(stones.length);for(int i = 0; i < stones.length; i++){int row = stones[i][0];int col = stones[i][1];if(!rowFirst.containsKey(row)){rowFirst.put(row, i);}else{union(rowFirst.get(row), i);}if(!colFirst.containsKey(col)){colFirst.put(col, i);}else{union(colFirst.get(col), i);}}return stones.length - sets;}}

最大的人工岛

本题注意的点就是, 首先我们二维的矩阵, 想要使用并查集, 需要先把二维的坐标转化为一维的坐标, 然后通过一维的坐标使用并查集, 首先把所有的岛进行合并, 然后来到一个 空位置 , 就尝试向四个方向进行扩展尝试进行岛屿的链接, 最后返回最大的连成一片的岛屿数量即可

/*** 本题我们是采用的并查集(轻量板子)的方法来做*  核心点就是首先使用 并查集 (二维下标转换一维下标) 进行人工岛的合并*  本题需要我们使用size的辅助信息, 因为 size 也相当于打标签的技巧*  然后枚举每一个位置进行岛屿的合并*/
class Solution {private static final int MAX_LEN = 501;private static final int MAX_SIZE = MAX_LEN * MAX_LEN;private static final int[] father = new int[MAX_SIZE];private static final int[] size = new int[MAX_SIZE];private static int len = 0;private static final int[] move = {-1, 0, 1, 0, -1};private static int find(int i){if(i != father[i]){father[i] = find(father[i]);}return father[i];}private static boolean isSameSet(int a, int b){return find(a) == find(b);}private static void union(int a, int b){if(!isSameSet(a, b)){int fa = find(a);int fb = find(b);if(size[fa] > size[fb]){father[fb] = fa;size[fa] += size[fb];}else{father[fa] = fb;size[fb] += size[fa];}}}// 初始化并查集private static void build(int[][] grid){len = grid.length;for(int i = 0; i < len; i++){for(int j = 0; j < len; j++){if(grid[i][j] == 1){int index = getIndex(i, j);father[index] = index;size[index] = 1;}}}}// 二维下标转换为一维下标private static int getIndex(int i, int j){return i * len + j;}public int largestIsland(int[][] grid) {build(grid);int res = 0;// 遍历矩阵进行合并for(int i = 0; i < len; i++){for(int j = 0; j < len; j++){if(grid[i][j] == 1){// 此时向四周进行扩展合并for(int k = 0; k < 4; k++){int nx = i + move[k];int ny = j + move[k + 1];if(nx >= 0 && nx < len && ny >= 0 && ny < len && grid[nx][ny] == 1){union(getIndex(i, j), getIndex(nx, ny));}}   // 尝试进行人工岛最大面积的更新res = Math.max(res, size[find(getIndex(i, j))]);}}}// 遍历所有的 0 位置, 尝试向四周进行枚举更新最大值// 创建一个map用来进行去重Set<Integer> set = new HashSet<>();for(int i = 0; i < len; i++){for(int j = 0; j < len; j++){if(grid[i][j] == 0){set.clear();int tempRes = 1;// 向四周进行扩展然后尝试进行岛屿链接for(int k = 0; k < 4; k++){int nx = i + move[k];int ny = j + move[k + 1];if(nx >= 0 && nx < len && ny >= 0 && ny < len && grid[nx][ny] == 1){int f = find(getIndex(nx, ny));if(!set.contains(f)){tempRes += size[f];set.add(f);}}}res = Math.max(res, tempRes); }}}return res;}
}

找出知晓秘密的所有专家

本题我们运用的是一种打标签的技巧, 还有就是注意的是并查集如何进行拆解, 其实就是修改一下father数组的内容, 然后把size数组的值置为1即可

/*** 本题主要就是涉及到并查集的打标签的技巧, 还有如何拆散一个并查集* 首先就是关于并查集打标签: 其实就是给集团领袖节点打上标签信息(类似size数组)* 关于拆散并查集: 其实就是把father数组重新设置为自身, size置为1(如果有的话)*/class Solution {// 并查集轻量化的板子private static final int MAXN = 100001;private static final int[] father = new int[MAXN];private static final boolean[] knowSecrets = new boolean[MAXN];private static int find(int i){if(i != father[i]){father[i] = find(father[i]);}return father[i];}private static boolean isSameSet(int a, int b){return find(a) == find(b);}private static void union(int a, int b){if(!isSameSet(a, b)){father[find(a)] = find(b);}}// 初始化并查集private static void build(int n, int firstPerson){for(int i = 0; i < n; i++){father[i] = i;knowSecrets[i] = false;}// 初始化知道秘密的集团(只需要给领袖节点打上标签就好了)union(0, firstPerson);knowSecrets[0] = true;knowSecrets[firstPerson] = true;}public List<Integer> findAllPeople(int n, int[][] meetings, int firstPerson) {// 首先初始化并查集build(n, firstPerson);// 把meetings进行排序便于处理Arrays.sort(meetings, (a, b) -> a[2] - b[2]);int l = 0;int r = 0;while(l < meetings.length){// 首先把r指针置为l的位置r = l;int tempL = l;// 向右侧扩充(结束的时候r指向的下一个不同的元素的边界位置)while(r < meetings.length && meetings[r][2] == meetings[l][2]){r++;}// 先便利一边并查集进行集合元素的合并while(l < r){union(meetings[l][0], meetings[l][1]);if(isSameSet(0, meetings[l][0])) knowSecrets[meetings[l][0]] = true;if(isSameSet(0, meetings[l][1])) knowSecrets[meetings[l][1]] = true;l++;}// 再次便利一边这个时间点的元素进行集合的拆解l = tempL;while(l < r){if(!isSameSet(meetings[l][0], 0) && !isSameSet(meetings[l][1], 0)){father[meetings[l][0]] = meetings[l][0];father[meetings[l][1]] = meetings[l][1];}l++;}l = r;}// 进行元素的收集List<Integer> res = new ArrayList<>();for(int i = 0; i < n; i++){if(isSameSet(0, i)){res.add(i);}}return res;}
}

建图及其拓扑排序篇

建图的方法有三种, 邻接表, 邻接矩阵, 以及链式前向星, 我们更推荐的是静态空间的链式前向星的建图法, 下面是链式前向星的板子

链式前向星建图板子

/*** 关于大厂笔试以及比赛中的建图方式的测试, 其实就是使用静态的数组空间进行建图* 我们设置 3 / 4 / 5 个静态数组空间* head数组(存储点对应的头边编号), next数组(边对应下一条边的编号), to数组(边去往的点), weight数组(边对应的权值), indegree数组(点对应的入度)* 关于拓扑排序(topoSort), 我们最常用的方法其实就是零入度删除法(使用队列, 必要的时候使用小根堆), 关于是否环化的判断我们使用计数器实现* 下面是我们提供的链式建图的板子, 以及拓扑排序的板子*/public class createGraphByLinkedProStar{// 设置点的最大数量private static final int MAXN = 10001;// 设置边的最大数量private static final int MAXM = 10001;// head数组private static final int[] head = new int[MAXN];// next数组private static final int[] next = new int[MAXM];// to数组private static final int[] to = new int[MAXM];// weight数组private static final int[] weight = new int[MAXM];// indegree数组(统计入度)private static final int[] indegree = new int[MAXN];// cnt统计边的数量private static int cnt = 1;// 添加边的方法(顺便统计入度)private static void addEdge(int u, int v, int w){next[cnt] = head[u];to[cnt] = v;weight[cnt] = w;head[u] = cnt++;indegree[v]++;}// 初始化静态空间(只需要清空head以及indegree数组)然后建图(这里是有向带权图)private static void build(int n, int[][] edges){cnt = 1;for(int i = 0; i <= n; i++){indegree[i] = 0;head[i] = 0;}for(int[] edge : edges){addEdge(edge[0], edge[1], edge[2]);}}// 拓扑排序(topoSort的板子)private static int[] topoSort(int n){// 首先创建一个队列(将来可以作为结果返回)int[] queue = new int[n];int l = 0;int r = 0;// 遍历入度表, 添加所有0入度的点进队列for(int i = 0; i < n; i++){if(indegree[i] == 0){queue[r++] = i;}}// 利用链式前向星的遍历开始跑拓扑排序int elemCnt = 0;while(l < r){int cur = queue[l++];elemCnt++;for(int ei = head[cur]; ei != 0; ei = next[ei]){if(--indegree[to[ei]] == 0){queue[r++] = to[ei];}}}return elemCnt == n ? queue : new int[0];}}

课程表

标准的使用拓扑排序的板子 + 加上链式前向星建图法直接打败 100 %

class Solution {// 设置点的最大数量private static final int MAXN = 10001;// 设置边的最大数量private static final int MAXM = 10001;// head数组private static final int[] head = new int[MAXN];// next数组private static final int[] next = new int[MAXM];// to数组private static final int[] to = new int[MAXM];// indegree数组(统计入度)private static final int[] indegree = new int[MAXN];// cnt统计边的数量private static int cnt = 1;// 添加边的方法(顺便统计入度)private static void addEdge(int u, int v){next[cnt] = head[u];to[cnt] = v;head[u] = cnt++;indegree[v]++;}// 初始化静态空间(只需要清空head以及indegree数组)然后建图(这里是有向带权图)private static void build(int n, int[][] edges){cnt = 1;for(int i = 0; i <= n; i++){indegree[i] = 0;head[i] = 0;}for(int[] edge : edges){addEdge(edge[1], edge[0]);}}// 拓扑排序(topoSort的板子)private static int[] topoSort(int n){// 首先创建一个队列(将来可以作为结果返回)int[] queue = new int[n];int l = 0;int r = 0;// 遍历入度表, 添加所有0入度的点进队列for(int i = 0; i < n; i++){if(indegree[i] == 0){queue[r++] = i;}}// 利用链式前向星的遍历开始跑拓扑排序int elemCnt = 0;while(l < r){int cur = queue[l++];elemCnt++;for(int ei = head[cur]; ei != 0; ei = next[ei]){if(--indegree[to[ei]] == 0){queue[r++] = to[ei];}}}return elemCnt == n ? queue : new int[0];}public int[] findOrder(int numCourses, int[][] prerequisites) {build(numCourses, prerequisites);return topoSort(numCourses);}
}