挑战程序竞赛系列（10）：2.4并查集

用户1147447

发布于 2019-05-26 09:52:18

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发布于 2019-05-26 09:52:18

文章被收录于专栏：机器学习入门机器学习入门

挑战程序竞赛系列（10）：2.4并查集

详细代码可以fork下Github上leetcode项目，不定期更新。

练习题如下：

POJ 2236: Wireless Network

写个Union，在距离范围内的电脑可以认为是同属于一个集合，进行合并。当然还需要注意电脑当前的状态是否修复，如果未修复，即使在指定距离内，也不能合并。

即使做了些状态记录，Java代码还是慢啊。

代码如下：

public class SolutionDay15_L2236 {

    static class Union{
        int[] id;
        int[] sz;

        public Union(int size){
            id = new int[size];
            sz = new int[size];

            for (int i = 0; i < size; ++i){
                id[i] = i;
                sz[i] = 1;
            }
        }

        public void union(int i, int j){
            int p = find(i);
            int q = find(j);

            if (p == q) return;

            if (sz[p] < sz[q]){
                id[p] = q;
                sz[q] += sz[p];
            }
            else{
                id[q] = p;
                sz[p] += sz[q];
            }
        }

        public int find(int i){
            while (id[i] != i){
                i = id[i];
            }
            return i;
        }

        public boolean connect(int i, int j){
            return find(i) == find(j);
        }
    }

    private static boolean distance(int[] c1, int[] c2, int d){
        int x = c1[0] - c2[0];
        int y = c1[1] - c2[1];
        return (x * x + y * y) <= d * d;
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        String[] n = in.nextLine().trim().split(" ");

        int N = Integer.parseInt(n[0]);
        int D = Integer.parseInt(n[1]);

        boolean[] status = new boolean[N];
        boolean[][] distance = new boolean[N][N];
        Union union = new Union(N);

        int[][] coord = new int[N][2];
        for (int i = 0; i < N; ++i){
            n = in.nextLine().trim().split(" ");
            coord[i][0] = Integer.parseInt(n[0]);
            coord[i][1] = Integer.parseInt(n[1]);
        }

        for (int i = 0; i < N; ++i){
            for (int j = i + 1; j < N; ++j){
                if(distance(coord[i], coord[j], D)){
                    distance[i][j] = true;
                    distance[j][i] = true;
                }
            }
        }

        while (in.hasNextLine()){
            String[] operates = in.nextLine().trim().split(" ");
            if (operates[0].equals("O")){
                int c = Integer.parseInt(operates[1]) - 1;
                status[c] = true;
                for (int j = 0; j < N; ++j){
                    if (j == c) continue;
                    if (status[j] && distance[j][c])
                        union.union(c, j);
                }
            }
            if (operates[0].equals("S")){
                int c1 = Integer.parseInt(operates[1]) - 1;
                int c2 = Integer.parseInt(operates[2]) - 1;
                System.out.println(union.connect(c1, c2) ? "SUCCESS" : "FAIL");
            }
        }
        in.close();
    }
}

POJ 1703: Find them, Catch them

思路：

copy一团罪犯，敌人的敌人是我们的朋友。

代码如下：

public class SolutionDay15_L1703 {

    static class Union{
        int[] id;
        int[] sz;

        public Union(int size){
            id = new int[size];
            sz = new int[size];
            for (int i = 0; i < size; ++i){
                id[i] = i;
                sz[i] = 1;
            }
        }

        public int find(int i){
            while (id[i] != i){
                i = id[i];
            }
            return i;
        }

        public boolean same(int i, int j){
            return find(i) == find(j);
        }

        public void union(int i, int j){
            int p = find(i);
            int q = find(j);

            if (p == q) return;
            if (sz[p] < sz[q]){
                id[p] = q;
                sz[q] += sz[p];
            }
            else{
                id[q] = p;
                sz[p] += sz[q];
            }
        }
    }
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int T = in.nextInt();
        for (int i = 0; i < T; ++i){
            int N = in.nextInt();
            int M = in.nextInt();
            Union u = new Union(2 * N);
            for (int j = 0; j < M; ++j){
                String opera = in.next();
                int c1 = in.nextInt() - 1;
                int c2 = in.nextInt() - 1;
                if (opera.equals("A")){
                    if (u.same(c1, c2)){
                        System.out.println("In the same gang.");
                    }
                    else if (u.same(c1, c2 + N)){
                        System.out.println("In different gangs.");
                    }
                    else{
                        System.out.println("Not sure yet.");
                    }
                }
                else{
                    u.union(c1, c2 + N);
                    u.union(c1 + N, c2);
                }
            }
        }
        in.close();
    }
}

AOJ 2170: Marked Ancestor

非常easy，dfs找父结点即可，如果mark，则把该结点的父亲改为自己即可。

代码如下：

    public static int find(int i){
        while (i != id[i]) i = id[i];
        return i;
    }

    static int[] id;
    public static void main(String[] args) throws IOException {
        Scanner in = new Scanner(System.in);
        while (true){
            int N = in.nextInt();
            int Q = in.nextInt();
            if (N == 0 && Q == 0) break;
            id = new int[N];

            for (int i = 1; i < N; ++i){
                int p = in.nextInt() - 1;
                id[i] = p;
            }
            long sum = 0;
            for (int i = 0; i < Q; ++i){
                String opera = in.next();
                int c = in.nextInt() - 1;
                if (opera.equals("Q")){
                    sum += (find(c) + 1);
                }
                else{
                    id[c] = c;
                }
            }
            System.out.println(sum);
        }

    }

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原始发表：2017年06月15日，如有侵权请联系 cloudcommunity@tencent.com 删除

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