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
• POJ 1703: Find them, Catch them
• AOJ 2170: Marked Ancestor

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);        }    }