挑战程序竞赛系列（74）：4.3强连通分量分解（1）

挑战程序竞赛系列（74）：4.3强连通分量分解（1）

传送门：POJ 2186: Popular Cows

题意：

每头牛都想成为牛群中的红人。给定N头牛的牛群和M个有序对(A,B)。(A, B)表示牛A认为牛B是红人。该关系具有传递性，所以如果A认为B是红人，B认为C是红人，则A认为C是红人。注意：给定的有序对中可能包含(A, B)和(B, C)，但不包含(A, C)。求被其他所有牛认为是红人的牛的总数。

强连通分量（采用kosarajuSCC算法）双DFS，具体参考《挑战》P321：

说说思路吧，最后一句话非常关键，意思是说从某个顶点出发的边只可能一条，所以被所有牛认为是红牛的群体一定构成了一个环（强连通分量），那么有多少个强连通分量符合这种情况呢？至多一个，因为假设存在两个强连通分量符合条件，那么必然从一个顶点出发，有两条边与之相连，与题意矛盾。

那么是哪一个强连通分量呢？在强连通分量分解时的最后一个，因为只有这一个才有更多的机会抵达先前的各个顶点，它最靠后。

代码如下：

import java.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.StringTokenizer;

public class Main{

    String INPUT = "./data/judge/201709/P2186.txt";

    public static void main(String[] args) throws IOException {
        new Main().run();
    }

    /******************强连通分量的求解*******************/
    static final int MAX_N = 10000 + 16;
    List<Edge>[] g  = new List[MAX_N];
    List<Edge>[] rg = new List[MAX_N];
    List<Integer> postOrder = new ArrayList<Integer>();
    boolean[] used = new boolean[MAX_N];
    int[] cmp = new int[MAX_N];
    int N;

    class Edge{
        int from;
        int to;
        Edge(int from, int to){
            this.from = from;
            this.to = to;
        }

        @Override
        public String toString() {
            return from + " " + to;
        }
    }

    void init(int n) {
        this.N = n;
        Arrays.fill(used, false);
        for (int i = 0; i < N; ++i) g[i]  = new ArrayList<Edge>();
        for (int i = 0; i < N; ++i) rg[i] = new ArrayList<Edge>(); 
    }

    void add(int from, int to) {
        g[from].add(new Edge(from, to));
        rg[to].add(new Edge(to, from));
    }

    void dfs(int v) {
        used[v] = true;
        for (Edge e : g[v]) {
            if (!used[e.to]) dfs(e.to);
        }
        postOrder.add(v);
    }

    void rdfs(int v, int k) {
        used[v] = true;
        cmp[v] = k;
        for (Edge e : rg[v]) {
            if (!used[e.to]) rdfs(e.to, k);
        }
    }

    int kosarajuSCC(int V) {
        for (int i = 0; i < V; ++i) {
            if (!used[i]) dfs(i);
        }
        Arrays.fill(used, false);
        int k = 0;
        for (int i = postOrder.size() - 1; i >= 0; --i) {
            if (!used[postOrder.get(i)]) rdfs(postOrder.get(i), k++);
        }
        return k;
    }



    void read() {
        int n = ni();
        int m = ni();
        init(n);
        for (int i = 0; i < m; ++i) {
            int from = ni();
            int to   = ni();
            from --;
            to   --;
            add(from, to);
        }
        int scc = kosarajuSCC(n);
        int u = 0, num = 0;
        for (int i = 0; i < n; ++i) {
            if (cmp[i] == scc - 1) {
                u = i;
                num ++;
            }
        }
        Arrays.fill(used, false);
        rdfs(u, 0);
        for (int v = 0; v < n; ++v) {
            if (!used[v]) {
                num = 0;
                break;
            }
        }
        out.println(num);
    }

    FastScanner in;
    PrintWriter out;

    void run() throws IOException {
        boolean oj;
        try {
            oj = ! System.getProperty("user.dir").equals("F:\\java_workspace\\leetcode");
        } catch (Exception e) {
            oj = System.getProperty("ONLINE_JUDGE") != null;
        }

        InputStream is = oj ? System.in : new FileInputStream(new File(INPUT));
        in = new FastScanner(is);
        out = new PrintWriter(System.out);
        long s = System.currentTimeMillis();
        read();
        out.flush();
        if (!oj){
            System.out.println("[" + (System.currentTimeMillis() - s) + "ms]");
        }
    }

    public boolean more(){
        return in.hasNext();
    }

    public int ni(){
        return in.nextInt();
    }

    public long nl(){
        return in.nextLong();
    }

    public double nd(){
        return in.nextDouble();
    }

    public String ns(){
        return in.nextString();
    }

    public char nc(){
        return in.nextChar();
    }

    class FastScanner {
        BufferedReader br;
        StringTokenizer st;
        boolean hasNext;

        public FastScanner(InputStream is) throws IOException {
            br = new BufferedReader(new InputStreamReader(is));
            hasNext = true;
        }

        public String nextToken() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (Exception e) {
                    hasNext = false;
                    return "##";
                }
            }
            return st.nextToken();
        }

        String next = null;
        public boolean hasNext(){
            next = nextToken();
            return hasNext;
        }

        public int nextInt() {
            if (next == null){
                hasNext();
            }
            String more = next;
            next = null;
            return Integer.parseInt(more);
        }

        public long nextLong() {
            if (next == null){
                hasNext();
            }
            String more = next;
            next = null;
            return Long.parseLong(more);
        }

        public double nextDouble() {
            if (next == null){
                hasNext();
            }
            String more = next;
            next = null;
            return Double.parseDouble(more);
        }

        public String nextString(){
            if (next == null){
                hasNext();
            }
            String more = next;
            next = null;
            return more;
        }

        public char nextChar(){
            if (next == null){
                hasNext();
            }
            String more = next;
            next = null;
            return more.charAt(0);
        }
    }
}