挑战程序竞赛系列（59）：4.6树上的分治法（2）

挑战程序竞赛系列（59）：4.6树上的分治法（2）

传送门：POJ 1741: Tree

思路：
首先还是求树的重心，这样可以把问题划分成多个子问题（树是天然的递归），之所以找重心考虑分治的效率问题，每次问题规模减少n/2，所以最终的时间复杂度为O(nlog2n)。

分为三种情况：

• 顶点 v， w属于同一子树的顶点对(v, w)
• 顶点 v，w 属于不同子树的顶点对(v, w)
• 顶点s和其他顶点v组成的顶点对(v, w)

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.Arrays;import java.util.StringTokenizer;public class Main{    String INPUT = "./data/judge/201709/P1741.txt";    public static void main(String[] args) throws IOException {        new Main().run();    }    static final int MAX_N = 10000 + 16;    static final int INF   = 0x3f3f3f3f;    class Edge{        int to;        int cost;        int next;        Edge(int to, int cost, int next){            this.to = to;            this.cost = cost;            this.next = next;        }        @Override        public String toString() {            return to + " " + cost + " " + next;        }    }    Edge[] edges = new Edge[MAX_N << 1];    int N, K, tot, cenv, minv, ans, point;    int[] head, son;    int[] dis;    boolean[] visited;    void init() {        head = new int[MAX_N];        son  = new int[MAX_N];        dis  = new int[MAX_N];        visited = new boolean[MAX_N];        cenv = point = tot = 0;        minv = INF;        Arrays.fill(head,  -1);    }    void add(int from, int to, int cost) {        edges[tot] = new Edge(to, cost, head[from]);        head[from] = tot++;    }    void dfs(int u, int fa) {        son[u] = 0;        int maxv = 0;        for (int i = head[u]; i != -1; i = edges[i].next) {            int v = edges[i].to;            if (v == fa || visited[v]) continue;            dfs(v, u);            son[u] += son[v] + 1;            maxv = Math.max(maxv, son[v] + 1);        }        maxv = Math.max(maxv, point - 1 - son[u]);        if (maxv < minv) {            minv = maxv;            cenv = u;        }    }    void getDis(int u, int fa, int dist) {        dis[tot++] = dist;        for (int i = head[u]; i != -1; i = edges[i].next) {            int v = edges[i].to;            if (v == fa || visited[v]) continue;            getDis(v, u, dist + edges[i].cost);        }    }    int count(int u, int d) {        tot = 0;        getDis(u, -1, d);        Arrays.sort(dis, 0, tot);        int l = 0, r = tot - 1, res = 0;        while (l < r) {            if (dis[l] + dis[r] <= K) {                res += r - l;                l ++;            }            else r --;        }        return res;    }    void solve(int u) {        minv = INF;        point = point != 0 ? son[u] + 1 : N;        dfs(u, -1);        int root = cenv;        visited[root] = true;        ans += count(root, 0);        for (int i = head[root]; i != -1; i = edges[i].next) {            int v = edges[i].to;            if (visited[v]) continue;            ans -= count(v, edges[i].cost);            solve(v);        }    }    void read() {        while (true) {            N = ni();            K = ni();            if (N == 0 && K == 0) break;            init();            for (int i = 1; i < N; ++i) {                int from = ni();                int to   = ni();                int cost = ni();                add(from, to, cost);                add(to, from, cost);            }            ans = 0;            solve(1);            out.println(ans);        }    }    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);        }    }    static class ArrayUtils {        public static void fill(int[][] f, int value) {            for (int i = 0; i < f.length; ++i) {                Arrays.fill(f[i], value);            }        }        public static void fill(int[][][] f, int value) {            for (int i = 0; i < f.length; ++i) {                fill(f[i], value);            }        }        public static void fill(int[][][][] f, int value) {            for (int i = 0; i < f.length; ++i) {                fill(f[i], value);            }        }    }}

count方法中找寻点对的方式还很独特，统计了所有经过重心且<=K的点对个数，这样在考虑子问题时，必须要把与重心相连的每个子问题的点对删掉（经过v但不经过重心的点对）。