前往小程序,Get更优阅读体验!
立即前往
首页
学习
活动
专区
工具
TVP
发布
社区首页 >专栏 >挑战程序竞赛系列(80):4.3 2-SAT(4)

挑战程序竞赛系列(80):4.3 2-SAT(4)

作者头像
用户1147447
发布2018-01-02 10:43:20
5690
发布2018-01-02 10:43:20
举报
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(80):4.3 2-SAT(4)

传送门:POJ 2749: Building roads

题意:

阳关路与独木桥:有N个农场,其中A对相互讨厌,不能碰面;B对相互喜欢,必须碰面。给定两个中转站S1和S2、各个农场的坐标,让每个农场连接到其中一个中转站。求最小化任意两个农场通过中转站的最大距离,若无法实现,输出-1。

最开始的想法是直接根据hate和friend关系推出矛盾,在可能的组合中找寻最大距离的最小值,但2-SAT的算法模型没法枚举出每条可行的路径,无奈只能采用二分。

此题的trick在于如何把距离的矛盾关系推出来,可以想象,如果农场A和农场B到S1的距离之和大于我们的猜测距离d时,则A只能选S1,B选S2,或者A选S2,B选S1。

而如果:

代码语言:javascript
复制
dist(A->S1) + dist(S1->S2) + dist(S2->B) > d

此时,能推出另一组矛盾关系,A和B只能同选S1或者S2

所以2-SAT模型的题,主要抓住矛盾关系即可。

代码如下:

代码语言:javascript
复制
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.Map;
import java.util.StringTokenizer;

public class Main{

    String INPUT = "./data/judge/201709/P2749.txt";

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

    int N, A, B;
    int[][] hates, likes;
    int[] s1, s2;
    int[][] vx;
    int[][] distance;
    int dist;
    int maxDistance;

    class SCC {
        int V;
        List<Integer>[] g, rg;
        List<Integer> po;
        int[] cmp;
        boolean[] used;

        SCC(int V) {
            this.V = V;
            g  = new ArrayList[V];
            rg = new ArrayList[V];
            po = new ArrayList<Integer>();
            used = new boolean[V];
            cmp = new int[V];
            for (int i = 0; i < V; ++i)
                g[i] = new ArrayList<Integer>();
            for (int i = 0; i < V; ++i)
                rg[i] = new ArrayList<Integer>();
        }

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

        void dfs(int v) {
            used[v] = true;
            for (int u : g[v]) {
                if (!used[u])
                    dfs(u);
            }
            po.add(v);
        }

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

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

    void read() {
        N = ni();
        A = ni();
        B = ni();
        s1 = new int[] {ni(), ni()};
        s2 = new int[] {ni(), ni()};

        dist = manhattanDistance(s1, s2);

        vx = new int[N][2];
        distance = new int[N][2];

        for (int i = 0; i < N; ++i) {
            vx[i] = new int[] {ni(), ni()};
            distance[i][0] = manhattanDistance(vx[i], s1);
            distance[i][1] = manhattanDistance(vx[i], s2);
        }

        for (int i = 0; i < N; ++i) {
            for (int j = i + 1; j < N; ++j) {
                maxDistance = Math.max(maxDistance, distance[i][0] + dist + distance[j][1]);
                maxDistance = Math.max(maxDistance, distance[i][1] + distance[j][1]);
                maxDistance = Math.max(maxDistance, distance[i][0] + distance[j][0]);
            }
        }

        hates = new int[A][2];
        for (int i = 0; i < A; ++i) {
            hates[i] = new int[] {ni() - 1, ni() - 1};
        }

        likes = new int[B][2];
        for (int i = 0; i < B; ++i) {
            likes[i] = new int[] {ni() - 1, ni() - 1};
        }

        solve();
    }

    int manhattanDistance(int[] x, int[] y) {
        return Math.abs(x[0] - y[0]) + Math.abs(x[1] - y[1]);
    }

    boolean check(int d) {
        SCC scc = new SCC(N * 2);
        for (int i = 0; i < N; ++i) {
            for (int j = i + 1; j < N; ++j) {
                for (int k = 0; k < 2; ++k) {
                    if (distance[i][k] + distance[j][k] > d) {  //连接同一个站点发生矛盾,不能同选s1或s2
                        scc.add(2 * i + k, 2 * j + 1 - k);
                        scc.add(2 * j + k, 2 * i + 1 - k);
                    }
                    if (distance[i][k] + dist + distance[j][1 - k] > d) { //连接不同站点发生矛盾,只能同选s1或s2
                        scc.add(2 * i + k, 2 * j + k);
                        scc.add(2 * j + 1 - k, 2 * i + 1 - k);
                    }
                }
            }
        }

        for (int i = 0; i < A; ++i) {
            for (int k = 0; k < 2; ++k) {
                scc.add(2 * hates[i][0] + k, 2 * hates[i][1] + 1 - k);
                scc.add(2 * hates[i][1] + k, 2 * hates[i][0] + 1 - k);
            }
        }

        for (int i = 0; i < B; ++i) {
            for (int k = 0; k < 2; ++k) {
                scc.add(2 * likes[i][0] + k, 2 * likes[i][1] + k);
                scc.add(2 * likes[i][1] + 1 - k, 2 * likes[i][0] + 1 - k);
            }
        }

        scc.kosarajuSCC();
        for (int i = 0; i < N; ++i) {
            if (scc.cmp[2 * i] == scc.cmp[2 * i + 1]) {
                return false;
            }
        }
        return true;
    }

    void solve() {
        maxDistance ++;
        int lf = 0, rt = maxDistance;
        while (lf < rt) {
            int mid = lf + (rt - lf) / 2;
            if (!check(mid)) {
                lf = mid + 1;
            }
            else rt = mid;
        }
        out.println(rt == maxDistance ? -1 :rt);
    }

    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 D {

        public static void pp(int[][] board, int row, int col) {
            StringBuilder sb = new StringBuilder();
            for (int i = 0; i < row; ++i) {
                for (int j = 0; j < col; ++j) {
                    sb.append(board[i][j] + (j + 1 == col ? "\n" : " "));
                }
            }
            System.out.println(sb.toString());
        }

        public static void pp(char[][] board, int row, int col) {
            StringBuilder sb = new StringBuilder();
            for (int i = 0; i < row; ++i) {
                for (int j = 0; j < col; ++j) {
                    sb.append(board[i][j] + (j + 1 == col ? "\n" : " "));
                }
            }
            System.out.println(sb.toString());
        }
    }

    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);
            }
        }
    }

    static class Num {
        public static <K> void inc(Map<K, Integer> mem, K k) {
            if (!mem.containsKey(k))
                mem.put(k, 0);
            mem.put(k, mem.get(k) + 1);
        }
    }
}
这里写图片描述
这里写图片描述
本文参与 腾讯云自媒体分享计划,分享自作者个人站点/博客。
原始发表:2017-09-26 ,如有侵权请联系 cloudcommunity@tencent.com 删除

本文分享自 作者个人站点/博客 前往查看

如有侵权,请联系 cloudcommunity@tencent.com 删除。

本文参与 腾讯云自媒体分享计划  ,欢迎热爱写作的你一起参与!

评论
登录后参与评论
0 条评论
热度
最新
推荐阅读
目录
  • 挑战程序竞赛系列(80):4.3 2-SAT(4)
领券
问题归档专栏文章快讯文章归档关键词归档开发者手册归档开发者手册 Section 归档