挑战程序竞赛系列（37）：3.4利用数据结构高效求解

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挑战程序竞赛系列（37）：3.4利用数据结构高效求解

详细代码可以fork下Github上leetcode项目，不定期更新。

练习题如下：

POJ 1769: Minimizing Maximizer
POJ 3171: Cleaning Shifts

POJ 1769: Minimizing Maximizer

思路：经历了一些波折，首先是考虑DP，DP[j]，表示在第j个位置所需要的最短的子序列的长度。

更新式还是比较好推的，因为DP[j]表示到第j个位置所需要的最短子序列长度，那么到下一个位置时，此时的区间长度为[si, ti]，那么，只要DP[j]在si和ti之间就可以更新为： DP[ti] =min{DP[ti], DP[j] + 1}, j >= si && j <= ti

物理含义：只要在区间内，[1, j]能够被排序，那么只要后续区间包含了j，那么自然也就能更新至[1,ti].

初始代码如下：

    void solve() {        int N = ni();        int M = ni();        Pair[] ps = new Pair[M];        for (int i = 0; i < M; ++i){            ps[i] = new Pair(ni(), ni());        }        int[] dp = new int[N + 16];        Arrays.fill(dp, INF);        dp[1] = 0;        for (int i = 0; i < M; ++i){            int s = ps[i].s;            int e = ps[i].e;            for (int j = 1; j <= N; ++j){                if (j >= s && j <= e){                    dp[e] = Math.min(dp[e], dp[j] + 1);                }            }        }        out.println(dp[ps[M - 1].e]);    }

可以观察下for循环中求最小值，此处可以用RMQ替代，无非就是求区间[s,e]中的最小值，这样时间复杂度降为O(nlogn)

哈哈，今天更新了最新的算法模版，拿来POJ实测一波，支持codeforce，uva，aoj哟，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/201708/P1769.txt";    public static void main(String[] args) throws IOException {        new Main().run();    }    class Pair{        int s;        int e;        public Pair(int s, int e){            this.s = s;            this.e = e;        }        @Override        public String toString() {            return s + " " + e;        }    }    static final int MAX_N = 50000 + 16;    static final int SIZE = (1 << 18) + 1;    static final int INF = 1 << 29;    int[] dat = new int[SIZE];    int[] dp;    void solve() {        int N = ni();        int M = ni();        Pair[] ps = new Pair[M];        for (int i = 0; i < M; ++i){            ps[i] = new Pair(ni(), ni());        }        dp = new int[MAX_N];        Arrays.fill(dp, INF);        init(N);        dp[1] = 0;        update (1, 0);        for (int i = 0; i < M; ++i){            int s = ps[i].s;            int e = ps[i].e;            int min = Math.min(dp[e],query(0, s, e + 1, 0, n_) + 1);            dp[e] = min;            update(e, min);        }        out.println(dp[N]);    }    /*********************以下是RMQ的实现*********************/    /**     * [L, r)     * @param k     * @param l     * @param r     */    int n_;    public void init(int N){        n_ = 1;        while (n_ < N) n_ *= 2;        for (int i = 0; i < 2 * n_ - 1; ++i) dat[i] = INF;    }    public void update(int k, int val){        k += (n_ - 1);        dat[k] = val;        while (k > 0){            k = (k - 1) / 2;            dat[k] = Math.min(dat[2 * k + 1], dat[k * 2 + 2]);        }    }    public int query(int k, int i, int j, int l, int r){        if (j <= l || i >= r) return INF;        else if (i <= l && j >= r){            return dat[k];        }        else{            int lch = 2 * k + 1;            int rch = 2 * k + 2;            int mid = (l + r) / 2;            int lf = query(lch, i, j, l, mid);            int rt = query(rch, i, j, mid, r);            return Math.min(lf, 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();        solve();        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);        }    }}

RMQ的实现有一个坑点，其实和《挑战》P188的递归实现还是有一些出入的，原因在与update更新是利用完全二叉树的性质来做的，而如果单纯的递归实现，如果不控制右边界r，那么会导致不匹配，找了很久错误都没找到。

所以init()中，把N转为最近的2次幂，这样一来，在query查询时，可以以完全二叉树的方式来遍历，这样就和update匹配了。

POJ 3171: Cleaning Shifts

感觉比上一题简单，可以用DIJKSTRA，建立点与边的关系，也可以用DP，同理DP[j]表示在第j个位置，所需要消耗的最小代价，把上题的dp[j] + 1，改成dp[j] + c，其他的都不变。传统DP超时，所以还是用RMQ吧！

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/201708/P3171.txt";    public static void main(String[] args) throws IOException {        new Main().run();    }    static final int INF = 1 << 29;    class Cow implements Comparable<Cow>{        int s;        int e;        int c;        public Cow(int s, int e, int c){            this.s = s;            this.e = e;            this.c = c;        }        @Override        public int compareTo(Cow that) {            return this.e == that.e ? this.s - that.s : this.e - that.e;        }    }    void solve() {        int N = ni();        int M = ni();        int E = ni();        Cow[] cows = new Cow[N];        for (int i = 0; i < N; ++i){            cows[i] = new Cow(ni(), ni(), ni());        }        int[] dp = new int[E + 16];        Arrays.fill(dp, INF);        dp[M] = 0;        init(E);        Arrays.sort(cows);        update(M, 0);        for (int i = 0; i < N; ++i){            int s = cows[i].s;            int e = cows[i].e;            s = s == 0 ? 0 : s - 1;            int min = Math.min(dp[e], query(0, s, e, 0, n_) + cows[i].c);            dp[e] = min;            update(e, min);        }        out.println(dp[E] >= INF ? -1 : dp[E]);    }    /*****************RMQ*******************/    static final int MAX_N = (1 << 18) - 1;    int n_;    int[] dat = new int[MAX_N];    public void init(int N){        n_ = 1;        while (n_ < N) n_ *= 2;        for (int i = 0; i < 2 * n_ - 1; ++i) dat[i] = INF;    }    public void update(int k, int val){        k += (n_ - 1);        dat[k] = val;        while (k > 0){            k = (k - 1) / 2;            dat[k] = Math.min(dat[2 * k + 1], dat[2 * k + 2]);        }    }    public int query(int k, int i, int j, int l, int r){        if (j <= l || i >= r) return INF;        else if (i <= l && j >= r) return dat[k];        else{            int lch = 2 * k + 1;            int rch = 2 * k + 2;            int mid = (l + r) / 2;            int lf = query(lch, i, j, l, mid);            int rt = query(rch, i, j, mid, r);            return Math.min(lf, 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();        solve();        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);        }    }}

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