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社区首页 >专栏 >POJ 刷题系列:1936. All in All

POJ 刷题系列:1936. All in All

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用户1147447
发布2019-05-26 07:51:03
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发布2019-05-26 07:51:03
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文章被收录于专栏:机器学习入门

版权声明:本文为博主原创文章,未经博主允许不得转载。 https://cloud.tencent.com/developer/article/1434482

POJ 刷题系列:1936. All in All

传送门:1936. All in All

题意:

给定两个字符串s和t,你需要判断s是否是t的“子列”。也就是说,如果你去掉t中的某些字符,剩下字符将连接而成为s。

思路:

动规or递归+记忆化,定义状态dp[i][j]表示T0…i - 1 及S0…j - 1能否构成子列,状态转移如下:

代码语言:javascript
复制
if S[j] == T[i]: dp[i][j] = dp[i - 1][j - 1]
else: dp[i][j] = dp[i][j - 1] 

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.Arrays;
import java.util.Map;
import java.util.StringTokenizer;

public class Main{

    String INPUT = "./data/judge/201801/P1936.txt";

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

    void read() {
        while (more()) {
            String a = ns();
            String b = ns();
            int l1 = a.length();
            int l2 = b.length();
            boolean[][] dp = new boolean[l1 + 1][l2 + 1];
            for (int i = 0; i <= l2; ++i) dp[0][i] = true;
            for (int i = 0; i < l1; ++i) {
                for (int j = 0; j < l2; ++j) {
                    if (a.charAt(i) == b.charAt(j)) {
                        dp[i + 1][j + 1] = dp[i][j];
                    }
                    else dp[i + 1][j + 1] = dp[i + 1][j];
                }
            }
            if (dp[l1][l2]) out.println("Yes");
            else out.println("No");
        }
    }

    FastScanner in;
    PrintWriter out;

    void run() throws IOException {
        boolean oj;
        try {
            oj = ! System.getProperty("user.dir").equals("F:\\oxygen_workspace\\Algorithm");
        } 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);
        }
    }
}
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原始发表:2018年01月14日,如有侵权请联系 cloudcommunity@tencent.com 删除

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