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1837. Balance

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用户1147447
发布2019-05-26 00:28:29
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发布2019-05-26 00:28:29
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文章被收录于专栏:机器学习入门机器学习入门

POJ 刷题系列:1837. Balance

题意:

一个天枰上有C个钩子,把M个砝码挂在这些C个钩子上,问多少种平衡的挂法。

思路: 首先考虑单个砝码在天枰的位置,可以得到C个状态,对应的平衡度为:G[0] * C[i],记录这些初始平衡度的个数。这样当放入第二个砝码时,可以遍历这些平衡度,从而得到一个新的平衡度。此时,不同位置,不同质量的砝码可能拥有相同平衡度,个数直接顺延给下一平衡度状态:当前平衡度+G[1]∗C[i] for i∈[−15,15]当前平衡度+G[1]∗C[i] for i∈[−15,15]当前平衡度+ G[1] * C[i] \space for \space i \in [-15, 15]

详见https://blog.csdn.net/lyy289065406/article/details/6648094/

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/201805/P2187.txt";

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

    void read() {
        int C = in.nextInt();
        int G = in.nextInt();
        int[] CS = new int[C];
        int[] GS = new int[G];
        for (int i = 0; i < C; ++i) CS[i] = in.nextInt();
        for (int j = 0; j < G; ++j) GS[j] = in.nextInt();
        int[][] dp = new int[21][15001];
        dp[0][7500] = 1;
        for (int i = 1; i <= G; ++i) { // 一个物品 每个 钩子 都放过一遍
            for (int j = 0; j <= 15000; ++j) {
                if (dp[i - 1][j] > 0) {
                    for (int k = 0; k < C; ++k) {
                        dp[i][j + GS[i - 1] * CS[k]] += dp[i - 1][j];
                    }
                }
            }
        }
        System.out.println(dp[G][7500]);
    }

    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);
        }
    }
}
这里写图片描述
这里写图片描述
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