O(n ^ 2)解决这个问题的速度不够快.更快的方法？

Question

我一直试图在ACM Timus上解决这个问题

http://acm.timus.ru/problem.aspx?space=1&num=1932

我的第一种方法是O(n ^ 2),它肯定不够快,无法通过所有测试.下面的O(n ^ 2)代码在测试10中给出TL.

import java.util.*;
import java.io.*;

public class testtest
{
    public static void main(String[] args) throws IOException
    {
        BufferedReader rr = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(rr.readLine());
        String[] p = new String[n];
        for (int i = 0; i < n; i ++)
        {
            p[i] = rr.readLine();
        }
        int[] diff = new int[]{0, 0, 0, 0};
        for (int i = 0; i < n - 1; i ++)
        {
            for (int j = i + 1; j < n; j ++)
            {
                int ans  = (p[i].charAt(0) == p[j].charAt(0) ? 0 : 1) +
                           (p[i].charAt(1) == p[j].charAt(1) ? 0 : 1) +
                           (p[i].charAt(2) == p[j].charAt(2) ? 0 : 1) +
                           (p[i].charAt(3) == p[j].charAt(3) ? 0 : 1);
                diff[ans - 1] ++;
            }
        }
        System.out.print(diff[0] + " " + diff[1] + " " + diff[2] + " " + diff[3]);
    }
}

有什么想让这种方法更快？我注意到输入中只允许一组有限的字符('0'..'9','a'..'f')所以我们可以创建数组(内存限制就足够了)来快速检查之前已输入过字符.

谢谢......我不需要实际的实施,只是快速的想法/想法会很棒. 编辑:谢谢你的好主意.我已尝试使用位逻辑对O(n ^ 2)进行改进,但仍超出时间限制.pascal代码如下.

program Project2;

{$APPTYPE CONSOLE}

var
  i, j, n, k, bits: integer;
  arr: array[1..65536] of integer;
  diff: array[1..4] of integer;
  a, b, c, d: char;

function g(c: char): integer; inline;
begin
  if ((c >= '0') and (c <= '9')) then
  begin
    Result := Ord(c) - 48;
  end
  else
  begin
    Result := Ord(c) - 87;
  end;
end;

begin
  Readln(n);
  for i := 1 to n do
  begin
    Read(a); Read(b); Read(c); Readln(d);
    arr[i] := g(a) * 16 * 16 * 16 + g(b) * 16 * 16 + g(c) * 16 + g(d);
    for j := 1 to i - 1 do
    begin
      bits := arr[i] xor arr[j];
      k := ((bits or (bits shr 1) or (bits shr 2) or (bits shr 3)) and $1111) mod 15;
      Inc(diff[k]);
    end;
  end;
  Write(diff[1], ' ', diff[2], ' ', diff[3], ' ', diff[4]);
{$IFNDEF ONLINE_JUDGE}
  Readln;
{$ENDIF}
end.

所以我想,我会尝试其他更好的建议..

编辑: 我已经尝试了丹尼尔的算法,这是非常有希望的,也许在下面的代码中有一个错误,它在测试10上一直得到错误的答案......有人可以看看吗？非常感谢...

import java.util.*;
import java.io.*;

public class testtest
{
    private static int g(char ch)
    {
        if ((ch >= '0') && (ch <= '9'))
        {
            return (int)ch - 48;
        }
        return (int)ch - 87;
    }

    public static void main(String[] args) throws IOException
    {
        BufferedReader rr = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(rr.readLine());
        int[] p = new int[n];
        int[] all = new int[65536];
        int[][] miss = new int[4][4096];
        int[] g12 = new int[256];
        int[] g13 = new int[256];
        int[] g14 = new int[256];
        int[] g23 = new int[256];
        int[] g24 = new int[256];
        int[] g34 = new int[256];
        int[][] gg = new int[4][16];
        int same3, same2, same1, same0, same4;
        for (int i = 0; i < n; i ++)
        {
            String s = rr.readLine();
            int x = g(s.charAt(0)) * 4096 + g(s.charAt(1)) * 256 + g(s.charAt(2)) * 16 + g(s.charAt(3));
            p[i] = x;
            all[x] ++;
            miss[0][x >> 4] ++;
            miss[1][(x & 0x000F) | ((x & 0xFF00) >> 4)] ++;
            miss[2][(x & 0x00FF) | ((x & 0xF000) >> 4)] ++;
            miss[3][x & 0x0FFF] ++;
            g12[x >> 8] ++;
            g13[((x & 0x00F0) >> 4) | ((x & 0xF000) >> 8)] ++;
            g14[(x & 0x000F) | ((x & 0xF000) >> 8)] ++;
            g23[(x & 0x0FF0) >> 4] ++;
            g24[(x & 0x000F) | ((x & 0x0F00) >> 4)] ++;
            g34[x & 0x00FF] ++;
            gg[0][x >> 12] ++;
            gg[1][(x & 0xF00) >> 8] ++;
            gg[2][(x & 0xF0) >> 4] ++;
            gg[3][x & 0xF] ++;
        }

        same4 = 0;
        for (int i = 0; i < 65536; i ++)
        {
            same4 += (all[i] - 1) * (all[i]) / 2;
        }

        same3 = 0;
        for (int i = 0; i < 4096; i ++)
        {
            same3 += (miss[0][i] - 1) * (miss[0][i]) / 2;
            same3 += (miss[1][i] - 1) * (miss[1][i]) / 2;
            same3 += (miss[2][i] - 1) * (miss[2][i]) / 2;
            same3 += (miss[3][i] - 1) * (miss[3][i]) / 2;
        }

        same2 = 0;
        for (int i = 0; i < 256; i ++)
        {
            same2 += (g12[i] - 1) * g12[i] / 2;
            same2 += (g13[i] - 1) * g13[i] / 2;
            same2 += (g14[i] - 1) * g14[i] / 2;
            same2 += (g23[i] - 1) * g23[i] / 2;
            same2 += (g24[i] - 1) * g24[i] / 2;
            same2 += (g34[i] - 1) * g34[i] / 2;
        }

        same1 = 0;
        for (int i = 0; i < 16; i ++)
        {
            same1 += (gg[0][i] - 1) * gg[0][i] / 2;
            same1 += (gg[1][i] - 1) * gg[1][i] / 2;
            same1 += (gg[2][i] - 1) * gg[2][i] / 2;
            same1 += (gg[3][i] - 1) * gg[3][i] / 2;
        }

        same3 -= 4 * same4;
        same2 -= 6 * same4 + 3 * same3;
        same1 -= 4 * same4 + 3 * same3 + 2 * same2;
        same0 = (int)((long)(n * (n - 1) / 2) - same4 - same3 - same2 - same1);
        System.out.print(same3 + " " + same2 + " " + same1 + " " + same0);
    }
}

编辑 终于得到了AC ...感谢Daniel这么棒的算法!

Answer 1

对于小型n,O(n²)当然,检查每对的蛮力算法更快,因此人们希望找到切换算法的良好截止点.在没有测量的情况下,由于不均匀的包络考虑,我预计值在200到3000之间.

通过将盗版ID int解析为十六进制数来将其转换为a.存储ID

int[] pirates = new int[n];

首先,计算具有相同ID的盗版对的数量(这里可以省略此步骤,因为问题陈述没有).

int[] allFour = new int[65536];
for(int i = 0; i < n; ++i) {
    allFour[pirate[i]] += 1;
}
int fourIdentical = 0;
for(int i = 0; i < 65536; ++i) {
    fourIdentical += allFour[i]*(allFour[i] - 1) / 2;
}

接下来,计算他们的ID中有三个相同半字节的海盗对,

int oneTwoThree(int p) {
    return p >> 4;
}
int oneTwoFour(int p) {
    return (p & 0x000F) | ((p & 0xFF00) >> 4);
}
int oneThreeFour(int p) {
    return (p & 0x00FF) | ((p & 0xF000) >> 4);
}
int twoThreeFour(int p) {
    return p & 0x0FFF;
}

int[] noFour  = new int[4096];
int[] noThree = new int[4096];
int[] noTwo   = new int[4096];
int[] noOne   = new int[4096];

for(int i = 0; i < n; ++i) {
    noFour[oneTwoThree(pirate[i])] += 1;
    noThree[oneTwoFour(pirate[i])] += 1;
    noTwo[oneThreeFour(pirate[i])] += 1;
    noOne[twoThreeFour(pirate[i])] += 1;
}

int threeIdentical = 0;
for(int i = 0; i < 4096; ++i) {
    threeIdentical += noFour[i]*(noFour[i]-1) / 2;
}
for(int i = 0; i < 4096; ++i) {
    threeIdentical += noThree[i]*(noThree[i]-1) / 2;
}
for(int i = 0; i < 4096; ++i) {
    threeIdentical += noTwo[i]*(noTwo[i]-1) / 2;
}
for(int i = 0; i < 4096; ++i) {
    threeIdentical += noOne[i]*(noOne[i]-1) / 2;
}

但是,每对具有四个相同半字节的盗贼4 choose 3 = 4在这里计算时间,对于每个可能选择的三个半字节,所以我们需要减去它(好吧,不是问题,但原则上):

threeIdentical -= 4*fourIdentical;

然后,计算他们的ID中有两个相同半字节的海盗对:

int oneTwo(int p) {
    return p >> 8;
}
int oneThree(int p) {
    return ((p & 0x00F0) >> 4) | ((p & 0xF000) >> 8);
}
int oneFour(int p) {
    return (p & 0x000F) | ((p & 0xF000) >> 8);
}
int twoThree(int p) {
    return (p & 0x0FF0) >> 4;
}
int twoFour(int p) {
    return (p & 0x000F) | ((p & 0x0F00) >> 4);
}
int threeFour(int p) {
    return p & 0x00FF;
}

分配的256个六个阵列intS和计数海盗与相应的半字节中的位数a和b,像

int twoIdentical = 0;
int[] firstTwo = new int[256];
for(int i = 0; i < n; ++i) {
    firstTwo[oneTwo(pirate[i])] += 1;
}
for(int i = 0; i < 256; ++i) {
    twoIdentical += firstTwo[i]*(firstTwo[i] - 1) / 2;
}
// analogous for the other five possible choices of two nibbles

但是,具有四个相同半字节的对在4 choose 2 = 6这里已被计数次数,并且具有三个相同半字节的对已被计数3 choose 2 = 3次数,因此我们需要减去

twoIdentical -= 6*fourIdentical + 3*threeIdentical;

接下来,具有一个相同半字节的对的数量.我相信你可以猜到需要四个函数和数组.然后我们将计算具有四个相同半字节4 choose 1 = 4时间的对,具有三个相同半字节3 choose 1 = 3时间的对,以及具有两个相同半字节2 choose 1 = 2时间的对,所以

oneIdentical -= 4*fourIdentical + 3*threeIdentical + 2*twoIdentical;

最后,没有相同半字节的对的数量是

int noneIdentical = (int)((long)n*(n-1) / 2) - oneIdentical - twoIdentical - threeIdentical - fourIdentical;

(施法者long要避免溢出n*(n-1)).