RMQ问题的Fischer-Heun算法（直接方法）
在普通RMQ问题的<O(n),O(1)>算法中，由于需要构造Cartesian Tree和得到Euler tour，两个2*n-1大小的数组E和L使得空间消耗增加O(4*n)。本文介绍的Fischer-Heun算法绕过构建Cartesian Tree的步骤，也不需要将普通RMQ转化为RMQ+1/-1问题再求解。该算法的基本思想也是采用RMQ+1/-1的分组处理, 分别生成组内的Lookup Table和组外的稀疏表。但是在每个组的大小缩小了一倍，成为s=(logn)/4。由于每个组对应一颗笛卡尔树，若两个不同的组对应的两棵笛卡尔树相同的话，那么这两个组的lookup table完全一样；另外由于大小为n的笛卡尔树的个数为Catalan数Cn，因此构建的基本组大小是Cn。在组大小为s=(logn)/4时，根据Catalan数Cn的近似公式得到基本组个数为O(4^s/s^(3/2))数量级，虽然Catalan数Cn增长的很快，但在实践中Cs是一个比较小的数：比如，当s=8时，C8=1430，可以处理的A数组大小为n=2^32，n已经很大了。类似于RMQ+1/-1算法的分析，组内预处理的复杂度为线性：O(4^s/s^(3/2) * s^2) = O(n)！组外预处理的复杂度为O(s*logs)=O(n)。查询方式也是采用RMQ+1/-1算法中的查询做法，复杂度也是O(1)。
Fischer-Heun算法的关键是如何确定每一个组的Cartesian Tree编号，即如何确定组的类型数组T。在J. Fischer和V. Heun 2006年发表的论文中，比较巧妙地使用ballot数来处理。ballot数对应的三角形中，设当前位置在(p,q)，那么ballot数C[p,q]相当于从(p,q)到(0,0)走法总数，如果p=q，C[p,q]就是Catalan数Cp，即对角线上的点对应Catalan数。确定组的类型数组的算法利用构造Cartesian Tree的过程，但是不真正构造树，而且堆栈中存放的是A[i]值，而不是下标值。首先在stack底部放置一个sentinel元素Integer.MIN_VALUE。设置Cartesian Tree编号N=0。在构造笛卡尔树的算法中，从ballot数三角形中的点(s,s)出发，每执行一次循环，对应往左走一步；如果某个元素A[i]小于栈顶元素，那么弹出栈顶元素，这时对应往上走一步，若当前位置在(p,q)，N的值需要增加ballot数C[p-1,q]；最后将A[i]入栈。由于构造Cartesian Tree算法的特点，按照这样的走法，可以保证在ballot数对应的三角形中不会超出边界范围。最后s个元素处理完时，N的值就是block的类型值，范围为0..Cs-1。
实现：
/**
*
* Fischer-Heun Algorithm
* A space-economic general RMQ algorithm without transforming into RMQ+1/-1
* time complexity: <O(n),O(1)>
*
* see also: (2006) J. Fischer, V. Heun, Theoretical and Practical Improvements on the RMQ-Problem, with Applications to LCA and LCE
*
* Copyright (c) 2011 ljs (http://blog.csdn.net/ljsspace/)
* Licensed under GPL (http://www.opensource.org/licenses/gpl-license.php)
*
*
* @author ljs
* 2011-08-07
*
*/
public class RMQ_FH {
private int blocksize;
private int blockcount;
//block type
private int[] T;
//block LookUp table
private int[][] P;
//out-of-block sparse table
//the first-dimension indices are block IDs (from 0...blockcount-1)
private int[][] M;
//Note: this method is not quite different from the one in class MinusOrPlusOne_RMQ:
//Except that blockTypesCount and computeBlockType are different, this method is
//almost the same as the one in class MinusOrPlusOne_RMQ.
public void preprocess(int[] A) throws Exception {
int n = A.length;
int paddingsize = 0;
//block size: (logn)/4
int s = (int) (Math.log(n) / Math.log(2)) >> 2;
if(s==0){
s = (n>4)?4:n; //small problem
}
int count = (int) Math.ceil(n / (double)s);
this.blocksize = s;
this.blockcount = count;
//compute the ballot numbers for determining each block's cartesian trees or block type
int[] ballotnums = RMQ_FH.makeBallotNumbersTable(s);
// padding the last block
int endblocksize = n - s * (count - 1);
if (endblocksize > 0 && endblocksize < s) {
paddingsize = s - endblocksize;
}
//step 1: in-block preprocess
//the size of LU table (one-dimensional)
int size = s*(s+1)/2;
int start = 0; // j is the index of A
int end = -1;
T = new int[count];
int blockTypesCount = ballotnums[(s+2)*(s+1)/2 - 1]; //catalan number
P = new int[blockTypesCount][size];
int[] B = new int[count]; //used in ST algorithm: the min-value array for each block
boolean[] pDone =new boolean[blockTypesCount];
int fullblockscnt = count;
if(paddingsize>0){
fullblockscnt = count - 1;
}
for (int i = 0; i < fullblockscnt; i++) {
start = end+1;
end = (i+1)*s-1;
//compute the type of the block
int type = computeBlockType(ballotnums,A,start,end);
T[i] = type;
if(!pDone[type]){
//if LU table is not done yet for this type of block
P[type] = makeLUTable(s,A,start,size);
pDone[type]=true;
}
B[i] = queryLUTable(s,start,start,end,type);
}
//the end block
if(paddingsize>0){
start = end+1;
end = count*s-1;
//extend the end block
int actualsize = n-start;
int[] D = new int[s];
System.arraycopy(A, start, D, 0, actualsize);
for (int k = actualsize; k < s; k++) {
D[k] = D[k - 1] + 1;
}
int type = computeBlockType(ballotnums,D,0,s-1);
T[count-1] = type;
if(!pDone[type]){
P[type] = makeLUTable(s,D,0,size);
pDone[type]=true;
}
//the min-index from start...n-1, not start...end
B[count-1] = queryLUTable(s,start,start,n-1,type);
}
//step 2: Sparse table algorithm applied to out-of-blocks
this.M = outBlockPreprocess(B,A,count);
}
//Note: this method is the same as the one in class MinusOrPlusOne_RMQ. It is copied from there.
//return the index
public int query(int[] A,int p,int q){
if(q<p){
//swap
int tmp=p;p=q;q=tmp;
}
int start = 0;
int end = -1;
int s = 0,t=0;
int startMin=-1,endMin=-1; //the start block and end block's min index
for (int i = 0; i < this.blockcount; i++) {
start = end+1;
end = (i+1)*this.blocksize-1;
if(p>=start && q<=end){
//within a block
return queryLUTable(blocksize,start,p,q,T[i]);
}else if(p>=start && p<=end){
startMin = queryLUTable(blocksize,start,p,end,T[i]);
s=i+1;
}else if(q<=end && q>=start){
endMin = queryLUTable(blocksize,start,start,q,T[i]);
t=i-1;
break;
}
}
int minIndex = startMin;
if(s<=t){
int outBlocksMin = outBlockQuery(A,s,t);
if(A[startMin]>A[outBlocksMin]){
minIndex = outBlocksMin;
}
}
if(A[minIndex]>A[endMin]){
minIndex = endMin;
}
return minIndex;
}
//Note: this method is the same as the one in class MinusOrPlusOne_RMQ. It is copied from there.
//ST: O(1) for querying
//precondition: s<=t
private int outBlockQuery(int[] A,int s,int t){
int k = (int)(Math.log(t-s+1)/Math.log(2));
//the first interval
int mina = M[s][k];
int minb = M[t-(1<<k)+1][k];
if(A[mina]<=A[minb])
return mina;
else
return minb;
}
//Note: this method is the same as the one in class MinusOrPlusOne_RMQ. It is copied from there.
private int[][] outBlockPreprocess(int[] B,int[] A,int count){
//floor value
int maxJ=(int)(Math.log(count)/Math.log(2));
int[][] M = new int[count][maxJ+1];
//initial condition for dynamic programming: the RMQ for interval length=1
for (int i = 0; i < count; i++)
M[i][0] = B[i];
//dynamic programming: compute values from smaller(j=1) to bigger intervals
for (int j = 1; j<=maxJ; j++){
for (int i = 0; i + (1 << j) - 1 < count; i++){
int nexti = i + (1 << (j - 1));
if (A[M[i][j - 1]] <= A[M[nexti][j - 1]])
M[i][j] = M[i][j - 1];
else
M[i][j] = M[nexti][j - 1];
}
}
return M;
}
//Note: this method is the same as the one in class MinusOrPlusOne_RMQ. It is copied from there.
//return the index
//precondition: i<=j
private int queryLUTable(int blocksize,int offset,int i,int j,int type){
i -= offset; j-=offset;
int[] L = P[type];
int index = blocksize*i - (i-1)*i/2 + (j-i);
return L[index]+offset;
}
//Note: this method is the same as the one in class MinusOrPlusOne_RMQ. It is copied from there.
//use naive method to compute the lookup table for a block
//the return index is relative to the block itself
private int[] makeLUTable(int blocksize, int[] D,int offset,int size) {
int[][] Q = new int[blocksize][blocksize];
for (int i = 0; i < blocksize; i++)
Q[i][i] = i;
for (int i = 0; i < blocksize; i++)
for (int j = i + 1; j < blocksize; j++)
if (D[Q[i][j - 1]+offset] <= D[j+offset])
Q[i][j] = Q[i][j - 1];
else
Q[i][j] = j;
//convert to one-dimension array
int[] L = new int[size];
int k=0;
for(int i=0;i<blocksize;i++){
for(int j=i;j<blocksize;j++,k++){
L[k] = Q[i][j];
}
}
return L;
}
/*
Ballot numbers:
see: Knuth's Art of Computer Programming Vol 4A section 7.2.1.6 "Generating all trees"
The diagonal numbers are Catalan numbers.
Catalan numbers:(0..16)
see: http://en.wikipedia.org/wiki/Catalan_number
1, 1, 2, 5, 14, 42, 132, 429, 1430,
4862, 16796, 58786, 208012, 742900,
2674440, 9694845, 35357670 ....
*/
public static int[] makeBallotNumbersTable(int s){
int k = (s+2)*(s+1)/2;
int[] ballotnums = new int[k];
ballotnums[0] = 1;
int i = 1;
for(int q=1;q<=s;q++){
//p=0, only top element is used, left element is 0
ballotnums[i] = ballotnums[i-q]; i++;
for(int p=1;p<q;p++){
ballotnums[i] = ballotnums[i-q] + ballotnums[i-1];
i++;
}
//p=q, only left element is used, top element is 0
ballotnums[i] = ballotnums[i-1]; i++;
}
return ballotnums;
}
private static void printBallotNumbersTriangle(int[] ballotnums,int s){
//print ballot numbers triangle
System.out.println("Ballot numbers: ");
int start = 0, end = -1;
for(int m=0;m<=s;m++){
start = end + 1;
end = (m+2)*(m+1)/2 - 1;
for(int n=start;n<=end;n++)
System.out.format(" %d", ballotnums[n]);
System.out.println();
}
}
private static void printCatalanNumbers(int[] ballotnums,int s){
//print catalan numbers:
System.out.println("Catalan numbers: ");
for(int m=0;m<=s;m++){
int j = (m+2)*(m+1)/2 - 1;
if(m % 5 == 0){
System.out.format("(%d):",m);
}
System.out.format(" %d", ballotnums[j]);
if((m+1) % 5 == 0)
System.out.println();
}
}
public static void testBallotNumberTable(){
int s = 16;
int[] ballotnums = RMQ_FH.makeBallotNumbersTable(s);
RMQ_FH.printCatalanNumbers(ballotnums, s);
System.out.println();
RMQ_FH.printBallotNumbersTriangle(ballotnums, s);
}
//compute the block type, ie. the sequence number of its cartesian tree
private int computeBlockType(int[] ballotnums, int[] D,int start,int end){
int s = end - start + 1;
//the stack is the rightmost path
int[] stack = new int[s+1];
stack[0] = Integer.MIN_VALUE; //the first element is a sentinel
int top = 0; //stack top
int N=0; //the sequence number of this block's cartesian tree
int q = s;
for(int i=0;i<s;i++){
int p = s-i-1;
while(stack[top]>D[i+start]){
//numbering: go up one level in the ballot number triangle
N += ballotnums[q*(q+1)/2+p];
top--; //remove the element
q--; //up one level
}
//when stack[top]<= D[i+start], we append the current number to the rightmost path
stack[++top] = D[i+start];
}
return N;
}
public void testComputeBlockType(int[] D){
int s = D.length; //block size
int[] ballotnums = RMQ_FH.makeBallotNumbersTable(s);
int nr = this.computeBlockType(ballotnums,D, 0, D.length-1);
System.out.format("block type(=cartesian tree sequence number): %d%n",nr);
}
private void reportLUTable(int[] A){
for(int x=0;x<A.length;x++){
System.out.format("%d..[%d-%d]",x,x,A.length-1);
for(int y=x;y<A.length;y++){
int p = query(A,x,y);
System.out.format(" %d/%d",A[p],p);
}
System.out.println();
}
}
public static void main(String[] args) throws Exception {
//test ballot numbers
RMQ_FH.testBallotNumberTable();
//test block types
System.out.println("*************");
int[] A = new int[]{1,2,3,4,5};
RMQ_FH rmq = new RMQ_FH();
rmq.testComputeBlockType(A);
System.out.println("*************");
A = new int[]{5,4,3,2,1};
rmq = new RMQ_FH();
rmq.testComputeBlockType(A);
//RMQ operations
System.out.format("***********************%n");
A=new int[]{1,7,3};
rmq = new RMQ_FH();
rmq.preprocess(A);
rmq.reportLUTable(A);
System.out.format("***********************%n");
A=new int[]{2,4,3,1,6,7,8,9,1,7};
rmq = new RMQ_FH();
rmq.preprocess(A);
rmq.reportLUTable(A);
System.out.format("***********************%n");
A=new int[]{10,15,34,20,7,5,18,68,29,40, //0..9
24,3,45,26,7,23,43,12,68,34, //10..19
26,34,33,12,80,57,24,42,77,27, //20..29
56,33,23,32,54,13,79,65,19,33, //30..39
15,24,43,73,55,13,63,8,23,17}; //40..49
rmq = new RMQ_FH();
rmq.preprocess(A);
int i=0;
int j=49;
int min = rmq.query(A,i,j);
System.out.format("RMQ for A[%d..%d]: A[%d]=%d%n", i,j,min,A[min]);
}
}
测试输出：
Catalan numbers:
(0): 1 1 2 5 14
(5): 42 132 429 1430 4862
(10): 16796 58786 208012 742900 2674440
(15): 9694845 35357670
Ballot numbers:
1
1 1
1 2 2
1 3 5 5
1 4 9 14 14
1 5 14 28 42 42
1 6 20 48 90 132 132
1 7 27 75 165 297 429 429
1 8 35 110 275 572 1001 1430 1430
1 9 44 154 429 1001 2002 3432 4862 4862
1 10 54 208 637 1638 3640 7072 11934 16796 16796
1 11 65 273 910 2548 6188 13260 25194 41990 58786 58786
1 12 77 350 1260 3808 9996 23256 48450 90440 149226 208012 208012
1 13 90 440 1700 5508 15504 38760 87210 177650 326876 534888 742900 742900
1 14 104 544 2244 7752 23256 62016 149226 326876 653752 1188640 1931540 2674440 2674440
1 15 119 663 2907 10659 33915 95931 245157 572033 1225785 2414425 4345965 7020405 9694845 9694845
1 16 135 798 3705 14364 48279 144210 389367 961400 2187185 4601610 8947575 15967980 25662825 35357670 35357670
*************
block type(=cartesian tree sequence number): 0
*************
block type(=cartesian tree sequence number): 41
***********************
0..[0-2] 1/0 1/0 1/0
1..[1-2] 7/1 3/2
2..[2-2] 3/2
***********************
0..[0-9] 2/0 2/0 2/0 1/3 1/3 1/3 1/3 1/3 1/3 1/3
1..[1-9] 4/1 3/2 1/3 1/3 1/3 1/3 1/3 1/3 1/3
2..[2-9] 3/2 1/3 1/3 1/3 1/3 1/3 1/3 1/3
3..[3-9] 1/3 1/3 1/3 1/3 1/3 1/3 1/3
4..[4-9] 6/4 6/4 6/4 6/4 1/8 1/8
5..[5-9] 7/5 7/5 7/5 1/8 1/8
6..[6-9] 8/6 8/6 1/8 1/8
7..[7-9] 9/7 1/8 1/8
8..[8-9] 1/8 1/8
9..[9-9] 7/9
***********************
RMQ for A[0..49]: A[11]=3
参考资料：
J. Fischer和V. Heun于2006年发表的论文“Theoretical and Practical Improvements on the RMQ-Problem, with Applications to LCA and LCE“