// for pairs
inline bool leq(int a1, int a2, int b1, int b2)
{
return (a1<b1 || a1==b1 && a2<b2);
}
// for triples
inline bool leq(int a1, int a2, int a3, int b1, int b2, int b3){
return (a1<b1 || a1==b1 && leq(a2,a3,b2,b3));
}
// stably sort a[0..n-1] to b[0..n-1] with keys in 0..K frmo r
static void radixPass(int *a, int *b, int *r, int n, int K){
int i,sum;
// count occurences
int *c = new int[K+1];
for(i=0; i<=K; i++) c[i]=0;
for(i=0; i<n; i++) c[r[a[i]]]++;
for(i=0, sum=0; i<=K; i++)
{ int t = c[i]; c[i] = sum; sum+=t;}
for(i=0; i<n; ++i) b[c[r[a[i]]]++] = a[i];
delete[] c;
}
//find the suffix array SA of s[0..n-1] in {1..K}^n
//requires s[n]=s[n+1]=s[n+2]=0, n>=2
void suffixArray(int *s, int *SA, int n, int K){
int i,j;
int n0 = (n+2)/3, n1=(n+1)/3, n2=n/3, n02=n0+n2;
int* s12 = new int[n02+3]; s12[n02]=s12[n02+1]=s12[n02+2]=0;
int* SA12 = new int[n02+3]; SA12[n02]=SA12[n02+1]=SA12[n02+2]=0;
int* s0 = new int[n0];
int* SA0 = new int[n0];
//generate positions of mod 1 and mod 2 suffixes
//the "+(n0-n1)" adds a dummy mod 1 suffix if n%3 == 1
for(i=0, j=0; i<n+(n0-n1); i++) if(i%3 != 0) s12[j++]=i;
//lsb radix sort the mod 1 and mod 2 triples
radixPass(s12 , SA12, s+2, n02, K);
radixPass(SA12, s12 , s+1, n02, K);
radixPass(s12 , SA12, s , n02, K);
//find lexicographic names of triples
int name = 0, c0 = -1, c1 = -1, c2 = -1;
for(i=0; i<n02; i++){
if(s[SA12[i]] != c0 || s[SA12[i]+1] != c1 || s[SA12[i]+2] != c2){
name++; c0 = s[SA12[i]]; c1 = s[SA12[i]+1]; c2 = s[SA12[i]+2];
}
if(SA12[i]%3 == 1) { s12[SA12[i]/3] = name; } //left half
else { s12[SA12[i]/3 + n0] = name; } //right half
}
//recurse if names are yet unique
if(name < n02) {
suffixArray(s12, SA12, n02, name);
// store unique names in s12 using the suffix array
for(i=0; i<n02; i++) s12[SA12[i]] = i+1;
}else
for(i=0; i<n02; i++) SA12[s12[i]-1] = i;
//stably sort the mod 0 suffixes from SA12 by their first chraccter
for(i=0, j=0; i<n02; ++i) if(SA12[i]<n0) s0[j++] = 3*SA12[i];
radixPass(s0, SA0, s, n0, K);
//merge sorted SA0 suffixes and sorted SA12 suffixes
for(int p=0, t=n0-n1, k=0; k<n; k++){
#define GetI() ( SA12[t] < n0 ? SA12[t] * 3 + 1 : (SA12[t] - n0) * 3 + 2)
i = GetI(); // pos of current offset 12 suffix
j = SA0[p]; // pos of current offset 0 suffix
if(SA12[t] < n0 ? // different compares for mod 1 and mod 2 suffixes
leq(s[i], s12[SA12[t] + n0], s[j], s12[j/3]) :
leq(s[i], s[i+1], s12[SA12[t]-n0+1], s[j], s[j+1], s12[j/3+n0] )){
//suffix from SA12 is smaller
SA[k] = i; t++;
if(t==n02) // done -- only SA0 suffixes left
for(k++; p<n0; p++, k++) SA[k] = SA0[p];
}else{
//suffix from SA0 is smaller
SA[k] = j; p++;
if(p==n0) // done -- only SA12 suffixes left
for(k++; t<n02; t++,k++) SA[k] = GetI();
}
}
delete[] s12;
delete[] SA12;
delete[] s0;
delete[] SA0;
}参考文献:
[1] IOI2009 后缀数组－处理字符串的有力工具
[2] Simple Linear Work Suffix Array Construction
posted on 2010-11-02 21:47
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