后缀数组（不相同的子串个数）——SPOJ 705

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Distinct Substrings

Time Limit: 159MS Memory Limit: Unknown 64bit IO Format: %lld & %llu

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Description

Given a string, we need to find the total number of its distinct substrings.

Input

T- number of test cases. T<=20;
Each test case consists of one string, whose length is <= 50000

Output

For each test case output one number saying the number of distinct substrings.

Example

Sample Input:
2
CCCCC
ABABA

Sample Output:
5
9

Explanation for the testcase with string ABABA:
len=1 : A,B
len=2 : AB,BA
len=3 : ABA,BAB
len=4 : ABAB,BABA
len=5 : ABABA
Thus, total number of distinct substrings is 9.

Source

题意：求给出的每个字符串的不相同子串数目

思路：后缀数组（不相同子串个数）的基本应用

#include <stdio.h>  #include <stdlib.h>  #include <string.h>  #define MS(x, y) memset(x, y, sizeof(x))  const int MAXN = 1000+10;  const int INF = 1<<30;    int wa[MAXN],wb[MAXN],wv[MAXN],ws[MAXN];  int rank[MAXN],r[MAXN],sa[MAXN],height[MAXN];  char str[MAXN];    int cmp(int *r, int a, int b, int l)  {      return r[a] == r[b] && r[a+l] == r[b+l];  }    void da(int *r, int *sa, int n, int m)  {      int i, j, p, *x = wa, *y = wb, *t;        for(i=0; i<m; i++) ws[i] = 0;      for(i=0; i<n; i++) ws[x[i] = r[i]]++;      for(i=1; i<m; i++) ws[i] += ws[i-1];      for(i=n-1; i>=0; i--) sa[--ws[x[i]]] = i;        for(j=1,p=1; p<n; j<<=1, m=p){            for(p=0,i=n-j; i<n; i++) y[p++] = i;          for(i=0; i<n; i++) if(sa[i] >= j) y[p++] = sa[i] - j;            for(i=0; i<n; i++) wv[i] = x[y[i]];          for(i=0; i<m; i++) ws[i] = 0;          for(i=0; i<n; i++) ws[wv[i]]++;          for(i=1; i<m; i++) ws[i] += ws[i-1];          for(i=n-1; i>=0; i--) sa[--ws[wv[i]]] = y[i];            for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1; i<n; i++)              x[sa[i]] = cmp(y, sa[i-1], sa[i], j) ? p-1 : p++;        }      return;  }    void calheight(int *r, int *sa, int n)  {      int i, j, k = 0;      for(i=1; i<n; i++) rank[sa[i]] = i;      for(i=0; i<n-1; height[rank[i++]] = k)          for(k ? k-- : 0,j=sa[rank[i]-1]; r[i+k] == r[j+k]; k++);      return;  }  int main(){//freopen("in.txt", "r", stdin);int T;scanf("%d", &T);while(T--){MS(r, 0);MS(sa, 0);MS(rank, 0);MS(height, 0);MS(ws, 0);MS(wv, 0);MS(wa, 0);MS(wb, 0);scanf("%s", str);int i;int len = strlen(str);for(i=0; i<len; i++)r[i] = (int)str[i] - 31;//可显示字符的范围是 32~126，r[len++] = 0;da(r, sa, len, 100);calheight(r, sa, len);int ans = 0;for(i=1; i<len; i++)ans += len - 1 - sa[i] - height[i];printf("%d\n", ans);}return 0;}

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