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

Submit Status

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;
}