SPOJ DISUBSTR 后缀数组

题目链接：http://www.spoj.com/problems/DISUBSTR/en/

题意：给定一个字符串，求不相同的子串个数。

思路：直接根据09年oi论文<<后缀数组——出来字符串的有力工具>>的解法。

还有另一种思想：总数为n*(n-1)/2,height[i]是两个后缀的最长公共前缀，所以用总数-height[i]的和就是答案

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<queue>
#include<vector>
#include<time.h>
#include<cmath>
using namespace std;
typedef long long int LL;
const int MAXN =  + ;
int cmp(int *r, int a, int b, int l){
    return r[a] == r[b] && r[a + l] == r[b + l];
}
int wa[MAXN], wb[MAXN], wv[MAXN], WS[MAXN];
void da(int *r, int *sa, int n, int m){
    int i, j, p, *x = wa, *y = wb, *t;
    for (i = ; i<m; i++) WS[i] = ;
    for (i = ; i<n; i++) WS[x[i] = r[i]]++;
    for (i = ; i<m; i++) WS[i] += WS[i - ];
    for (i = n - ; i >= ; i--) sa[--WS[x[i]]] = i;
    for (j = , p = ; p<n; j *= , m = p)
    {
        for (p = , i = n - j; i<n; i++) y[p++] = i;
        for (i = ; i<n; i++) if (sa[i] >= j) y[p++] = sa[i] - j;
        for (i = ; i<n; i++) wv[i] = x[y[i]];
        for (i = ; i<m; i++) WS[i] = ;
        for (i = ; i<n; i++) WS[wv[i]]++;
        for (i = ; i<m; i++) WS[i] += WS[i - ];
        for (i = n - ; i >= ; i--) sa[--WS[wv[i]]] = y[i];
        for (t = x, x = y, y = t, p = , x[sa[]] = , i = ; i<n; i++)
            x[sa[i]] = cmp(y, sa[i - ], sa[i], j) ? p -  : p++;
    }
    return;
}
int Rank[MAXN], height[MAXN],sa[MAXN];
void calheight(int *r, int *sa, int n){
    int i, j, k = ;
    for (i = ; i <= n; i++) Rank[sa[i]] = i;
    for (i = ; i < n; height[Rank[i++]] = k)
        for (k ? k-- : , j = sa[Rank[i] - ]; r[i + k] == r[j + k]; k++);
    return;
}
void solve(int n){
    int ans = ;
    for (int i = ; i <= n; i++){
        ans += ((n - ) - sa[i] +  - height[i]);
        //(n-1)的原因，因为在最后加了一个原理字符串不存在的最小值
        //0，所以字符串原始长度为n-1
    }
    printf("%d\n", ans);
}
int t, len,r[MAXN];
char str[MAXN];
int main(){
//#ifdef kirito
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
//#endif
//    int start = clock();
    scanf("%d", &t);
    while (t--){
        scanf("%s", str); len = strlen(str);
        for (int i = ; i <= len; i++){
            if (i == len){ r[i] = ; continue; } //字符串最后添加一个
            //小于所以字符的值
            r[i] = (int)str[i];
        }
        da(r, sa, len+, );
        calheight(r, sa, len);
        solve(len);
    }
//#ifdef LOCAL_TIME
//    cout << "[Finished in " << clock() - start << " ms]" << endl;
//#endif
    return ;
}