SPOJ SUBST1 后缀数组

题目链接：http://www.spoj.com/problems/SUBST1/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。

此题和SPOJ DISUBSTR一样，至少数据范围变大了。

#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 = 50000 + 5;
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 = 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 *= 2, 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;
}
int Rank[MAXN], height[MAXN], sa[MAXN];
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; height[Rank[i++]] = k)
        for (k ? k-- : 0, j = sa[Rank[i] - 1]; r[i + k] == r[j + k]; k++);
    return;
}
void solve(int n){
    int ans = 0;
    for (int i = 1; i <= n; i++){
        ans += ((n - 1) - sa[i] + 1 - height[i]);
    }
    printf("%d
", 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 = 0; i <= len; i++){
            if (i == len){ r[i] = 0; continue; }
            r[i] = (int)str[i];
        }
        da(r, sa, len + 1, 256);
        calheight(r, sa, len);
        solve(len);
    }
    //#ifdef LOCAL_TIME
    //    cout << "[Finished in " << clock() - start << " ms]" << endl;
    //#endif
    return 0;
}