spoj Distinct Substrings

题目链接

分析：
sa后缀数组题。正着不好做可以反着想。我们可以思考重复的串有多少个，这就对应着后缀数组求完后求的height数组。每个$height[i]$长度的串可以构成$height[i]$个重复的串，通过总的减去就行了。

刘汝佳的大白sa模板有一些小问题，在此进行了修正。

代码：

/*****************************************************/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <map>
#include <set>
#include <ctime>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>

using namespace std;

#define   offcin        ios::sync_with_stdio(false)
#define   sigma_size    26
#define   lson          l,m,v<<1
#define   rson          m+1,r,v<<1|1
#define   slch          v<<1
#define   srch          v<<1|1
#define   sgetmid       int m = (l+r)>>1
#define   ll            long long
#define   ull           unsigned long long
#define   mem(x,v)      memset(x,v,sizeof(x))
#define   lowbit(x)     (x&-x)
#define   bits(a)       __builtin_popcount(a)
#define   mk            make_pair
#define   pb            push_back
#define   fi            first
#define   se            second

const int    INF    = 0x3f3f3f3f;
const ll     INFF   = 1e18;
const double pi     = acos(-1.0);
const double inf    = 1e18;
const double eps    = 1e-9;
const ll     mod    = 1e9+7;
const int    maxmat = 10;
const ull    BASE   = 133333331;

/*****************************************************/
inline void RI(int &x) {
      char c;
      while((c=getchar())<'0' || c>'9');
      x=c-'0';
      while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0';
}
/*****************************************************/

const int maxn = 1e3 + 5;

char tmp[maxn];
int s[maxn];
int rank_sa[maxn], height[maxn];
int sa[maxn], t[maxn], t2[maxn], c[maxn];
int N;

// 以字符值数组s构造sa，字符值从0-m-1
void build_sa(int n, int m) {
    int *x = t, *y = t2;
    for (int i = 0; i < m; i ++) c[i] = 0;                
    for (int i = 0; i < n; i ++) c[x[i] = s[i]] ++;
    for (int i = 1; i < m; i ++) c[i] += c[i - 1];          
    for (int i = n - 1; i >= 0; i --) sa[-- c[x[i]]] = i;
    for (int k = 1; k < n; k <<= 1) {
        int p = 0;              
        for (int i = n - k; i < n; i ++) y[p ++] = i;
        for (int i = 0; i < n; i ++) if (sa[i] >= k) y[p ++] = sa[i] - k;
        for (int i = 0; i < m; i ++) c[i] = 0; 
        for (int i = 0; i < n; i ++) c[x[y[i]]] ++;
        for (int i = 1; i < m; i ++) c[i] += c[i - 1];
        for (int i = n - 1; i >= 0; i --) sa[-- c[x[y[i]]]] = y[i];
        swap(x, y);                              
        p = 1; x[sa[0]] = 0;
        for (int i = 1; i < n; i ++)
            x[sa[i]] = (y[sa[i - 1]] == y[sa[i]] 
                && y[sa[i - 1] + k] == y[sa[i] + k]) ? p - 1 : p ++;
        if (p >= n) break;
        m = p;
    }
}

void getHeight(int n) {
    int k = 0;
    height[0] =  0;  
    for (int i = 0; i < n; i ++) rank_sa[sa[i]] = i;
    for(int i = 0; i < n - 1; i ++) {  
        int j = sa[rank_sa[i] - 1];  
        while(i + k < n && j + k < n && s[i + k] == s[j + k]) k++;
        height[rank_sa[i]] = k;
        k = max(0, k - 1);  
    }  
}

void Debug() {
    // cout<<N<<endl;
    for (int i = 0; i <= N; i ++)
        cout<<sa[i]<<" ";
    cout<<endl;
    for (int i = 0; i <= N; i ++)
        cout<<height[i]<<" ";
    cout<<endl;
}

int main(int argc, char const *argv[]) {
    int T; cin>>T;
    while (T --) {
        scanf("%s", tmp);
        N = strlen(tmp);
        memset(s, 0, sizeof(s));
        for (int i = 0; i <= N; i ++) s[i] = tmp[i];
        s[N] = 0;

        build_sa(N + 1, 100);
        getHeight(N + 1);

        // Debug();

        ll ans = N * (N + 1) / 2;
        for (int i = 2; i <= N; i ++) ans -= height[i];
        printf("%I64d\n", ans);
    }
    return 0;
}