HDU3518 Boring counting

后缀数组

要求统计出现两次以上不重叠的子串，其实就是统计出现两次以上不重叠的后缀的前缀，不难想到height数组。

相邻的后缀只有包含和被包含关系，或者不在同一块内，重复出现的前缀肯定在这块的height里，所以我们统计连续块内最小和最大的sa相减判断是否满足条件即可。

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
#define __fastIn ios::sync_with_stdio(false), cin.tie(0)
#define pb push_back
using namespace std;
using LL = long long;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int ret = 0, w = 0; char ch = 0;
    while(!isdigit(ch)){
        w |= ch == '-', ch = getchar();
    }
    while(isdigit(ch)){
        ret = (ret << 3) + (ret << 1) + (ch ^ 48);
        ch = getchar();
    }
    return w ? -ret : ret;
}
template <typename A>
inline A __lcm(A a, A b){ return a / __gcd(a, b) * b; }
template <typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}
const int N = 2000;
string s;
LL ans;

namespace SuffixArray{

    string s;
    int sa[N], t[N], t2[N], c[N], rank[N], height[N];

    void build(int m, int n){
        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 = 0; 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 = 0; 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]] && sa[i - 1] + k < n &&
                        sa[i] + k < n && y[sa[i - 1] + k] == y[sa[i] + k] ? p - 1 : p ++;
            }
            if(p >= n) break;
            m = p;
        }

        int k = 0;
        for(int i = 0; i < n; i ++) rank[sa[i]] = i;
        for(int i = 0; i < n; i ++){
            if(!rank[i]) continue;
            if(k) k --;
            int j = sa[rank[i] - 1];
            while(s[i + k] == s[j + k]) k ++;
            height[rank[i]] = k;
        }
    }
}

using SuffixArray::build;
using SuffixArray::sa;
using SuffixArray::height;

bool solve(int k){
    bool good = false;
    int l = INF, r = 0;
    for(int i = 1; i < s.size(); i ++){
        if(height[i] >= k){
            l = min(l, min(sa[i], sa[i - 1]));
            r = max(r, max(sa[i], sa[i - 1]));
        }
        else{
            if(r - l >= k) ans ++, good = true;
            l = INF, r = 0;
        }
    }
    if(r - l >= k) ans ++, good = true;
    return good;
}

int main(){

    __fastIn;

    while(cin >> s && s != "#"){
        SuffixArray::s = s;
        build(128, (int)s.size());
        ans = 0;
        for(int i = 1; i <= (s.size() + 1) / 2; i ++){
            if(!solve(i)) break;
        }
        cout << ans << endl;
    }
    return 0;
}