hdu3518 Boring counting(后缀数组)

Boring counting

题目传送门

解题思路

后缀数组。枚举每种长度，对于每个字符串，记录其最大起始位置和最小起始位置，比较是否重合。

代码如下

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;

const int N = 1005;

char s[N];
int sa[N], x[N], y[N], c[N];
int n, m;

void get_sa()
{
    for(int i = 1; i <= m; i ++) c[i] = 0;
    for(int i = 1; i <= n; i ++) c[x[i] = s[i]] ++;
    for(int i = 1; i <= m; i ++) c[i] += c[i - 1];
    for(int i = n; i >= 1; i --) sa[c[x[i]] --] = i;
    for(int k = 1; k <= n; k <<= 1){
        int num = 0;
        for(int i = n - k + 1; i <= n; i ++) y[++num] = i;
        for(int i = 1; i <= n; i ++) if(sa[i] > k) y[++num] = sa[i] - k;
        for(int i = 1; i <= m; i ++) c[i] = 0;
        for(int i = 1; i <= n; i ++) c[x[i]] ++;
        for(int i = 1; i <= m; i ++) c[i] += c[i - 1];
        for(int i = n; i >= 1; i --) sa[c[x[y[i]]] --] = y[i], y[i] = 0;
        swap(x, y);
        num = 1;
        x[sa[1]] = num;
        for(int i = 2; i <= n; i ++){
            if(sa[i] + k <= n && sa[i - 1] + k <= n)
                x[sa[i]] = (y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k])? num: ++num;
            else 
                x[sa[i]] = ++num;
        }
        if(num == n)
            break;
        m = num;
    }
}

int height[N], rk[N];

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

int main()
{
    while(scanf("%s", s + 1) != EOF && s[1] != '#'){
        n = strlen(s + 1);
        m = 'z';
        get_sa();
        get_h();
        ll ans = 0;
        for(int k = 1; k <= n / 2; k ++){
            int maxx, minn;
            maxx = sa[1], minn = sa[1];
            for(int i = 2; i <= n; i ++){
                if(height[i] < k){
                    if(maxx - minn >= k) ++ans;
                    maxx = sa[i], minn = sa[i];
                }
                else {
                    maxx = max(maxx, sa[i]);
                    minn = min(minn, sa[i]);
                }
            }
            if(maxx - minn >= k) ++ ans;
        }
        printf("%lld\n", ans);
    }
    return 0;
}