CSL 的密码(后缀数组)

CSL 的密码

题目传送门

解题思路

后缀数组。对于每一个后缀\(k\)都有\(n - k + 1\)个前缀，把所有不和前一个(排序后的)公共且长度大于\(m\)的前缀个数加起来。

代码如下

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

const int N = 100005;

int x[N], y[N], c[N], sa[N];
char s[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;
        num = 1;
        swap(x, y);
        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()
{
    int len;
    scanf("%d%d", &n, &len);
    m = 'z';
    scanf("%s", s + 1);
    get_SA();
    get_h();
    ll ans = 0;
    for(int i = 1; i <= n; i ++){
        if(sa[i] > n - len + 1) continue;
        int num = n - sa[i] + 1 - len + 1;
        ans += num - max(0, height[i] - len + 1);
    }
    printf("%lld\n", ans);
    return 0;
}