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  • SPOJ:NSUBSTR

    题面

    字符串$ S (最多包含) 25 (万个小写拉丁字母。我们将) F(x) (定义为长度为) x (的某些字符串出现在) s (中的最大次数。例如,对于字符串) “ababa”(,)F(3) (将为) 2(,因为存在两次出现的字符串) “aba”(。您的任务是为每个) i $输出 (F(i)),以便$ 1 <= i < = |S|$

    Sol

    (sam)
    直接求一下每个(endpos(right))集合的子串出现次数
    然后就没了

    # include <bits/stdc++.h>
    # define IL inline
    # define RG register
    # define Fill(a, b) memset(a, b, sizeof(a))
    using namespace std;
    typedef long long ll;
    
    template <class Int>
    IL void Input(RG Int &x){
        RG int z = 1; RG char c = getchar(); x = 0;
        for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
        for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
        x *= z;
    }
    
    const int maxn(5e5 + 5);
    
    int n, trans[26][maxn], fa[maxn], len[maxn], tot = 1, last = 1;
    int id[maxn], t[maxn], size[maxn], ans[maxn];
    char s[maxn];
    
    IL void Extend(RG int c){
        RG int p = last, np = ++tot; last = tot;
        len[np] = len[p] + 1, size[np] = 1;
        while(p && !trans[c][p]) trans[c][p] = np, p = fa[p];
        if(!p) fa[np] = 1;
        else{
            RG int q = trans[c][p];
            if(len[q] == len[p] + 1) fa[np] = q;
            else{
                RG int nq = ++tot;
                fa[nq] = fa[q], len[nq] = len[p] + 1;
                for(RG int i = 0; i < 26; ++i) trans[i][nq] = trans[i][q];
                fa[q] = fa[np] = nq;
                while(p && trans[c][p] == q) trans[c][p] = nq, p = fa[p];
            }
        }
    }
    
    int main(RG int argc, RG char* argv[]){
        scanf(" %s", s), n = strlen(s);
        for(RG int i = 0; i < n; ++i) Extend(s[i] - 'a');
        for(RG int i = 1; i <= tot; ++i) ++t[len[i]];
        for(RG int i = 1; i <= tot; ++i) t[i] += t[i - 1];
        for(RG int i = 1; i <= tot; ++i) id[t[len[i]]--] = i;
        for(RG int i = tot; i; --i){
            size[fa[id[i]]] += size[id[i]];
            ans[len[id[i]]] = max(ans[len[id[i]]], size[id[i]]);
        }
        for(RG int i = tot; i; --i) ans[i] = max(ans[i], ans[i + 1]);
        for(RG int i = 1; i <= n; ++i) printf("%d
    ", ans[i]);
        return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/cjoieryl/p/8900950.html
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