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SPOJ8222 NSUBSTR - Substrings(后缀自动机)

2018-06-29 06:14:37来源：博客园阅读 ()

You are given a string S which consists of 250000 lowercase latin letters at most. We define F(x) as the maximal number of times that some string with length x appears in S. For example for string 'ababa' F(3) will be 2 because there is a string 'aba' that occurs twice. Your task is to output F(i) for every i so that 1<=i<=|S|.

Input

String S consists of at most 250000 lowercase latin letters.

Output

Output |S| lines. On the i-th line output F(i).

Example

Input:
ababa

Output:
3
2
2
1
1

Submit solution!

为什么我的后缀自动机写的这么奇怪qwq。。

为什么泥萌都在写桶排？？！！

这题不是dfs一下求出$right$集合来就完了么qwq、、

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
const int MAXN = 250001 << 1, INF = 1e9 + 10;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
char s[MAXN];
int N;
int f[MAXN], siz[MAXN], fa[MAXN], len[MAXN], ch[MAXN][26], root = 1, last = 1, tot = 1;
void insert(int x) {
    int now = ++tot, pre = last; last = now;
    siz[now] = 1;
    len[now] = len[pre] + 1;
    for(; pre && !ch[pre][x]; pre = fa[pre]) ch[pre][x] = now;
    if(!pre) fa[now] = root;
    else {
        int q = ch[pre][x];
        if(len[q] == len[pre] + 1) fa[now] = q;
        else {
            int nows = ++tot; 
            memcpy(ch[nows], ch[q], sizeof(ch[q]));
            fa[nows] = fa[q]; fa[q] = fa[now] = nows;
            len[nows] = len[pre] + 1;
            for(; pre && ch[pre][x] == q; pre = fa[pre]) ch[pre][x] = nows;
        }
    }
}
vector<int> v[MAXN];
void dfs(int x) {
    for(int i = 0; i < v[x].size(); i++) {
        dfs(v[x][i]);
        siz[x] += siz[v[x][i]];
    }
    f[len[x]] = max(f[len[x]], siz[x]);
}
int main() {
#ifdef WIN32
    freopen("a.in", "r", stdin);
#else
    //freopen("pow.in", "r", stdin);
    //freopen("pow.out", "w", stdout); 
#endif    
    scanf("%s", s + 1);
    N = strlen(s + 1);
    for(int i = 1; i <= N; i++) insert(s[i] - 'a');
    for(int i = 2; i <= tot; i++) v[fa[i]].push_back(i);
    dfs(root);
    for(int i = 1; i <= N; i++) 
        printf("%d\n", f[i]);
    return 0;
}