BZOJ3277: 串(广义后缀自动机)

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
const int MAXN = 1e6 + 10;
string s[MAXN];
int N, K;
int fa[MAXN], len[MAXN], ch[MAXN][26], root = 1, last = 1, tot = 1, times[MAXN];
void insert(int x) {
    int now = ++tot, pre = last; last = now; 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; len[nows] = len[pre] + 1;
            memcpy(ch[nows], ch[q], sizeof(ch[q]));
            fa[nows] = fa[q]; fa[q] = fa[now] = nows;
            for(; pre && ch[pre][x] == q; pre = fa[pre]) ch[pre][x] = nows;
        } 
    }
}
int vis[MAXN], sum[MAXN];
void GetTimes() {//求出每一个状态在几个字符串出现过 
    for(int i = 1; i <= N; i++) {
        int now = root;
        for(int j = 0; j < s[i].length(); j++) {
            now = ch[now][s[i][j] - 'a'];//枚举每一个前缀 
            int t = now;
            while(t && vis[t] != i) vis[t] = i, times[t]++, t = fa[t];//枚举每一个后缀 
        }
    }
}
void dfs(int x) {
    if(x == root || vis[x]) return ;
    vis[x] = 1; 
    dfs(fa[x]); 
    sum[x] += sum[fa[x]];
}
int main() {
#ifdef WIN32
    freopen("a.in", "r", stdin);
#endif
    ios::sync_with_stdio(0);
    cin >> N >> K;
    for(int i = 1; i <= N; i++) cin >> s[i];
    for(int i = 1; i <= N; i++) {
        last = 1;
        for(int j = 0; j < s[i].length(); j++)
            insert(s[i][j] - 'a');
    }
    
    GetTimes();
    
    for(int i = 1; i <= tot; i++) sum[i] = (times[i] >= K) * (len[i] - len[fa[i]]);//i状态所表示的子串集合对答案的贡献
    memset(vis, 0, sizeof(vis));
    for(int i = 1; i <= tot; i++) dfs(i);
    for(int i = 1; i <= N; i++) {
        int ans = 0, now = root;
        for(int j = 0; j < s[i].length(); j++)
            now = ch[now][s[i][j] - 'a'], ans += sum[now];
        //枚举前缀，算出每一个前缀所包含的后缀对答案啊的贡献 
        printf("%d ", ans);
    }
    
    return 0;
}