bzoj3473 字符串

双倍经验：bzoj3277

题意：给定n个字符串，对于每个字符串，求有多少个非空子串是其中至少k个字符串的子串。

解：有一种毒瘤后缀数组解法...不过后缀自动机吊打后缀树组！耶~

广义后缀自动机。

先建出来，然后对于每个串，跑一遍后缀自动机。我们试图把该串的所有子串都打上标记(sum++)。

跑到的每个节点，都代表一个前缀，我们跳fail就能得到该前缀的所有后缀。用vis数组避免重复标记。

所有sum >= k的节点，贡献为len - len[fail]，否则为0。

之后对于每个串再跑后缀自动机，跑到的每个节点的贡献为它到根的贡献，可以预处理出来，加起来就是答案。

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>

const int N = 500010;

struct Edge {
    int nex, v;
}edge[N << 1]; int top;

int tr[N][26], len[N], cnt[N], sum[N], fail[N], vis[N];
int last = 1, tot = 1, Time, k, e[N];
std::string str[100010];
char s[N];

inline void add(int x, int y) {
    top++;
    edge[top].v = y;
    edge[top].nex = e[x];
    e[x] = top;
    return;
}

inline int split(int p, int f) {
    int Q = tr[p][f], nQ = ++tot;
    len[nQ] = len[p] + 1;
    fail[nQ] = fail[Q];
    fail[Q] = nQ;
    memcpy(tr[nQ], tr[Q], sizeof(tr[Q]));
    while(tr[p][f] == Q) {
        tr[p][f] = nQ;
        p = fail[p];
    }
    return nQ;
}

inline int insert(int p, char c) {
    int f = c - 'a';
    if(tr[p][f]) {
        int Q = tr[p][f];
        if(len[Q] == len[p] + 1) {
            return Q;
        }
        return split(p, f);
    }
    int np = ++tot;
    len[np] = len[p] + 1;
    while(p && !tr[p][f]) {
        tr[p][f] = np;
        p = fail[p];
    }
    if(!p) {
        fail[np] = 1;
    }
    else {
        int Q = tr[p][f];
        if(len[Q] == len[p] + 1) {
            fail[np] = Q;
        }
        else {
            fail[np] = split(p, f);
        }
    }
    return np;
}

void DFS(int x) {
    if(vis[x] == Time || !x) {
        return;
    }
    sum[x]++;
    vis[x] = Time;
    DFS(fail[x]);
    return;
}

void DFS_1(int x) {
    cnt[x] = cnt[fail[x]] + (len[x] - len[fail[x]]) * (sum[x] >= k);
    //printf("cnt %d = %d + %d * %d 
", x, cnt[fail[x]], len[x] - len[fail[x]], sum[x] >= k);
    for(int i = e[x]; i; i = edge[i].nex) {
        int y = edge[i].v;
        DFS_1(y);
    }
    return;
}

int main() {
    int n;
    scanf("%d%d", &n, &k);
    for(int i = 1; i <= n; i++) {
        scanf("%s", s);
        int t = strlen(s);
        last = 1;
        for(int j = 0; j < t; j++) {
            last = insert(last, s[j]);
        }
        str[i] = (std::string)(s);
    }

    for(int i = 1; i <= n; i++) {
        int p = 1, t = str[i].size();
        Time = i;
        for(int j = 0; j < t; j++) {
            int f = str[i][j] - 'a';
            p = tr[p][f];
            DFS(p);
        }
    }

    for(int i = 2; i <= tot; i++) {
        add(fail[i], i);
        //printf("add %d %d len = %d %d 
", fail[i], i, len[fail[i]], len[i]);
    }
    DFS_1(1);

    for(int i = 1; i <= n; i++) {
        int p = 1, t = str[i].size(), ans = 0;
        //printf("i = %d 
", i);
        for(int j = 0; j < t; j++) {
            int f = str[i][j] - 'a';
            p = tr[p][f];
            ans += cnt[p];
            //printf(" > p = %d ans += %d 
", p, cnt[p]);
        }
        printf("%d ", ans);
    }

    return 0;
}