BZOJ 3277/3473 广义后缀自动机

说实话没啥难的.

建一棵广义后缀自动机，暴力自底向上更新即可.

时间复杂度非常玄学，但据说是可以过的.

要注意每个串中相同的子串的贡献是都要加进去的，开始因为这个被坑了好久 QAQ

Code:

#include <cstdio>
#include <algorithm>
#include <vector>
#include <cstring>
#include <string>
#define setIO(s) freopen(s".in","r",stdin)
#define maxn 300000
#define N 30
using namespace std;
int m,k,n,length[maxn];
char str[maxn];
string s[maxn];
struct SAM{
    int last,tot;
    int ch[maxn][N], f[maxn],cnt[maxn],len[maxn],C[maxn],rk[maxn],mk[maxn];
    long long sumv[maxn];
    void init() { last = tot = 1; }
    void ins(int c){
        int p=last,np,nq;
        if(ch[p][c]){
            int q=ch[p][c];
            if(len[q]==len[p]+1) last=q;
            else
            {
                nq=++tot,last=nq;
                f[nq]=f[q],f[q]=nq,len[nq]=len[p]+1;
                memcpy(ch[nq],ch[q],sizeof(ch[q]));
                while(p&&ch[p][c]==q)ch[p][c]=nq,p=f[p];
            }
        }
        else {
            np=++tot,last=np,len[np]=len[p]+1;
            while(p&&!ch[p][c]) ch[p][c]=np,p=f[p];
            if(!p) f[np]=1;
            else {
                int q=ch[p][c];
                if(len[q]==len[p]+1) f[np]=q;
                else
                {
                    nq=++tot;
                    f[nq]=f[q],f[q]=f[np]=nq,len[nq]=len[p]+1;
                    memcpy(ch[nq],ch[q],sizeof(ch[q]));
                    while(p&&ch[p][c]==q) ch[p][c]=nq,p=f[p];
                }
            }    

        }
    }
    void Solve(){
        for(int i = 1;i <= m; ++i) {
            int p = 1,c,u;
            for(int j = 0;j < length[i]; ++j) {
                c = s[i][j] - 'a'; p = ch[p][c]; u = p;
                while(u && mk[u] != i) ++cnt[u],mk[u] = i,u = f[u];
            }
        }
        for(int i = 1;i <= tot; ++i) C[len[i]]++;
        for(int i = 1;i <= tot; ++i) C[i] += C[i - 1];
        for(int i = 1;i <= tot; ++i) rk[C[len[i]]--] = i;
        for(int i = 1;i <= tot; ++i)
        {
            int t = rk[i];
            sumv[t] = cnt[t] >= k ? sumv[f[t]] + (len[t] - len[f[t]]) : sumv[f[t]];
        }
        for(int i = 1;i <= m; ++i)
        {
            int p = 1;
            long long ans = 0;
            for(int j = 0;j < length[i]; ++j) {
                p = ch[p][s[i][j]-'a'];
                ans += sumv[p];
            }
            printf("%lld ",ans);
        }
    }
}T;
int main() {
    //setIO("input");
    scanf("%d%d",&m,&k),T.init();
    for(int i = 1;i <= m; ++i) {
        T.last = 1;
        scanf("%s",str),s[i] = string(str),length[i] = strlen(str);
        for(int j = 0;j < length[i]; ++j) T.ins(s[i][j] - 'a');
    }
    T.Solve();
    return 0;
}