題意

給定若幹組由數字構成的字符串，求所有不重復子串的和（把他們看成十進制），答案mod(1e9+7)

題解：

類似後綴數組的做法，把字符串之間用‘:‘連接，這裏用‘:‘是因為‘:‘的ascii碼恰好是9的下一個

然後建立後綴自動機。

之後把其實只要把其中的所有‘:‘邊刪去，就可以進行轉移了

如果x連向了y，邊權是c，那麽有轉移

dp[y] += dp[x]*10 + c*sz[x]

所以只要拓撲排序一下就好

（寫這題wa了好幾次，主要是在刪邊建立新圖的過程出了問題）

#include <iostream>
#include <cstring>
#include 
 
<cstdio>
#include <queue>
using namespace std;
typedef long long LL;
int n = 0, len, st;
const int maxL = 2e6 + 100;
const int MOD = 1e9 + 7;
int maxlen[2*maxL], minlen[2*maxL], trans[2*maxL][12], slink[2*maxL], col[2*maxL];
LL in[2*maxL], dp[2*maxL], sz[2*maxL];
int new_state(int _maxlen, int
 
 _minlen, int *_trans, int _slink){
    maxlen[n] = _maxlen;
    minlen[n] = _minlen;
    for(int i = 0; i < 11; i++){
        if(_trans == NULL)
            trans[n][i] = -1;
        else
            trans[n][i] = _trans[i];
    }
    slink[n] = _slink;
    return n++;
}

int add_char(char ch, int
 
 u){
    int c = ch - ‘0‘;
    int z = new_state(maxlen[u]+1, -1, NULL, -1);
    int v = u;
    while(v != -1 && trans[v][c] == -1){
        trans[v][c] = z;
        v = slink[v];
    }
    if(v == -1){
        minlen[z] = 1;
        slink[z] = 0;
        return z;
    }
    int x = trans[v][c];
    if(maxlen[v] + 1 == maxlen[x]){
        minlen[z] = maxlen[x] + 1;
        slink[z] = x;
        return z;
    }
    int y = new_state(maxlen[v] + 1, -1, trans[x], slink[x]);
    slink[y] = slink[x];
    minlen[x] = maxlen[y] + 1;
    slink[x] = y;
    minlen[z] = maxlen[y] + 1;
    slink[z] = y;
    int w = v;
    while(w != -1 && trans[w][c] == x){
        trans[w][c] = y;
        w = slink[w];
    }
    minlen[y] = maxlen[slink[y]] + 1;
    return z;
}

char str[maxL];
queue<int> Q;
int main()
{
    freopen("a.txt", "r", stdin);
    int N;
    cin>>N;
    st = new_state(0, 0, NULL, -1);
    while(N--){
        cin>>str;
        int len = strlen(str);
        for(int i = 0; i < len; i++) {
            st = add_char(str[i], st);
        }
        st = add_char(‘:‘, st);
    }
    Q.push(0); col[0] = 1;
    while(!Q.empty()){
        int x = Q.front(); Q.pop();
        for(int c = 0; c < 10; c++){
            int y = trans[x][c];
            if(y == -1) continue;
            if(!col[y]) Q.push(y); col[y] = 1;
            in[y]++;
        }
    }
    Q.push(0); sz[0] = 1;
    while(!Q.empty()){
        int x = Q.front(); Q.pop();
        for(int c = 0; c < 10; c++){
            int y = trans[x][c];
            if(y == -1) continue;
            in[y]--;
            if(in[y] == 0) Q.push(y);
            (sz[y] += sz[x]) %= MOD;
            (dp[y] += dp[x]*10 + c*sz[x]) %= MOD;
        }
    }
    LL ans = 0;
    /*
    for(int i = 0; i <  n; i++){
        for(int c = 0; c < 10; c++){
            if(trans[i][c] == -1) continue;
            cout<<i<<" "<<trans[i][c]<<" "<<c<<endl;
        }
    }*/
    //for(int i = 0; i < n; i++) cout<<dp[i]<<" "; cout<<endl;
    for(int i = 0; i < n; i++) (ans += dp[i]) %= MOD;
    cout<<ans<<endl;
    return 0;
}

hihocoder 1457（後綴自動機+拓撲排序）