Yuta has n01 strings si, and he wants to know the number of 01 antisymmetric strings of length 2L which contain all given strings si as continuous substrings.
A 01 string s is antisymmetric if and only if s[i]≠s[|s|−i+1] for all i∈[1,|s|].
It is too difficult for Rikka. Can you help her?
In the second sample, the strings which satisfy all the restrictions are 000111,001011,011001,100110
.
Input The first line contains a number
t(1≤t≤5),
the number of the testcases. For each testcase, the first line contains two numbers n,L(1≤n≤6,1≤L≤100).
Then n lines follow, each line contains a 01 string si(1≤|si|≤20). Output For each testcase, print a single line with a single number -- the answer modulo 998244353.Sample Input
2 2 2 011 001 2 3 011 001Sample Output
1
分析:現在要求的是包含所有模板的情況,因爲n很小所以我們可以容斥做,每次枚舉一個子集然後求不包含這個子集的任何一個模板的情況,把模板全部翻轉後再插入就可以
表示後綴了，dp[i][j][k]表示當前匹配到第i位其中前綴匹配到自動機的第j個點,後綴匹配到第k個點的方案數,最後枚舉一下dp[L][j][k]中合法的j,k再統計答案就可了,這一步可以
暴力沿着fail指針遍歷判斷前綴後綴會不會拼出新的模板.
#include<bits/stdc++.h>
using namespace std;
const int MOD = 998244353;
const int MAXNODE = 125;
int T,n,l,ans,rc[1<<6],dp[105][125][125];
char s[7][25],rs[7][25];
struct ACautomata
{
int ch[MAXNODE][2];
int num[MAXNODE];
int deep[MAXNODE];
int f[MAXNODE]; // fail函數
int val[MAXNODE]; // 是否爲單詞結尾
int tot; // trie 單詞總數
int last[MAXNODE];
void init()
{
tot = 1;
memset(ch,0,sizeof(ch));
memset(num,0,sizeof(num));
memset(last,0,sizeof(last));
memset(f,0,sizeof(f));
}
int idx(char c)
{
return c - '0';
}
void insert(char *s,int v)
{
int u = 0,n = strlen(s);
for(int i = 0;i < n;i++)
{
int c = idx(s[i]);
if(!ch[u][c])
{
val[tot] = 0;
deep[tot] = deep[u] + 1;
ch[u][c] = tot++;
}
num[u] |= 1<<(v-1);
u = ch[u][c];
}
val[u] = true;
num[u] |= 1<<(v-1);
}
void getFail()
{
queue<int> q;
f[0] = 0;
for(int c = 0;c < 2;c++)
{
int u = ch[0][c];
if(u)
{
f[u] = 0;
q.push(u);
last[u] = 0;
}
}
while(!q.empty())
{
int r = q.front();q.pop();
for(int c = 0;c < 2;c++)
{
int u = ch[r][c];
if(!u)
{
ch[r][c] = ch[f[r]][c];
continue;
}
q.push(u);
f[u] = ch[f[r]][c];
last[u] = val[f[u]] ? f[u] : last[f[u]];
}
}
}
}tree1,tree2;
bool check(int x,int y)
{
if(tree1.val[x] || tree2.val[y] || tree1.last[x] || tree2.last[y]) return false;
for(int i = 1;i <= n;i++)
{
int len = strlen(s[i]);
for(int u = x;u;u = tree1.f[u])
if(tree1.num[u] & (1<<(i-1)))
for(int v = y;v;v = tree2.f[v])
if(tree2.num[v] & (1<<(i-1)))
{
if(tree1.deep[u] + tree2.deep[v] < len) break;
if(tree1.deep[u] + tree2.deep[v] == len) return false;
}
}
return true;
}
int main()
{
cin.sync_with_stdio(false);
cin>>T;
while(T--)
{
ans = 0;
cin>>n>>l;
for(int i = 1;i <= n;i++)
{
cin>>s[i];
int m = strlen(s[i]);
for(int j = 0;j < m;j++) rs[i][j] = s[i][m-1-j];
rs[i][m] = s[i][m];
}
rc[0] = 1;
for(int i = 1;i < (1<<n);i++) rc[i] = rc[i - (i & -i)] * -1;
for(int mask = 0;mask < (1<<n);mask++)
{
tree1.init(),tree2.init();
memset(dp,0,sizeof(dp));
for(int i = 1;i <= n;i++)
if(mask & (1<<(i-1))) tree1.insert(s[i],i),tree2.insert(rs[i],i);
tree1.getFail();
tree2.getFail();
dp[0][0][0] = 1;
for(int i = 0;i < l;i++)
for(int j = 0;j < tree1.tot;j++)
for(int k = 0;k < tree2.tot;k++)
if(dp[i][j][k] && !tree1.val[j] && !tree2.val[k] && !tree1.last[j] && !tree2.last[k])
for(int c = 0;c < 2;c++)
{
int next1 = tree1.ch[j][c],next2 = tree2.ch[k][c^1];
dp[i+1][next1][next2] += dp[i][j][k];
dp[i+1][next1][next2] %= MOD;
}
for(int j = 0;j < tree1.tot;j++)
for(int k = 0;k < tree2.tot;k++)
if(dp[l][j][k] && check(j,k))
{
ans += (MOD + dp[l][j][k]*rc[mask] % MOD) % MOD,ans %= MOD;
if(ans < 0) cout<<233<<endl;
}
}
cout<<ans<<endl;
}
}