uvalive 6343

一开始完全看不出这是AC自动机啊。

求长度为T且不包含长度大于K的回文串的二进制串有多少种。

枚举长度为K和K+1的全部回文串,保存在tire里面,然后构造自动机DP

因为长度>k+1的肯定也包含这些长度为K或K+1的。

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<cmath>
#include<climits>
#include<string>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
#define pb(a) push_back(a)
#define INF 0x1f1f1f1f
#define lson idx<<1,l,mid
#define rson idx<<1|1,mid+1,r
#define PI  3.1415926535898
template<class T> T min(const T& a,const T& b,const T& c) {
    return min(min(a,b),min(a,c));
}
template<class T> T max(const T& a,const T& b,const T& c) {
    return max(max(a,b),max(a,c));
}
void debug() {
#ifdef ONLINE_JUDGE
#else

    freopen("d:\in.txt","r",stdin);
   // freopen("d:\out1.txt","w",stdout);
#endif
}
int getch() {
    int ch;
    while((ch=getchar())!=EOF) {
        if(ch!=' '&&ch!='
')return ch;
    }
    return EOF;
}
const int MAX_NODE=1<<11;
const int SIGMA_SIZE=2;
const int mod=1000000007;

int ch[MAX_NODE][SIGMA_SIZE];
int val[MAX_NODE];
int fail[MAX_NODE];
int sz;

int dp[MAX_NODE][444];

void init()
{
    memset(ch[0],0,sizeof(ch[0]));
    val[0]=0;
    sz=1;
}

int idx(char c){return c-'0';}
void insert(const char *s)
{
    int u=0;
    for(int i=0;s[i]!='';i++)
    {
        int v=idx(s[i]);
        if(!ch[u][v])
        {
            memset(ch[sz],0,sizeof(ch[sz]));
            val[sz]=0;
            ch[u][v]=sz++;
        }
        u=ch[u][v];
    }
    val[u]=1;
}

void construct()
{
    fail[0]=0;
    queue<int> q;
    for(int c=0;c<SIGMA_SIZE;c++)
    {
        int u=ch[0][c];
        if(u){fail[u]=0;q.push(u);}
    }
    while(!q.empty())
    {
        int r=q.front();
        q.pop();
        for(int c=0;c<SIGMA_SIZE;c++)
        {
            int u=ch[r][c];
            if(!u)
            {
                ch[r][c]=ch[fail[r]][c];
                continue;
            }
            q.push(u);
            int v=fail[r];
            while(v&&!ch[v][c])v=fail[v];
            fail[u]=ch[v][c];
            val[u]|=val[fail[u]];
        }
    }
}

int f(int u,int n)
{
    if(n==0)return 1;
    if(dp[u][n]>=0)return dp[u][n];

    int &res=dp[u][n];
    res=0;
    for(int c=0;c<SIGMA_SIZE;c++)
    {
        if(!val[ch[u][c]])
            res=(res+f(ch[u][c],n-1))%mod;
    }
    return res;
}

int check(int st,int len,char *s)
{
    int i=0;
    while(i<len)
    {
        s[i++]=st%2+'0';
        st/=2;
    }
    s[len]='';
    for(int i=0,j=len-1;i<j;i++,j--)
    {
        if(s[i]!=s[j])return 0;
    }
    return 1;
}
void prework(int k)
{
    for(int len=k;len<=k+1;len++)
    {
        for(int st=0;st<(1<<len);st++)
        {
            char s[12];
            if(check(st,len,s))
            {
                insert(s);
            }
        }
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    for(int ca=1;ca<=t;ca++)
    {
        init();
        int n,k;
        scanf("%d%d",&n,&k);
        memset(dp,-1,sizeof(dp));
        prework(k);
        construct();
        int num=f(0,n);
        printf("%d
",num);
    }
    return 0;
}
View Code
原文地址:https://www.cnblogs.com/BMan/p/3415177.html