题意:给你两个串s,p,问你把s分开顺序不变,能不能用最多k段合成p.

题解:dp[i][j]表示s到了前i项,用了j段的最多能合成p的前缀是哪里,那么转移就是两种,\(dp[i+1][j]=dp[i][j],dp[i+lcp][j+1]=dp[i][j]+lcp\),这里的lcp是dp[i][j]和i的lcp,然后sa预处理一下st表就行了

//#pragma GCC optimize(2)

//#pragma GCC optimize(3)

//#pragma GCC optimize(4)

//#pragma GCC optimize("unroll-loops")

//#pragma comment(linker, "/stack:200000000")

//#pragma GCC optimize("Ofast,no-stack-protector")

//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#include<bits/stdc++.h>

#define fi first

#define se second

#define db double

#define mp make_pair

#define pb push_back

#define pi acos(-1.0)

#define ll long long

#define vi vector<int>

#define mod 1000000009

#define ld long double

//#define C 0.5772156649

//#define ls l,m,rt<<1

//#define rs m+1,r,rt<<1|1

#define pll pair<ll,ll>

#define pil pair<int,ll>

#define pli pair<ll,int>

#define pii pair<int,int>

#define ull unsigned long long

//#define base 1000000000000000000

#define fin freopen("a.txt","r",stdin)

#define fout freopen("a.txt","w",stdout)

#define fio ios::sync_with_stdio(false);cin.tie(0)

inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}

inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}

template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}

template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}

inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}

inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const ull ba=233;

const db eps=1e-5;

const ll INF=0x3f3f3f3f3f3f3f3f;

const int N=200000+10,maxn=1000000+10,inf=0x3f3f3f3f;

char s[N],p[N];

int sa[N], t[N], t2[N], c[N], rk[N], height[N];

void buildSa(int n, int m) {

    int i, j = 0, k = 0, *x = t, *y = t2;

    for(i = 0; i < m; i++) c[i] = 0;

    for(i = 0; i < n; i++) c[x[i] = s[i]]++;

    for(i = 1; i < m; i++) c[i] += c[i - 1];

    for(i = n - 1; i >= 0; i--) sa[--c[x[i]]] = i;

    for(int k = 1; k < n; k <<= 1) {

        int p = 0;

        for(i = n - k; i < n; i++) y[p++] = i;

        for(i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i] - k;

        for(i = 0; i < m; i++) c[i] = 0;

        for(i = 0; i < n; i++) c[x[y[i]]]++;

        for(i = 1; i < m; i++) c[i] += c[i - 1];

        for(i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];

        swap(x, y);

        p = 1; x[sa[0]] = 0;

        for(int i = 1; i < n; i++) {

            if(y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + k] == y[sa[i] + k])

                x[sa[i]] = p - 1;

            else x[sa[i]] = p++;

        }

        if(p >= n) break;

        m = p;

     }

     for(i = 1; i < n; i++) rk[sa[i]] = i;

     for(i = 0; i < n - 1; i++) {

        if(k) k--;

        j = sa[rk[i] - 1];

        while(s[i + k] == s[j + k]) k++;

        height[rk[i]] = k;

     }

}

int Log[N];

struct ST {

    int dp[N][20],ty;

    void build(int n, int b[], int _ty) {

        ty = _ty;

        for(int i = 1; i <= n; i++) dp[i][0] = ty * b[i];

        for(int j = 1; j <= Log[n]; j++)

            for(int i = 1; i+(1<<j)-1 <= n; i++)

                dp[i][j] = max(dp[i][j-1], dp[i+(1<<(j-1))][j-1]);

    }

    int query(int x, int y) {

        int k = Log[y - x + 1];

        return ty * max(dp[x][k], dp[y-(1<<k)+1][k]);

    }

}st;

int n,m,x;

int lcp(int x,int y)

{

    x=rk[x],y=rk[y];

    if(x>y)swap(x,y);x++;

    return st.query(x,y);

}

int dp[100010][33];

int main()

{

    for(int i = -(Log[0]=-1); i < N; i++)

        Log[i] = Log[i - 1] + ((i & (i - 1)) == 0);

    scanf("%d%s%d%s%d",&n,s,&m,p,&x);

    s[n]='z'+1;

    for(int i=n+1;i<n+1+m;i++)s[i]=p[i-n-1];

    buildSa(n+m+2,258);

    st.build(n+m+1,height,-1);

    dp[0][0]=0;

    for(int i=0;i<=n;i++)

    {

        for(int j=0;j<=x;j++)

        {

            if(dp[i][j]==m)return 0*puts("YES");

            dp[i+1][j]=max(dp[i+1][j],dp[i][j]);

            if(j<x)

            {

                int lc=lcp(i,n+1+dp[i][j]);

                dp[i+lc][j+1]=max(dp[i+lc][j+1],dp[i][j]+lc);

            }

        }

    }

    puts("NO");

    return 0;

}

/********************

9

hloyaygrt

6

loyyrt

3

********************/