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