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
- ********************/
来源: http://www.bubuko.com/infodetail-3028928.html