大意: 给定串$s,t$, 给定整数$x$, 求判断$t$是否能划分为至多$x$段, 使这些段在$s$中按顺序,不交叉的出现.
设$dp_{i,j}$表示$s$匹配到$i$位, 划分了$j$段, 匹配到$t$中的最大位置
每次取一个极长的lcp转移即可, lcp可以二分哈希或者用后缀数组+RMQ求
#include #include #include #include #include #include #include #include #include #include #include #include #include #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl '\n' #define DB(a) ({REP(__i,1,n) cout< pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;} //head const int N = 1e6+50; int n,m,x; char s[N],t[N]; int dp[N][40]; void chkmax(int &a, int b) {a>1]+1; REP(j,1,19) for (int i=0;i+(1<k) y[++p]=sa[i]-k; for(i=1;i<=m;i++) c[i]=0; for(i=1;i<=n;i++) c[x[y[i]]]++; for(i=1;i<=m;i++) c[i]+=c[i-1]; for(i=n;i;i--) sa[c[x[y[i]]]--]=y[i]; swap(x,y); x[sa[1]]=1; p=1; for(i=2;i<=n;i++) x[sa[i]]=(y[sa[i-1]]==y[sa[i]] && y[sa[i-1]+k]==y[sa[i]+k])?p:++p; if(p==n) break; m=p; } for(i=1;i<=n;i++) rk[sa[i]]=i; for(int i=1,j,k=0;i<=n;i++) { if(k) k--; j=sa[rk[i]-1]; while(a[i+k]==a[j+k]) k++; h[rk[i]] = k; } } int lcp(int x, int y) { x = rk[x], y = rk[y]; if (x>y) swap(x,y); return RMQ(x+1,y); } int a[N]; int main() { scanf("%d%s%d%s%d",&n,s+1,&m,t+1,&x); REP(i,1,n) a[i]=s[i]-'a'+1; a[n+1]=30; REP(i,n+2,n+m+1) a[i]=t[i-n-1]-'a'+1; build(a,n+m+1,100); init(h,n+m+1); //dp[i][j] = s的前i位,t中分j段的最长匹配位置 //dp[i][j] <- dp[i-1][j] //dp[i+lcp(i,dp[i-1][j]+1)-1][j+1] <- dp[i-1][j] REP(i,1,n) REP(j,0,x) { int &r = dp[i-1][j]; chkmax(dp[i][j],r); if (r!=m) { int t = lcp(i,n+r+2); chkmax(dp[i+t-1][j+1],r+t); } } REP(i,0,x) if (dp[n][i]==m) return puts("YES"),0; puts("NO"); }
