大意: 给定字符串$s$, $p$, 对于$0le xle |s|$, 求$s$删除$x$个字符后, $p$在$s$中的最大出现次数.
显然答案是先递增后递减的, 那么问题就转化求最大出现次数为$y$时, 求$S$所需要删除的最少字符数.
先暴力O(n^2)求出以每个字符开头匹配完一个$p$后的最小右端点, 然后$dp$即可.
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <math.h> #include <set> #include <map> #include <queue> #include <string> #include <string.h> #include <bitset> #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 ' ' #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, P2 = 998244353, 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 = 2e3+10; int n, m; char s[N], p[N]; int nxt[N], ans[N], dp[N][N]; int main() { scanf("%s%s", s+1, p+1); n = strlen(s+1), m = strlen(p+1); REP(i,1,n) { int now = 1; nxt[i] = N-1; REP(j,i,n) { if (s[j]==p[now]) ++now; if (now>m) {nxt[i]=j;break;} } } memset(dp,0x3f,sizeof dp); REP(i,0,n) dp[i][0] = 0; REP(i,1,n) { REP(j,1,n) dp[i][j] = min(dp[i][j], dp[i-1][j]); REP(j,1,n) if (dp[i-1][j-1]!=INF&&nxt[i]<=n) { dp[nxt[i]][j] = min(dp[nxt[i]][j], dp[i-1][j-1]+nxt[i]-i+1-m); } } REP(i,1,n) if (dp[n][i]!=INF) ans[dp[n][i]] = i; REP(i,1,n) ans[i] = max(ans[i], ans[i-1]); REP(i,1,n) ans[i] = min(ans[i], (n-i)/m); REP(i,0,n) printf("%d ", ans[i]);hr; }