• Dreamoon and Strings CodeForces


    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;
    }
    
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  • 原文地址:https://www.cnblogs.com/uid001/p/10919801.html
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