正文
Maximum Questions CodeForces - 900E (字符串,dp)
提示：扫一扫查出行【扫一扫了解最新限行尾号】
复制提示
大意:给定长$n$的字符串$s$, 只含'a','b','?', '?'可以替换为任意字符, 在给定长$t$的字符串, "ababab...", 求替换尽量少的'?', 使得$s$能匹配最多的不相交的$t$.
先不考虑最少替换的限制, 要尽量多的匹配$t$, 可以先预处理出可以匹配的位置, 然后$dp$.
要求最小的话, 每次$dp$转移时可能有多个转移点, 对每个dp值维护一个前缀最小值即可.
#include <iostream>
#include <iostream>
#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 '\n'
#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, 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#ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 111;
#endifint n, m;
char s[N];
int a[2][N], b[2][N], f[N], dp[N], w[N], c[N];
void init() {
REP(i,1,n) {
a[0][i]=a[0][i-1],a[1][i]=a[1][i-1];
b[0][i]=b[0][i-1],b[1][i]=b[1][i-1];
if (s[i]=='a') a[i&1][i]=max(a[i&1][i],i);
else if (s[i]=='b') b[i&1][i]=max(b[i&1][i],i);
w[i]=w[i-1]+(s[i]=='?');
}
}int main() {
scanf("%d%s%d", &n, s+1, &m);
init();
REP(i,m,n) {
if (m&1) f[i]=b[i&1][i]<=i-m&&a[i&1^1][i]<=i-m;
else f[i]=a[i&1][i]<=i-m&&b[i&1^1][i]<=i-m;
}
REP(i,m,n) {
if (f[i]) {
dp[i] = dp[i-m]+1;
c[i] = c[i-m]+w[i]-w[i-m];
if (dp[i]==dp[i-1]) c[i]=min(c[i],c[i-1]);
}
else c[i]=c[i-1];
dp[i] = max(dp[i-1], dp[i]);
}
int ans = INF;
REP(i,1,n) if (dp[i]==dp[n]) ans=min(ans,c[i]);
printf("%d\n", ans);
}
