传送门
题目描述
给定长度为n的串S和长度为2的串T,和一个整数k,
一次操作你可以将S串中任意一个字符修改为其他字符,最多进行k次操作,
问最后S串中最多有多少个子序列T。
分析
我们分类讨论一下第i位字符修改,不修改的情况,令f[i][j][l]表示前i个字母中修改j次,包含有l个op[0]的情况下,答案总数,然后进行状态转移即可
代码
#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
#define _CRT_SECURE_NO_WARNINGS
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef vector<int> VI;
const int INF = 0x3f3f3f3f;
const int N = 220;
const ll mod= 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a){
char c=getchar();T x=0,f=1;while(!isdigit(c)){
if(c=='-')f=-1;c=getchar();}
while(isdigit(c)){
x=(x<<1)+(x<<3)+c-'0';c=getchar();}a=f*x;}
int gcd(int a,int b){
return (b>0)?gcd(b,a%b):a;}
char str[N],op[3];
int f[N][N][N];
int n,k;
int main(){
read(n),read(k);
scanf("%s %s",str,op);
memset(f,-0x3f,sizeof f);
f[0][0][0] = 0;
int s = (op[0] == op[1]);
for(int i = 0;i < n;i++){
int x = (str[i] == op[0]),y = (str[i] == op[1]);
for(int j = 0;j <= k;j++)
for(int p = 0;p <= i;p++){
f[i + 1][j][p + x] = max(f[i + 1][j][p + x],f[i][j][p] + y * p);
f[i + 1][j + 1][p + 1] = max(f[i + 1][j + 1][p + 1],f[i][j][p] + s * p);
f[i + 1][j + 1][p + s] = max(f[i + 1][j + 1][p + s],f[i][j][p] + p);
}
}
int ans = 0;
for(int i = 0;i <= k;i++)
for(int j = 0;j <= n;j++)
ans = max(ans,f[n][i][j]);
di(ans);
return 0;
}
/**
* ┏┓ ┏┓+ +
* ┏┛┻━━━┛┻┓ + +
* ┃ ┃
* ┃ ━ ┃ ++ + + +
* ████━████+
* ◥██◤ ◥██◤ +
* ┃ ┻ ┃
* ┃ ┃ + +
* ┗━┓ ┏━┛
* ┃ ┃ + + + +Code is far away from
* ┃ ┃ + bug with the animal protecting
* ┃ ┗━━━┓ 神兽保佑,代码无bug
* ┃ ┣┓
* ┃ ┏┛
* ┗┓┓┏━┳┓┏┛ + + + +
* ┃┫┫ ┃┫┫
* ┗┻┛ ┗┻┛+ + + +
*/