D - Avoid K Palindrome

D - Avoid K Palindrome

https://atcoder.jp/contests/abc359/tasks/abc359_d

 

思路

https://atcoder.jp/contests/abc359/submissions/54822869

状压DP

以 K二进制位表示 K字符串（由AB组成）， 判断并记录是否为回文。

dp[i][j]  -- 前i个字符，如果以j（k字符状压表示）结尾，是good string的可能字符串个数。

初始化前K位dp

计算后续dp。

code

https://atcoder.jp/contests/abc359/submissions/55346351

typedef long long ll;
const ll mod = 998244353;ll n, k;
string s;const ll nsize = 1005;
const ll kstatesize = 1<<10 + 5;
bool mirror[kstatesize];ll dp[nsize][kstatesize];/*kseq is represented by bitseq of k bit length10101011111111this function is to check if it is of mirror structuremirror structure has central symmetry, for example10111011
*/
bool checkmirror(ll kseq){/*iterate from 0 positon to half position k>>11111111111111111111111111111l                  r*/for(int i=0; i<(k>>1); i++){ll l = i;ll r = k - i -1;ll lbit = (kseq>>l)&1;ll rbit = (kseq>>r)&1;if (lbit != rbit){return false;}}return true;
}int main()
{cin >> n >> k;cin >> s;for(int i=0; i<(1<<k); i++){mirror[i] = checkmirror(i);}/*initialize the first k seq of dp*/for(int i=0; i<(1<<k); i++){// only non-mirror seq takes effectif (mirror[i]){continue;}/*suppose the first k seq follow the i caseincrease dp*/dp[k-1][i]++;/*then detect if any break with i case,if yes, decrease dp*/for(int j=0; j<k; j++){/*in i case, for each bit,0    --    A1   --  B*/ll jbitpos = k - j - 1;ll jbit = (i>>jbitpos)&1;if (s[j]=='A' && jbit==1){dp[k-1][i]--;break;}if (s[j]=='B' && jbit==0){dp[k-1][i]--;break;}}
//        cout<<dp[k-1][i];
    }/*now iterate from k to n-1 to calculate the following dp*/for(int i=k; i<n; i++){/*iterate each non-mirror cases*/for(int j=0; j<(1<<k); j++){// as of this case j, the previous state i-1 is not a good string// i.e. dp == 0// for the new added char of i index, the new string is still not a good string// so skipif (dp[i-1][j] == 0){continue;}// if the new added char of i index is not A,// the possible value is B or ?// let's make the possible state transferif (s[i]!='A'){// newk is appended by B, and removed the left-most charll newk = ((j<<1)|1)&((1<<k)-1);
//                cout<<newk;// if newk is not mirror, the new string is a good string// newk is not mirror, make state stranferif (!mirror[newk]){dp[i][newk] = (dp[i][newk] + dp[i-1][j]) % mod;}}// if the new added char of i index is not B,// the possible value is A or ?// let's make the possible state transferif (s[i]!='B'){// newk is appended by A, and removed the left-most charll newk = (j<<1)&((1<<k)-1);
//                cout<<newk;// if newk is not mirror, the new string is a good string// newk is not mirror, make state stranferif (!mirror[newk]){dp[i][newk] = (dp[i][newk] + dp[i-1][j]) % mod;}}}
//        for(int j=0;j<(1<<k);j++)cout<<dp[i][j];
//        cout<<'\n';
    }// now we get dp[n-1],// let calculate total number of all statesll ans = 0;for(int i=0; i<(1<<k); i++){ans = (ans + dp[n-1][i]) % mod;
//        cout<<dp[n-1][i];
    }cout << ans << endl;return 0;
}