思路：dp

dp[i][j][k]：a的前i个和b的前j个和c的前k个能构成的最前面的位置

删字符时状态不用改变，加字符时只会改变1*250*250个状态

代码：

#pragma GCC optimize(2)#pragma GCC optimize(3)#pragma GCC optimize(4)#include
     
      using namespace std;#define y1 y11#define fi first#define se second#define pi acos(-1.0)#define LL long long#define ls rt<<1, l, m#define rs rt<<1|1, m+1, r//#define mp make_pair#define pb push_back#define ULL unsigned LL#define pll pair
      
       #define pli pair
       
        #define pii pair
        
         #define piii pair
         
          #define pdi pair
          
           #define pdd pair
           
            #define mem(a, b) memset(a, b, sizeof(a))#define debug(x) cerr << #x << " = " << x << "\n";#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);//headconst int N = 1e5 + 5;char s[N];int nxt[N][26], n, q, x;char op[10], cc[10];int dp[255][255][255];string a, b, c;int main() { scanf("%d %d", &n, &q); scanf("%s", s+1); for (int i = 0; i < 26; ++i) nxt[n+1][i] = n+1, nxt[n+2][i] = n+1; for (int i = n; i >= 1; --i) { for (int j = 0; j < 26; ++j) { if(j == s[i]-'a') nxt[i][j] = i; else nxt[i][j] = nxt[i+1][j]; } } while(q--) { scanf("%s %d", op, &x); if(op[0] == '+') { scanf("%s", cc); if(x == 1){ a.push_back(cc[0]); for (int i = 0; i <= b.size(); ++i) { for (int j = 0; j <= c.size(); ++j) { dp[a.size()][i][j] = n+1; if(a.size() > 0) dp[a.size()][i][j] = min(dp[a.size()][i][j], nxt[dp[a.size()-1][i][j]+1][a.back()-'a']); if(i > 0) dp[a.size()][i][j] = min(dp[a.size()][i][j], nxt[dp[a.size()][i-1][j]+1][b[i-1]-'a']); if(j > 0) dp[a.size()][i][j] = min(dp[a.size()][i][j], nxt[dp[a.size()][i][j-1]+1][c[j-1]-'a']); } } } else if(x == 2) { b.push_back(cc[0]); for (int i = 0; i <= a.size(); ++i) { for (int j = 0; j <= c.size(); ++j) { dp[i][b.size()][j] = n+1; if(b.size() > 0) dp[i][b.size()][j] = min(dp[i][b.size()][j], nxt[dp[i][b.size()-1][j]+1][b.back()-'a']); if(i > 0) dp[i][b.size()][j] = min(dp[i][b.size()][j], nxt[dp[i-1][b.size()][j]+1][a[i-1]-'a']); if(j > 0) dp[i][b.size()][j] = min(dp[i][b.size()][j], nxt[dp[i][b.size()][j-1]+1][c[j-1]-'a']); } } } else if(x == 3) { c.push_back(cc[0]); for (int i = 0; i <= a.size(); ++i) { for (int j = 0; j <= b.size(); ++j) { dp[i][j][c.size()] = n+1; if(c.size() > 0) dp[i][j][c.size()] = min(dp[i][j][c.size()], nxt[dp[i][j][c.size()-1]+1][c.back()-'a']); if(i > 0) dp[i][j][c.size()] = min(dp[i][j][c.size()], nxt[dp[i-1][j][c.size()]+1][a[i-1]-'a']); if(j > 0) dp[i][j][c.size()] = min(dp[i][j][c.size()], nxt[dp[i][j-1][c.size()]+1][b[j-1]-'a']); } } } } else { if(x == 1) a.pop_back(); else if(x == 2) b.pop_back(); else if(x == 3) c.pop_back(); } if(dp[a.size()][b.size()][c.size()] > n) printf("NO\n"); else printf("YES\n"); } return 0;}