Three Religions CodeForces

大意: 给定字符串S, 要求维护三个串, 支持在每个串末尾添加或删除字符, 询问S是否能找到三个不相交的子序列等于三个串.

暴力DP, 若不考虑动态维护的话, 可以直接$O(len^3)$处理出最少需要S中前多少位能匹配.

考虑添加删除的话, DP刷表, $O(len^2q)$

#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 '
'
#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



const int N = 1e5+10, M = 300;
int n, q;
char s[N], a[4][M];
int dp[M][M][M], nxt[N][27], len[4];
void upd(int x) {
	REP(i,x==1?len[1]:0,len[1]) {
		REP(j,x==2?len[2]:0,len[2]) {
			REP(k,x==3?len[3]:0,len[3]) {
				int &r = dp[i][j][k] = n+2;
				if (i&&dp[i-1][j][k]+1<=n) r=min(r,nxt[dp[i-1][j][k]+1][a[1][i]-'a']);
				if (j&&dp[i][j-1][k]+1<=n) r=min(r,nxt[dp[i][j-1][k]+1][a[2][j]-'a']);
				if (k&&dp[i][j][k-1]+1<=n) r=min(r,nxt[dp[i][j][k-1]+1][a[3][k]-'a']);
			}
		}
	}
}

int main() {
	scanf("%d%d%s", &n, &q, s+1);
	REP(i,'a','z') {
		nxt[n+1][i-'a']=n+2;
		PER(j,1,n) { 
			if (s[j]==i) nxt[j][i-'a']=j;
			else nxt[j][i-'a']=nxt[j+1][i-'a'];
		}
	}
	REP(i,1,q) {
		char op, c;
		int x;
		scanf(" %c%d", &op, &x);
		if (op=='+') {
			scanf(" %c", &c);
			a[x][++len[x]]=c;
			upd(x);
		}
		else --len[x];
		puts(dp[len[1]][len[2]][len[3]]<=n?"YES":"NO");
	}
}