QOJ.ac

QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#130685#5066. String-dle CountPetroTarnavskyiTL 0ms0kbC++172.3kb2023-07-24 19:43:132023-07-24 19:43:16

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-07-24 19:43:16]
  • 评测
  • 测评结果:TL
  • 用时:0ms
  • 内存:0kb
  • [2023-07-24 19:43:13]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;

#define SZ(a) (int)a.size()
#define ALL(a) a.begin(), a.end()
#define FOR(i, a, b) for (int i = (a); i<(b); ++i)
#define RFOR(i, b, a) for (int i = (b)-1; i>=(a); --i)
#define MP make_pair
#define PB push_back
#define F first
#define S second
#define FILL(a, b) memset(a, b, sizeof(a))

typedef long long LL;
typedef pair<int, int> PII;
typedef vector<int> VI;

const int mod = 1e9 + 7;
void upd(int& a, int b){
	a += b;
	if(a >= mod)
		a -= mod;
}


const int ALP = 26;
const int K = 19;

int L[ALP], R[ALP];
bool can[ALP][K];

int dp[2][1 << K][K + 1][K + 1];

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n, k;
	cin >> n >> k;
	FOR(i, 0, ALP){
		FOR(j, 0, k)
			can[i][j] = 1;
		R[i] = k;
		L[i] = 0;
		R[i] = k;
	}
	int ans = 1;
	FOR(i, 0, n){
		string q, a;
		cin >> q >> a;
		VI cnt(ALP);
		VI wasX(ALP);
		FOR(j, 0, k){
			cnt[q[j] - 'A'] += (a[j] != 'x');
			if(a[j] == '-'){
				if(wasX[q[j] - 'A'])
					ans = 0;
			}
			wasX[q[j] - 'A'] |= a[j] == 'x';
		}
		FOR(j, 0, k){
			int pos = q[j] - 'A';
			L[pos] = max(L[pos], cnt[pos]);
			
			if(a[j] == 'x'){
				can[pos][j] = 0;
				if(L[pos] > cnt[pos])
					ans = 0;
				if(R[pos] < cnt[pos])
					ans = 0;
				
				L[pos] = R[pos] = cnt[pos];
				continue;
			}
			
			if(a[j] == '-')
				can[pos][j] = 0;
			else{
				FOR(bad, 0, ALP)
					if(bad != pos)
						can[bad][j] = 0;
			}	
		}
	}
	
	if(ans == 0){
		cout << 0 << "\n";
		return 0;
	}
	dp[0][(1 << k) - 1][0][0] = 1;
	FOR(a, 0, ALP){		
		int t = a % 2;
		int nt = (a + 1) % 2;
		//clear
		//FOR(mask, 0, 1 << k)
		//	FOR(bit, 0, k + 1)
		//		FOR(cnt, 0, k + 1)
		//			dp[nt][mask][bit][cnt] = 0;
		FILL(dp[nt], 0);



		RFOR(mask, 1 << k, 0){
			FOR(bit, 0, k){
				FOR(cnt, 0, k + 1){
					
					upd(dp[t][mask][bit + 1][cnt], dp[t][mask][bit][cnt]);
					
					if((mask & (1 << bit)) == 0)
						continue;
					if(can[a][bit] == 0)
						continue;
					if(cnt + 1 > R[a])
						continue;

					upd(dp[t][mask - (1 << bit)][bit + 1][cnt + 1], dp[t][mask][bit][cnt]);
				}
			}
			FOR(cnt, L[a], R[a] + 1)
				upd(dp[nt][mask][0][0], dp[t][mask][k][cnt]);
		}
	}
	cout << dp[ALP % 2][0][0][0] << endl;
	
	return 0;
}


Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 0
Time Limit Exceeded

input:

2 5
CRANE
xx--x
NASAL
OOxOO

output:


result: