[C++] 20340번 Lost Map

https://www.acmicpc.net/problem/20340

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26/02/16

 

 

FFT를 이용한 문자열 매칭 문제이다.

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문제 접근 방식:

 

 

이전에 14958번 Rock Paper Scissors문제를 푼 적 있었는데 그거랑 매우 유사하고, 그거랑 비교했을 때 약간 한번 더 생각하면 풀리는 문제이다.

일단 경우의 수가 4개가 있다.

1. 두 문자 모두 ?가 아니고 서로 같음
2. 두 문자 모두 ?가 아니고 서로 다름
3. 두 문자 중 하나가 ?임
4. 두 문자 모두 ?임

우리가 구하고자 하는 것은 1번 + 3번 + 4번 경우이다.

즉, 1번 + 3번 + 4번 경우가 $m$이라면 문자열이 매칭된다고 볼 수 있다.

다른 말로, 2번 경우가 $0$이라면 문자열이 매칭된다고 볼 수 있다.

'?'일때는 $0$, '?'가 아닐땐 $1$로 배열 $A$를 만들어주고, $A$끼리 곱하면 1번 + 2번 경우의 수를 구할 수 있다.

그리고 14958번에서 했던 것처럼 나머지 $28$개의 문자들(n, e, s, w 1~7)에 대해 $0/1$배열 만들어주고, 각 배열끼리 FFT를 돌려서 모두 더해주면 1번 경우의 수를 구할 수 있다.

이제 위에서 아래를 빼면 2번 경우의 수를 쉽게 구할 수 있다.

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아래는 내가 위의 접근 방식과 같이 작성한 C++ 코드이다. 더보기를 누르면 확인할 수 있다.

// 20340번 Lost Map
// NTT
/*
접근 방법:
14958번이랑 동일하게 접근
*/
#include <iostream>
#include <vector>

using namespace std;
#define fastio ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl '\n'

/*
Notes:
MOD = 998'244'353 = 119*2^23 + 1
MOD = 1'004'535'809 = 479*2^21 + 1 / G = 3 / Works for lengths up to 2^21. / Use with CRT and 998'244'353
MOD = 469'762'049 = 7*2^26 + 1 / G = 3 / Works for lengths up to 2^26.
MOD = 167'772'161 = 5*2^25 + 1 / G = 3 / Works for lengths up to 2^25.
MOD = 1'224'736'769 = 73*2^24 + 1 / G = 3 / Works for lengths up to 2^24.
Combine above 3-NTT primes with CRT -> 75bits fast product ok.
*/
template<int MOD, int G>
struct NTT {
private :
    // HELPER FUNCTION
    inline int mul_mod(int a, int b) const {return (long long)a * b % MOD;}
    inline int pow_mod(int a, int e) const {
        int r = 1;
        while(e > 0){
            if(e & 1) r = mul_mod(r, a);
            a = mul_mod(a, a);
            e >>= 1;
        }
        return r;
    }
    inline int mod_inv(int a) const {return pow_mod(a, MOD-2);}
public : 
    void iterative_NTT(vector<int>& a, int invert){
        int n = (int)a.size();
        // 1) bit-reversal permutation
        for (int i = 1, j = 0; i < n; i++){
            int bit = n >> 1;
            while (j & bit){
                j ^= bit;
                bit >>= 1;
            }
            j ^= bit;
            if (i < j) swap(a[i], a[j]);
        }
        // 2) butterflies by length = 2, 4, 8, ...
        for (int len = 2; len <= n; len <<= 1){
            // primitive len-th root of unity
            int w_len = pow_mod(G, (MOD - 1) / len);
            if (invert) w_len = mod_inv(w_len);
            for (int i = 0; i < n; i += len){
                int w = 1;
                for (int j = 0; j < len / 2; j++){
                    int u = a[i+j];
                    int v = mul_mod(w, a[i+j+len/2]);
                    int x = u+v; if (x >= MOD) x -= MOD;
                    int y = u-v; if (y < 0) y += MOD;
                    a[i+j] = x;
                    a[i+j+len/2] = y;
                    w = mul_mod(w, w_len);
                }
            }
        }
        // 3) Divide by n (multiply by inverse) for inverse transform
        if (invert){
            int inv_n = mod_inv(n);
            for (int i = 0; i < n; ++i){
                a[i] = mul_mod(a[i], inv_n);
            }
        }
    }
    // Input : Coefficient vector {a_0, a_1, ...}, {b_0, b_1, ...}
    // Output : Convolution of two coefficient vector
    vector<int> convolution(vector<int> a, vector<int> b) {
        if (a.empty() || b.empty()) return {};
        int S_a = (int)a.size(), S_b = (int)b.size();
        // Make vector size easy to dnc(2^n).
        int n = 1;
        while (n < S_a + S_b - 1){
            n <<= 1;
        }
        a.resize(n); b.resize(n);
        // Normalize to [0, MOD)
        for (int i = 0; i < n; ++i){
            a[i] %= MOD; if (a[i] < 0) a[i] += MOD;
            b[i] %= MOD; if (b[i] < 0) b[i] += MOD;
        }
        // NTT
        iterative_NTT(a, 0); iterative_NTT(b, 0);
        // Pointwise product
        for (int i = 0; i < n; i++) {
            a[i] = mul_mod(a[i], b[i]);
        }
        // INTT
        iterative_NTT(a, 1);
        a.resize(S_a + S_b - 1);
        return a;
    }
};

int main(void){
    fastio

    int n, m; cin >> n >> m;
    vector<vector<int>> firstMap(28, vector<int>(n, 0));
    vector<int> firstMapReveal(n, 1);
    vector<vector<int>> secondMap(28, vector<int>(m, 0));
    vector<int> secondMapReveal(m, 1);
    char c; int step;
    for (int i = 0; i < n; ++i){
        cin >> c;
        switch (c) {
            case 'n':
                cin >> step; firstMap[step-1][i] = 1; break;
            case 's':
                cin >> step; firstMap[7+step-1][i] = 1; break;
            case 'e':
                cin >> step; firstMap[14+step-1][i] = 1; break;
            case 'w':
                cin >> step; firstMap[21+step-1][i] = 1; break;
            case '?':
                firstMapReveal[i] = 0; break;
        }
    }
    for (int i = 0; i < m; ++i){
        cin >> c;
        switch (c) {
            case 'n':
                cin >> step; secondMap[step-1][m-i-1] = 1; break;
            case 's':
                cin >> step; secondMap[7+step-1][m-i-1] = 1; break;
            case 'e':
                cin >> step; secondMap[14+step-1][m-i-1] = 1; break;
            case 'w':
                cin >> step; secondMap[21+step-1][m-i-1] = 1; break;
            case '?':
                secondMapReveal[m-i-1] = 0; break;
        }
    }
    NTT<998'244'353, 3> ntt;
    auto bothReveal = ntt.convolution(firstMapReveal, secondMapReveal);
    vector<vector<int>> revealedAndMatch(28);
    for (int i = 0; i < 28; ++i){
        revealedAndMatch[i] = ntt.convolution(firstMap[i], secondMap[i]);
    }
    vector<int> revealAndMatch(n+m-1, 0);
    for (int i = 0; i < n+m-1; ++i){
        for (int j = 0; j < 28; ++j){
            revealAndMatch[i] += revealedAndMatch[j][i];
        }
    }
    vector<int> conflict(n+m-1);
    for (int i = 0; i < n+m-1; ++i){
        conflict[i] = bothReveal[i] - revealAndMatch[i];
    }
    int ans = 0;
    for (int i = m-1; i < n; ++i){
        if (conflict[i] == 0) ++ans;
    }
    cout << ans;
    return 0;
}