제출 #1304806

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1304806		anhkiet	Snake Escaping (JOI18_snake_escaping)	C++20	75 / 100	362 ms	110940 KiB

snake_escaping

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 5e4;

#define ll long long
string s;
int l, q;
int n, ans;
char b[maxn];
int c[maxn];
string qry[maxn];

void xuly() {
    for (int i = 1; i <= l; i++) {
        if (c[i] != (b[i] - '0') && b[i] != '?') return;
    }
    int cur = 0;
    int t = 0;
    for (int i = l; i >= 1; i--) {
        if (c[i] == 1) t += (1 << cur);
        cur++;
    }
    ans += (s[t] - '0');
}

void backtrack(int i) {
    for (int j = 0; j <= 1; j++) {
        c[i] = j;
        if (i == l) xuly();
        else backtrack(i + 1);
    }
}

namespace subtask1 {
    void solve() {
        for (int i = 1; i <= q; i++) {
            for (int j = 1; j <= l; j++) b[j] = qry[i][j - 1];
            backtrack(1);
            cout << ans << '\n';
            ans = 0;
        }
    }
}

// phần sửa: an toàn cấp phát theo S và kiểu phù hợp
namespace subtaskbip {
    void solve() {
        int S = (1 << l); // beware: đảm bảo l <= 30 trong problem constraints
        vector<long long> base(S), dp1(S), dp0(S);

        // fill base
        for (int mask = 0; mask < S; ++mask) {
            base[mask] = s[mask] - '0';
            dp1[mask] = base[mask];
            // mirror into dp0: dp0[(S-1)^mask] = base[mask]
            dp0[(S - 1) ^ mask] = base[mask];
        }

        // subset-sum (zeta) for dp1 and dp0
        for (int i = 0; i < l; ++i) {
            for (int mask = 0; mask < S; ++mask) {
                if (mask & (1 << i)) {
                    dp1[mask] += dp1[mask ^ (1 << i)];
                    dp0[mask] += dp0[mask ^ (1 << i)];
                }
            }
        }

        for (int qi = 1; qi <= q; ++qi) {
            string st = qry[qi];
            reverse(st.begin(), st.end());
            int x = 0, y = 0, z = 0;
            int d0 = 0, d1 = 0, d2 = 0;
            for (int j = 0; j < l; ++j) {
                if (st[j] == '0') { x |= (1 << j); d0++; }
                else if (st[j] == '1') { y |= (1 << j); d1++; }
                else { z |= (1 << j); d2++; }
            }

            long long ans = 0;
            if (d2 <= 6) {
                // enumerate subsets of z
                for (int mask = z; ; mask = (mask - 1) & z) {
                    ans += base[mask | y];
                    if (mask == 0) break;
                }
            }
            else if (d1 <= 6) {
                for (int mask = y; ; mask = (mask - 1) & y) {
                    if (__builtin_popcount(mask ^ y) & 1) ans -= dp1[mask | z];
                    else ans += dp1[mask | z];
                    if (mask == 0) break;
                }
            }
            else {
                for (int mask = x; ; mask = (mask - 1) & x) {
                    if (__builtin_popcount(mask ^ x) & 1) ans -= dp0[mask | z];
                    else ans += dp0[mask | z];
                    if (mask == 0) break;
                }
            }
            cout << ans << '\n';
        }
    }
}


signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    #define NAME "test"
    if (fopen(NAME".INP", "r")) {
        freopen(NAME".INP", "r",stdin);
        freopen(NAME".OUT", "w",stdout);
    }
    cin >> l >> q;
    cin >> s;
    for (int i = 1; i <= q; i++) cin >> qry[i];
    subtaskbip::solve();
}

컴파일 시 표준 에러 (stderr) 메시지

snake_escaping.cpp: In function 'int main()':
snake_escaping.cpp:113:16: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
  113 |         freopen(NAME".INP", "r",stdin);
      |         ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~
snake_escaping.cpp:114:16: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
  114 |         freopen(NAME".OUT", "w",stdout);
      |         ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~

• Subtask 1: 5 / 5
• Subtask 2: 7 / 7
• Subtask 3: 10 / 10
• Subtask 4: 53 / 53
• Subtask 5: 0 / 25