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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#451310#8770. Comparatorhenryx#RE 0ms3616kbC++203.4kb2024-06-23 05:17:332024-06-23 05:17:33

Judging History

你现在查看的是最新测评结果

  • [2024-06-23 05:17:33]
  • 评测
  • 测评结果:RE
  • 用时:0ms
  • 内存:3616kb
  • [2024-06-23 05:17:33]
  • 提交

answer

#include <bits/stdc++.h>
const char nl = '\n';
using namespace std;
using ll = long long;
using ld = long double;
const int K = 10;

int f(char op, int x, int y) {
  if (op == '&') return x & y;
  else if (op == '|') return x | y;
  else if (op == '^') return x ^ y;
  else if (op == '=') return x == y;
  assert(0);
  return 0;
}

const map<char, int> pri = {
  {'!', 5},
  {'=', 4},
  {'&', 3},
  {'|', 2},
  {'^', 1}
};

int eval(const string& s, int x, int y) {
  stack<int> exprs, ops;
  auto resolve_op = [&] {
    if (ops.empty() || ops.top() == '(') return;
    if (ops.top() == '!') {
      ops.pop();
      int e = exprs.top(); exprs.pop();
      exprs.push(!e);
    } else {
      char op = ops.top(); ops.pop();
      /*
      if (!ops.empty()) {
        char op2 = ops.top();
        if (pri.at(op2) >= pri.at(op1))
      }*/
      int e1 = exprs.top(); exprs.pop();
      int e2 = exprs.top(); exprs.pop();
      //cerr << e1 << op << e2 << nl;
      exprs.push(f(op, e1, e2));
    }
  };
  //cerr << "parsing " << s << nl;
  for (char c : s) {
    //cerr << c << nl;
    if (c == 'x') {
      exprs.push(x);
      //resolve_op();
    } else if (c == 'y') {
      exprs.push(y);
      //resolve_op();
    } else if (c == '0') {
      exprs.push(0);
      //resolve_op();
    } else if (c == '1') {
      exprs.push(1);
      //resolve_op();
    } else if (c == '(') {
      ops.push('(');
    } else if (c == ')') {
      while (!ops.empty()) {
        if (ops.top() == '(') {
          ops.pop();
          break;
        }
        resolve_op();
      }
    } else {
      while (!ops.empty() && ops.top() != '(' && pri.at(ops.top()) >= pri.at(c)) {
        resolve_op();
      }
      ops.push(c);
    }
  }
  while (!ops.empty()) resolve_op();
  //cerr << exprs.size() << " " << exprs.top() << nl;
  assert(exprs.size() == 1);
  return exprs.top();
}

char vis[K][K][16];
bitset<1 << K> tt[1 << K];

int main() {
  cin.tie(0)->sync_with_stdio(0);
  int n, k; cin >> n >> k;
  vector<tuple<int, int, int, int>> exprs;
  for (int i = 0; i < n; i++) {
    int a, b; string expr; int r; cin >> a >> b >> expr >> r; a--; b--;
    int val = 0;
    for (int x = 0; x < 2; x++) {
      for (int y = 0; y < 2; y++) {
        val |= (eval(expr, x, y) << (2*x+y));
      }
    }
    //cerr << expr << " -> " << bitset<4>(val) << nl;
    if (vis[a][b][val]) continue;
    vis[a][b][val] = 1;
    if (val != 0) {
      exprs.emplace_back(a, b, val, r);
    }
  }
  int fr; cin >> fr;
  for (int x = 0; x < (1 << k); x++) {
    for (int y = 0; y < (1 << k); y++) {
      bool bad = 1;
      for (auto [a, b, val, r] : exprs) {
        int xi = (x >> a) & 1, yi = (y >> b) & 1;
        if ((val >> (2*xi + yi)) & 1) {
          tt[x][y] = r;
          bad = 0;
          break;
        }
      }
      if (bad) tt[x][y] = fr;
    }
  }
  // reflexive
  int ans = 0;
  for (int x = 0; x < (1 << k); x++) {
    if (tt[x][x]) ans++;
  }
  cout << ans << " ";
  ans = 0;
  for (int x = 0; x < (1 << k); x++) {
    for (int y = 0; y < (1 << k); y++) {
      if (tt[x][y] && tt[y][x]) ans++;
    }
  }
  cout << ans << " ";
  ans = 0;
  for (int x = 0; x < (1 << k); x++) {
    auto nx = ~tt[x];
    for (int y = 0; y < (1 << k); y++) {
      if (!tt[x][y]) continue;
      ans += (nx & tt[y]).count();
    }
  }
  cout << ans << nl;
}

Details

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Test #1:

score: 100
Accepted
time: 0ms
memory: 3616kb

input:

3 2
1 1 (x=0)&(y=1) 1
1 1 (x=1)&(y=(x^x)) 0
2 2 (x=1)|(y=0) 0
1

output:

0 0 0

result:

ok single line: '0 0 0'

Test #2:

score: 0
Accepted
time: 0ms
memory: 3540kb

input:

4 3
2 1 x=0&(y=1) 1
1 2 !x^!!y 0
2 3 ((x|1)=y)&1&1 1
3 1 !x&!x&!x 0
1

output:

3 25 52

result:

ok single line: '3 25 52'

Test #3:

score: -100
Runtime Error

input:

1413 3
1 3 0 0
3 3 !x 0
2 2 x=0 1
1 2 !y^0 1
2 3 (x^1) 0
3 2 ((!0)) 1
1 1 !!1=(y) 0
2 2 !(1^x)&y 1
3 2 (y)&1|!!1 0
3 1 !x=(y&y=y) 0
2 1 (((!1)^!x)) 1
2 3 !0=(0&y)=1&y 0
1 2 ((((!0)))|!1) 0
3 1 !(y=!1=x|(!x)) 0
1 1 ((((y=!y)))&!0) 0
2 3 ((y=1)^!1^!!1|0) 1
2 3 1|(!x)&!x|1|(x=1) 1
2 3 !0^!!!!y&0=(!1&!0...

output:


result: