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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#638564#9457. Prime Sethos_lyricAC ✓49ms6552kbC++144.8kb2024-10-13 16:14:292024-10-13 18:24:54

Judging History

你现在查看的是测评时间为 2024-10-13 18:24:54 的历史记录

  • [2024-10-13 19:28:32]
  • 自动重测本题所有获得100分的提交记录
  • 测评结果:AC
  • 用时:49ms
  • 内存:6232kb
  • [2024-10-13 19:27:24]
  • hack成功,自动添加数据
  • (/hack/967)
  • [2024-10-13 18:39:23]
  • 管理员手动重测本题所有得分≥97分的提交记录
  • 测评结果:100
  • 用时:49ms
  • 内存:7388kb
  • [2024-10-13 18:24:54]
  • 自动重测本题所有获得100分的提交记录
  • 测评结果:100
  • 用时:49ms
  • 内存:6552kb
  • [2024-10-13 18:24:51]
  • hack成功,自动添加数据
  • (/hack/965)
  • [2024-10-13 18:17:59]
  • 自动重测本题所有获得100分的提交记录
  • 测评结果:100
  • 用时:49ms
  • 内存:7076kb
  • [2024-10-13 18:17:54]
  • hack成功,自动添加数据
  • (/hack/963)
  • [2024-10-13 16:14:32]
  • 评测
  • 测评结果:100
  • 用时:49ms
  • 内存:6248kb
  • [2024-10-13 16:14:29]
  • 提交

answer

#include <cassert>
#include <cmath>
#include <cstdint>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <functional>
#include <iostream>
#include <limits>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <sstream>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>

using namespace std;

using Int = long long;

template <class T1, class T2> ostream &operator<<(ostream &os, const pair<T1, T2> &a) { return os << "(" << a.first << ", " << a.second << ")"; };
template <class T> ostream &operator<<(ostream &os, const vector<T> &as) { const int sz = as.size(); os << "["; for (int i = 0; i < sz; ++i) { if (i >= 256) { os << ", ..."; break; } if (i > 0) { os << ", "; } os << as[i]; } return os << "]"; }
template <class T> void pv(T a, T b) { for (T i = a; i != b; ++i) cerr << *i << " "; cerr << endl; }
template <class T> bool chmin(T &t, const T &f) { if (t > f) { t = f; return true; } return false; }
template <class T> bool chmax(T &t, const T &f) { if (t < f) { t = f; return true; } return false; }
#define COLOR(s) ("\x1b[" s "m")


namespace bm {
constexpr int LIM_N0 = 3010;
constexpr int LIM_N1 = 3010;
int n0, n1;
bitset<LIM_N0> yet0, layers0[LIM_N0 + 1];
bitset<LIM_N1> yet1, layers1[LIM_N0], adj[LIM_N0];
int to[LIM_N0], fr[LIM_N1], tof;
int used[LIM_N0], lev[LIM_N0];
void init(int n0_, int n1_) {
  n0 = n0_; n1 = n1_;
  for (int u = 0; u < n0; ++u) adj[u].reset();
}
void ae(int u, int v) {
  adj[u].set(v);
}
int augment(int u) {
  used[u] = tof;
  const auto sub = adj[u] & layers1[lev[u]];
  for (int v = sub._Find_first(); v < n1; v = sub._Find_next(v)) {
    const int w = fr[v];
    if (!~w || (used[w] != tof && lev[u] < lev[w] && augment(w))) {
      to[u] = v; fr[v] = u; return 1;
    }
  }
  return 0;
}
int run() {
  memset(to, ~0, n0 * sizeof(int));
  memset(fr, ~0, n1 * sizeof(int));
  memset(used, ~0, n0 * sizeof(int));
  for (tof = 0; ; ) {
    yet0.set();
    yet1.set();
    layers0[0].reset();
    for (int u = 0; u < n0; ++u) if (!~to[u]) layers0[0].set(u);
    yet0 ^= layers0[0];
    for (int d = 0; layers0[d].any(); ++d) {
      layers1[d].reset();
      for (int u = layers0[d]._Find_first(); u < n0; u = layers0[d]._Find_next(u)) {
        lev[u] = d;
        layers1[d] |= adj[u];
      }
      layers1[d] &= yet1; yet1 ^= layers1[d];
      layers0[d + 1].reset();
      for (int v = layers1[d]._Find_first(); v < n1; v = layers1[d]._Find_next(v)) {
        if (~fr[v]) layers0[d + 1].set(fr[v]);
      }
      layers0[d + 1] &= yet0; yet0 ^= layers0[d];
    }
    int f = 0;
    for (int u = 0; u < n0; ++u) if (!~to[u]) f += augment(u);
    if (!f) return tof;
    tof += f;
  }
}

// s: true, t: false (s: reachable from unmatched left)
// vertex cover: (0: false, 0: true)
// independent set: (0: true, 1: false)
bool side0[LIM_N0], side1[LIM_N1];
void dfs(int u) {
  if (!side0[u]) {
    side0[u] = true;
    for (int v = adj[u]._Find_first(); v < n1; v = adj[u]._Find_next(v)) {
      if (!side1[v]) {
        side1[v] = true;
        const int w = fr[v];
        if (~w) dfs(w);
      }
    }
  }
}
void minCut() {
  memset(side0, 0, n0 * sizeof(bool));
  memset(side1, 0, n1 * sizeof(bool));
  for (int u = 0; u < n0; ++u) if (!~to[u]) dfs(u);
}
}  // namespace bm


/*
  matching: (1 + 1) or (even + odd = prime)
  M: max matching
  B: max matching as bipartite
  floor(B/2) = M
    (>=)
      {u,v} -> (u,v),(v,u)
      M >= 2B
    (<=)
      graph on [N]
      path: can take >=half of edges
      even cycle: can take half of edges
      odd cycle: must contain 1, connect all odd cycles at 1, can take floor(len/2)
*/

constexpr int LIM = 2'000'010;
bitset<LIM> isp;

int N, K;
vector<int> A;

int main() {
  isp.set();
  isp.reset(0);
  isp.reset(1);
  for (int p = 2; p < LIM; ++p) if (isp[p]) {
    for (int n = 2 * p; n < LIM; n += p) isp.reset(n);
  }
  
  for (int numCases; ~scanf("%d", &numCases); ) { for (int caseId = 1; caseId <= numCases; ++caseId) {
    scanf("%d%d", &N, &K);
    A.resize(N);
    for (int u = 0; u < N; ++u) {
      scanf("%d", &A[u]);
    }
    
    int n = 0;
    bm::init(N, N);
    for (int u = 0; u < N; ++u) {
      bool any = false;
      for (int v = 0; v < N; ++v) if (isp[A[u] + A[v]]) {
        if (u != v) any = true;
        bm::ae(u, v);
      }
      if (any) ++n;
    }
    const int b = bm::run();
    const int m = b / 2;
// cerr<<"n = "<<n<<", b = "<<b<<", m = "<<m<<endl;
    
    int ans = K;
    ans += min(K, m);
    chmin(ans, n);
    printf("%d\n", ans);
  }
#ifndef LOCAL
  break;
#endif
  }
  return 0;
}

这程序好像有点Bug,我给组数据试试?

详细

Test #1:

score: 100
Accepted
time: 9ms
memory: 4236kb

input:

4
4 2
2 3 4 5
5 3
3 4 12 3 6
6 3
1 3 6 8 1 1
1 0
1

output:

4
3
6
0

result:

ok 4 lines

Test #2:

score: 0
Accepted
time: 49ms
memory: 6552kb

input:

110
3 3
2 2 2
3 1
1 1 1
3 2
1 1 1
3 3
1 1 1
4 2
1 1 1 1
10 7
1 1 1 2 4 6 8 10 12 14
18 1
10 5 9 4 10 7 2 4 9 1 6 1 4 9 2 4 9 7
19 5
1 6 1 8 8 1 5 10 9 2 10 2 1 9 6 10 4 3 6
14 4
6 6 8 9 5 3 4 9 2 2 1 10 10 1
15 10
5 4 2 4 10 1 8 4 10 3 10 3 7 10 4
17 5
10 7 2 4 3 1 10 8 2 8 3 4 1 1 1 1 1
20 6
8 4 6 ...

output:

0
2
3
3
4
8
2
10
8
15
10
12
10
10
12
16
10
10
12
16
14
13
13
14
17
0
19
12
12
11
16
10
19
2
12
10
5
14
10
10
13
2
18
2
4
4
11
8
11
8
0
10
10
0
10
0
8
15
12
11
13
18
10
17
9
8
20
19
13
6
4
6
11
9
13
11
6
2
8
12
17
14
6
2
20
2
18
17
6
2
10
0
7
16
2
2
0
2
18
14
11
8
4
6
0
19
890
204
2746
2372

result:

ok 110 lines

Extra Test:

score: 0
Extra Test Passed