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#259926#7749. A Simple MST Problemucup-team1508WA 303ms148416kbC++203.3kb2023-11-21 16:17:042023-11-21 16:17:05

Judging History

你现在查看的是最新测评结果

  • [2023-11-21 16:17:05]
  • 评测
  • 测评结果:WA
  • 用时:303ms
  • 内存:148416kb
  • [2023-11-21 16:17:04]
  • 提交

answer

#include<bits/stdc++.h>
#define F(i, a, b) for(int i = a; i <= b; i ++)
#define Fd(i, a, b) for(int i = a; i >= b; i --)
#define int long long
#define pii pair<int, int>
#define fi first
#define se second
#define pb push_back
using namespace std;
const int N = 1e6 + 10;
const int inf = 1e18;
typedef array<int, 3> a3;
int L, R, l[N], r[N], p[N], pr[N], idx = 0, is_prime[N], dis[N], vst[N], cnt[N], fa[N];
int mp1[N], mp2[N];
vector<pii> ed[N];
vector<int> a[N];
vector<a3> q;
int st;
int find(int x) {return x == fa[x] ? x : find(fa[x]);}
bool merge(int x, int y) {
    int p = find(x), q = find(y);
    if(p == q) return false;
    fa[p] = q;
    return true;
}
void init() {
    int MAX = 1e6;
    p[1] = 1;
    F(i, 2, MAX) {
        if(! p[i]) pr[++ idx] = i, p[i] = i, is_prime[i] = 1, a[i].pb(i);
        for(int j = 1; j <= idx && pr[j] * i <= MAX; j ++) {
            p[i * pr[j]] = pr[j];
            a[i * pr[j]] = a[i];
            if(p[i] == pr[j]) break;
            else a[i * pr[j]].pb(pr[j]);
        }
    }
    F(i, 1, MAX) {
        l[i] = 0;
        int sz = a[i].size();
        F(j, 0, (1 << sz) - 1) {
            int tmp = 1;
            F(k, 0, sz - 1) if((j >> k) & 1) tmp *= a[i][k];
            l[i] = max(l[i], mp1[tmp]);
        }
        int now = 1;
        for(auto x : a[i]) now *= x;
        mp1[now] = i;
    }
    F(i, 1, MAX) mp2[i] = MAX + 1;
    Fd(i, MAX, 1) {
        r[i] = MAX + 1;
        int sz = a[i].size();
        F(j, 0, (1 << sz) - 1) {
            int tmp = 1;
            F(k, 0, sz - 1) if((j >> k) & 1) tmp *= a[i][k];
            r[i] = min(r[i], mp2[tmp]);
        }
        int now = 1;
        for(auto x : a[i]) now *= x;
        mp2[now] = i;        
    }
}
int prim(int l, int r) {
    int res = 0;
    F(i, l, r) dis[i] = inf, vst[i] = 0, ed[i].clear();
    dis[l] = 0;
    F(i, l, r) F(j, i + 1, r) {
        int w = 0;
        for(auto x : a[i]) w += ! cnt[x] ++;
        for(auto x : a[j]) w += ! cnt[x] ++;
        for(auto x : a[i]) cnt[x] --;
        for(auto x : a[j]) cnt[x] --;
        ed[i].pb({j, w}), ed[j].pb({i, w});
    } 
    F(i, l, r) {
        int u = 0, now = inf;
        F(j, l, r) if(! vst[j] && dis[j] < now) u = j, now = dis[j];
        vst[u] = 1, res += now;
        for(auto [v, w] : ed[u]) dis[v] = min(dis[v], w);
    }
    return res;
}
void sol() {
    cin >> L >> R;
    int flag = 0;
    F(i, L, R) if(is_prime[i]) flag = i;
    if(! flag) {
        cout << prim(L, R) << "\n";
        return;
    }
    q.clear();
    F(i, L, R) {
        if(l[i] >= L) q.pb({(int)a[i].size(), i, l[i]});
        if(r[i] <= R) q.pb({(int)a[i].size(), i, r[i]});
        int w = a[i].size() + 1;
        if(is_prime[i]) {
            F(j, L, R) if(j != i) {
                q.pb({w, i, j});
                break;
            }
        } else q.pb({w, i, flag});
    }
    F(i, L, R) fa[i] = i;
    int res = 0, tot = 0;
    sort(q.begin(), q.end());
    for(auto [w, u, v] : q) {
        if(! merge(u, v)) continue;
        res += w;
        if(++ tot == R - L) break;
    }
    cout << res << "\n";
}
signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    init();
    int t; cin >> t;
    F(i, 1, t) sol();
    return 0;
}

詳細信息

Test #1:

score: 100
Accepted
time: 199ms
memory: 147144kb

input:

5
1 1
4 5
1 4
1 9
19 810

output:

0
2
3
9
1812

result:

ok 5 lines

Test #2:

score: -100
Wrong Answer
time: 303ms
memory: 148416kb

input:

2
27 30
183704 252609

output:

8
223091

result:

wrong answer 2nd lines differ - expected: '223092', found: '223091'