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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
#97753#6192. Interval Problemcardinal_cityRE 3ms4408kbC++203.3kb2023-04-18 07:50:422023-04-18 07:50:46

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-04-18 07:50:46]
  • 评测
  • 测评结果:RE
  • 用时:3ms
  • 内存:4408kb
  • [2023-04-18 07:50:42]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
 
typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector< vector <int> > vvi;
typedef pair<int, int> pii;
typedef pair < pair < int, int >, int > piii;
typedef pair < pair <int, int > , pair <int, int> > piiii;
typedef pair<ll, ll> pll;
typedef vector<bool> vb;
typedef vector<char> vc;
typedef vector<string> vs;
 
#define FOR(i,a,b) for(int i = a; i < b; i ++)
#define RFOR(i,a,b) for(int i = a-1; i >= b; i --)
#define all(a) a.begin(), a.end()
#define endl '\n';
#define sz(x) (int)(x).size()
 
#define mp make_pair
#define pb push_back
#define ff first
#define ss second
 
template <typename T>
void pr(vector<T> &v) {
    FOR(i, 0, sz(v)) cout << v[i] << " ";
    cout << endl;
}
template <typename T>
void pr(vector<vector<T> > &v) {
    FOR(i, 0, sz(v)) { pr(v[i]); }
}
template <typename T>
void re(T &x) {
    cin >> x;
}
template <typename T>
void re(vector<T> &a) {
    FOR(i, 0, sz(a)) re(a[i]);
}
template <class Arg, class... Args>
void re(Arg &first, Args &... rest) {
    re(first);
    re(rest...);
}
template <typename T>
void pr(T x) {
    cout << x << endl;
}
template <class Arg, class... Args>
void pr(const Arg &first, const Args &... rest) {
    cout << first << " ";
    pr(rest...);
    cout << endl;
}
void ps() { cout << endl; }
template<class T, class... Ts>
void ps(const T& t, const Ts&... ts) {
    cout << t; if (sizeof...(ts)) cout << " "; ps(ts...);
}
 
const ll MOD  =  998244353;
#define inf 1e18;
#define INF INT_MAX

long double PI = 4*atan(1);
long double eps = 1e-16;
int MAXN = 100005;
vector<pii> ivs(MAXN);
vector<ll> ans(MAXN);
int n;

void solve(int n) {
    vi p(2*n+1);
    vi d(2*n+1);
    vector<ll> sum(2*n+1);
    vi numc(2 * n + 1);
    vi lft(2 * n + 1, 1e8);

    FOR(i,0,n) {
        numc[ivs[i].ss]++;
        lft[ivs[i].ss] = ivs[i].ff;
    }
    FOR(i,1,2*n + 1) numc[i] += numc[i-1];
    RFOR(i,2*n, 1) lft[i] = min(lft[i], lft[i+1]);

    FOR(i,1, 2 * n + 1) {
        if(lft[i] >= i) {
            p[i] = i;
            d[i] = 0;
            sum[i] = 0;
        } else {
            p[i] = p[lft[i]];
            d[i] = d[lft[i]] + 1;
            sum[i] = sum[lft[i]] + numc[i];
        }
    }

    FOR(i,0,n) {
        int l = ivs[i].ff;
        int par = p[l];
        ans[i] += sum[l] - ((ll) d[l] + 1) * (numc[par]);
    }
}


int main() {
    auto start = chrono::high_resolution_clock::now();
    // ios_base::sync_with_stdio(0);cin.tie(0);
    //ofstream cout("output.txt");
    //ifstream cin("input.txt");
    #ifdef DEBUG
      freopen("input.txt", "r", stdin);
      freopen("output.txt", "w", stdout);
    #endif  
    cin >> n;
    ivs.resize(n);

    FOR(i,0,n) {
        int l,r; cin >> l >> r;
        ivs[i] = {l,r};
    }

    solve(n);

    FOR(i,0,n) {
        pii nw = {2 *n + 1 - ivs[i].ss, 2 * n + 1 - ivs[i].ff};
        ivs[i] = nw;
    }
    solve(n);

    FOR(i,0,n) {
        cout << ans[i] + n - 1 << endl;
    }




    
    

    
    // auto stop = chrono::high_resolution_clock::now();
    // auto duration = chrono::duration_cast<chrono::microseconds>(stop - start);
    // cout << duration.count() << endl;
    //cin.close();
    //cout.close();
}

详细

Test #1:

score: 100
Accepted
time: 3ms
memory: 4408kb

input:

5
2 3
6 7
1 9
5 10
4 8

output:

7
5
4
5
5

result:

ok 5 number(s): "7 5 4 5 5"

Test #2:

score: -100
Runtime Error

input:

200000
342504 342505
248314 248317
259328 259334
234817 234821
225726 225732
371424 371425
260562 260565
34752 34753
132098 132100
262130 262134
7254 7255
228491 228492
43488 43492
188408 188410
11691 11692
350334 350335
327350 327352
360684 360685
182051 182052
72131 72135
194666 194668
61303 61313...

output:


result: