E. The Best among Equals

Posted 2020-08-17 stupid_one

http://codeforces.com/gym/101149/problem/E

这题的话，关键是注意到一定是要max score

然后就可以选出一个L最大优先，并且R最大的区间，

扫一次就能得到答案了。

3
1 3
1 3
4 5

这组数据，只能是1

因为max score优先，要选[4，5]这段区间的人。

 

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#define ios ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;

#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
const int maxn = 1e6 + 20;
int arr[maxn];
int L[maxn];
int R[maxn];
int book[maxn];
void work() {
    int n;
    cin >> n;
    int lenarr = 0;
    for (int i = 1; i <= n; ++i) {
        cin >> L[i] >> R[i];
        if (L[i] > R[i]) swap(L[i], R[i]);
        arr[++lenarr] = L[i];
        arr[++lenarr] = R[i];
    }
    int mx = -inf;
    int id = inf;
    for (int i = 1; i <= n; ++i) {
        if (mx < L[i]) {
            mx = L[i];
            id = i;
        } else if (mx == L[i]) {
            if (R[id] < R[i]) {
                id = i;
            }
        }
    }
    int ans = 0;
    for (int i = 1; i <= n; ++i) {
        if (R[i] < L[id]) continue;
        ans++;
    }
    cout << ans << endl;
//    sort(arr + 1, arr + 1 + lenarr);
//    for (int i = 1; i <= n; ++i) {
//        int hash_L = lower_bound(arr + 1, arr + 1 + lenarr, L[i]) - arr;
//        int hash_R = lower_bound(arr + 1, arr + 1 + lenarr, R[i]) - arr;
////        cout << hash_L << " " << hash_R << endl;
//        book[hash_L]++;
//        book[hash_R + 1]--;
//    }
//    int ans = 0;
//    int begin = lower_bound(arr + 1, arr + 1 + lenarr, L[id]) - arr;
//    int end = lower_bound(arr + 1, arr + 1 + lenarr, R[id]) - arr;
//    for (int i = 1; i <= lenarr + 1; ++i) {
//        book[i] += book[i - 1];
//        if (i >= begin && i <= end)
//            ans = max(book[i], ans);
//    }
//    cout << ans << endl;
}

int main() {
#ifdef local
    freopen("data.txt","r",stdin);
#endif
    IOS;
    work();
    return 0;
}

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