D. Fedor and coupons 二分暴力

http://codeforces.com/contest/754/problem/D

给出n条线段，选出k条，使得他们的公共部分长度最大。

公共部分的长度，可以二分出来，为val。那么怎么判断有k条线段有共同的这个长度，而且选他们出来呢？

可以把右端点减去val - 1，那么以后就只需要k条线段至少有一个交点就可以了。

那么怎么确定这个交点呢？

我的做法是直接离散，然后暴力找出覆盖次数>=k的那个点。

复杂度好像有点高，

log2e10 * nlogn

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#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>
#include <bitset>
const int maxn = 6e5 + 20;
int L[maxn], R[maxn];
int pos[maxn];
int book[maxn];
int n, k;
map<LL, int>save;
bool check(LL val) {
    if (val == 0) return true;
    int lenpos = 0;
    for (int i = 1; i <= n; ++i) {
        if (R[i] - L[i] + 1 < val) continue;
        pos[++lenpos] = L[i];
        pos[++lenpos] = R[i] - val + 1;
    }
    int mx = -inf;
    if (lenpos == 0) return false;
    sort(pos + 1, pos + 1 + lenpos);
    for (int i = 1; i <= n; ++i) {
        if (R[i] - L[i] + 1 < val) continue;
        int t1 = lower_bound(pos + 1, pos + 1 + lenpos, L[i]) - pos;
        int t2 = lower_bound(pos + 1, pos + 1 + lenpos, R[i] - val + 1) - pos;
        book[t1]++;
        book[t2 + 1]--;
        mx = max(mx, t2 + 1);
    }
    bool flag = false;
    for (int i = 1; i <= mx; ++i) {
        book[i] += book[i - 1];
        if (book[i] >= k) {
//            assert(save[val] == 0);
            save[val] = pos[i];
            flag = true;
            break;
        }
    }
    for (int i = 1; i <= mx; ++i) book[i] = 0;
    return flag;
}
void work() {
    cin >> n >> k;
    for (int i = 1; i <= n; ++i) {
        cin >> L[i] >> R[i];
    }
    LL be = 0, en = 2e10;
    while (be <= en) {
        LL mid = (be + en) >> 1;
        if (check(mid)) {
            be = mid + 1;
        } else en = mid - 1;
    }
    cout << en << endl;
    if (en == 0) {
        for (int i = 1; i <= k; ++i) {
            cout << i << " ";
        }
    } else {
        int use = 0;
        int tpoint = save[en];
//        cout << tpoint << endl;
        for (int i = 1; i <= n && use < k; ++i) {
            if (R[i] - tpoint + 1 < en) continue;
            if (tpoint >= L[i] && tpoint <= R[i]) {
                use++;
                cout << i << " ";
            }
        }
//        assert(use == k);
    }
}

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

其实这题关键要把判断k条线段相交于同一个区间，化简为相交于同一个点