Codeforces-754D Fedor and coupons

传送门

给你n(<=3e5)个区间，让你从中选出k(<=n)个，使这些区间的交集长度最大。区间范围[-1e9, 1e9]

要求输出区间长度和选取的区间编号

典型的优先队列问题。先把区间按左值从小到大排序，这样做保证了后面取出来的区间的左值大于先前所有的，然后我门就只需要比较右值了。注意到最终答案的区间是由k个区间交，所以我们要维护k个区间的信息。每次由这k个右值里的最小值来减去k个左值里的最大值再加一去更新答案。由于最后取出的区间的左值最大（排序过），所以我们维护k个右值即可。选取优先队列维护

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long LL;
int N, K;
const int maxn = 3e5 + 10;
struct node {
    int left, right, num;
} seg[maxn];
int left, right;

bool cmp(const node &A, const node &B) {
    return A.left < B.left;
}

int main() {
    scanf("%d%d", &N, &K);
    for (int i = 0; i < N; i++) {
        scanf("%d%d", &seg[i].left, &seg[i].right);
        seg[i].num = i + 1;
    }
     int tmp = 1e9 + 10;;
     sort(seg, seg + N, cmp);
    priority_queue<int, vector<int>, greater<int> > que;
     for (int i = 0; i < K; i++) {
         que.push(seg[i].right);
         tmp = min(tmp, seg[i].right);
     }    
     int ans = 0;
     if (!(tmp < seg[K - 1].left)) {
         ans = tmp - seg[K - 1].left + 1;
         left = seg[K - 1].left;
         right = tmp;
     }
     tmp = que.top();
     for (int i = K; i < N; i++) {
         if (seg[i].right > tmp) {
             que.pop();
             que.push(seg[i].right);
             tmp = que.top();
             if (tmp - seg[i].left + 1 > ans) {
                 ans = tmp - seg[i].left + 1;
                 right = tmp;
                 left = seg[i].left;
             }
         }
     }
     printf("%d
", ans);
     if (ans == 0) {
         for (int i = 1; i <= K; i++) {
             printf("%d%c", i, i == K ? '
' : ' ');
         }
     } else {
         for (int i = 0; i < N && K; i++) {
             if (seg[i].left <= left && right <= seg[i].right) {
                 printf("%d%c", seg[i].num, K == 1 ? '
' : ' ');
                 --K;
             }
         }
      }
    return 0;
}