Codeforces 754D Fedor and coupons (priority queue)

Test instructions: given n coupons, each has a certain range of concessions, and then to choose K Zhang, to ensure that K Zhang Common preferential range the largest.

First, all coupons are sorted by the left endpoint, then a priority queue with a capacity of k is maintained, the minimum value in the priority queue is updated each time, and the current right endpoint is

The distance between them. The priority queue is just as good as storing the right endpoint.

The code is as follows:

#pragma COMMENT (linker, "/stack:1024000000,1024000000") #include <cstdio> #include <string> #include < cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include < queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include < cmath> #include <stack> #include <sstream> #include <unordered_map> #include <unordered_set > #define DEBUG () puts ("++++"), #define FREOPENR freopen ("In.txt", "R", stdin) #define FREOPENW freopen ("OUT.txt", "W", STDOUT) using namespace Std;typedef long long ll;typedef unsigned long long ull;typedef pair<int, int> p;const int in F = 0x3f3f3f3f;const LL LNF = 1LL << 60;const Double inf = 0x3f3f3f3f3f3f;const double PI = ACOs ( -1.0); const DOUBLE EPS = 1e-8;const int maxn = 3e5 + 5;const int mod = 2000;const int dr[] = {-1, 1, 0, 0};const int dc[] = {0, 0, 1, -1};con St Char *de[] = {"0000", "0001", "0010", "0011", "0100", "01"0110", "0111", "" "," "1001", "1010", "1011", "1100", "1101", "1110", "1111"};int N, m;const int mon[] = {0, 31, 28 ,,,,,,,,,,,,,,,,,,,,,,,, 31};const int monn[] = {0, ~, ~,------ OL is_in (int r, int c) {return R >= 0 && r < n && C >= 0 && C < m;}    struct node{int L, r, id;    BOOL operator < (const Node &p) const{return L < P.L; }}; Node A[maxn];int Main () {while (scanf ("%d%d", &n, &m) = = 2) {for (int i = 0; i < n; ++i) {s            CANF ("%d%d", &AMP;A[I].L, &AMP;A[I].R);        A[i].id = i + 1;        } sort (A, a + N);        Priority_queue<int, Vector<int>, greater<int> >pq;        int ans = 0;        int t = 0;            for (int i = 0; i < n; ++i) {Pq.push (A[I].R);            if (Pq.size () > M) pq.pop ();            int tmp = Pq.top ()-A[I].L + 1; if (pq.size () = = m && ans <TMP) {ans = tmp;            t = A[I].L;        }} printf ("%d\n", ans);            if (!ans) {printf ("1");            for (int i = 2; I <= m; ++i) printf ("%d", I);        Continue            } else{int cnt = 0;  for (int i = 0; i < n && m; ++i) if (a[i].l <= t && a[i].r >= T + ans-1) {if (CNT)                Putchar (");                printf ("%d", a[i].id);                --m;            ++cnt;    }} printf ("\ n"); } return 0;}

Codeforces 754D Fedor and coupons (priority queue)