Final Exam Arrangement

Final Exam Arrangement
In Zhejiang University, there are N different courses labeled from 1 to N. each course has its own time slot during the week. we can represent the time slot of a course by an left-closed right-open interval [s, t ).
Now we are going to arrange the final exam time of all the courses.
The final exam period will contain in multiple days. in each day, multiple final exams will be held simultaneously. if two courses 'time slots are not overlapped, there may be students who are attending both of them, so we cannot arrange their final exams at the same day.
Now you're to arrange the final exam period, to make the total days as small as possible.
Input
There are multiple test cases separated by blank lines.
For each rows, the 1st line contains one integer N (1 <= N <= 100000 ).
Then N lines, the I + 1th line contains s and t of the interval [s, t) for the ith course. (0 <= s There is a blank line after each test case.
Output
For each case, the 1st line contains the days P in the shortest final exam period.
Next P lines, the I + 1th line contains the numbers of courses whose final exam is arranged on the ith day separated by one space.
Output a blank line after each test case.
Sample Input
4
0 1
1 2
2 3
3 4

4
0 2
1 3
2 4
3 5

4
0 4
1 5
2 4
3 6

Sample Output
4
1
2
3
4

2
1 2
3 4

1

1 2 3 4

Scanning line

Sorts the time points. If the time is the same, the right endpoint is prioritized because it is an open interval.

Scan each vertex sequentially. Before the first right endpoint that has not been processed, the left endpoint that has been scanned indicates that the exam can be placed in one day.

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            using namespace std;//LOOP#define FE(i, a, b) for(int i = (a); i <= (b); ++i)#define FED(i, b, a) for(int i = (b); i>= (a); --i)#define REP(i, N) for(int i = 0; i < (N); ++i)#define CLR(A,value) memset(A,value,sizeof(A))//STL#define PB push_back//INPUT#define RI(n) scanf("%d", &n)#define RII(n, m) scanf("%d%d", &n, &m)#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)#define RS(s) scanf("%s", s)#define FF(i, a, b) for(int i = (a); i < (b); ++i)#define FD(i, b, a) for(int i = (b) - 1; i >= (a); --i)#define CPY(a, b) memcpy(a, b, sizeof(a))#define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++)#define EQ(a, b) (fabs((a) - (b)) <= 1e-10)#define ALL(c) (c).begin(), (c).end()#define SZ(V) (int)V.size()#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)#define WI(n) printf("%d\n", n)#define WS(s) printf("%s\n", s)#define debugt(a) cout << (#a) << "=" << a << " ";#define debugI(a) debugt(a) cout << endl#define debugII(a, b) debugt(a) debugt(b) cout << endl#define debugIII(a, b, c) debugt(a) debugt(b) debugt(c) cout << endl#define debugIV(a, b, c, d) debugt(a) debugt(b) debugt(c) debugt(d) cout << endl#define debugV(a, b, c, d, e) debugt(a) debugt(b) debugt(c) debugt(d) debugt(e) cout << endltypedef long long LL;const int INF = 0x3f3f3f3f;const int maxn = 100010;typedef vector 
            
              VI;typedef unsigned long long ULL;const double eps = 1e-10;const LL MOD = 1e9 + 7;struct course{ int w, id; bool l; bool operator < (const course& a) const{ if (w == a.w) return l < a.l; return w < a.w; }}c[maxn * 2 + 10];VI ans[maxn];bool use[maxn * 2];int main(){ //freopen("0.txt", "r", stdin); int n; while (~RI(n)) { REP(i, n) { RII(c[i * 2].w, c[i * 2 + 1].w); c[i * 2].id = c[i * 2 + 1].id = i + 1; c[i * 2].l = 1, c[i * 2 + 1].l = 0; } sort(c, c + 2 * n); REP(i, n + 1) ans[i].clear(); CLR(use, 0); int cnt = 0, j; REP(i, 2 * n) { if (use[c[i].id]) continue; j = i; while (j < 2 * n) { if (use[c[j].id]) { j++; continue; } if (!use[c[j].id] && c[j].l == 0) break; j++; } FF(k, i, j) if (c[k].l) { ans[cnt].PB(c[k].id); use[c[k].id] = 1; } cnt++; i = j; } WI(cnt); REP(i, cnt) { sort(ans[i].begin(), ans[i].end()); REP(j, ans[i].size()) { if (j) printf(" "); printf("%d", ans[i][j]); } puts(""); } puts(""); } return 0;}