HDU 4343 Interval Query (multiplication thought + greed)



Test instructions: given n (n<=100000) intervals (left closed right open) and M (m<=100000) queries [L, R], ask all the maximum number of 22 disjoint intervals within the [L, R] interval. ，

Idea: First greedy thought, to remove the large interval containing other intervals, this will certainly not affect the results.

Then, for all intervals, the left endpoint is sorted in ascending order, so that when all the intervals are not mutually inclusive, their right endpoints are incremented.

Then for each inquiry, must be from left to right to go as many intervals, this greedy easy to think of.

Discretization of data, recording the nearest distance from each point through the I interval, this step uses the idea of multiplication, because if a point order is found, then the time complexity and space complexity will not be able to bear.

In particular, using f[i][j] to record the nearest point through the 2^j interval from I, a recursive relationship can be derived

F[I][J] = f[f[i][j-1]][j-1];

So for each inquiry, we will ask QL,QR also discretization, just find out how many intervals from the QL so that it does not exceed the QR.

PS: Sleep in the middle of the night is really heartache, up the problem ...

#include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> #include <iostream > #include <algorithm> #include <vector> #include <map> #include <queue> #include <stack& Gt #include <string> #include <map> #include <set> #define EPS 1e-6 #define LL Long Long #define PII (pair<  int, int>)//#pragma comment (linker, "/stack:1024000000,1024000000") using namespace std; const int MAXN = 100000 + 100;const int INF = 0x3f3f3f3f;int N, m;struct Node {int L, R;bool operator < (const NODE&AMP ; A) Const {if (L = = A.L) return r > A.r;return l < a.l;}} NODE[MAXN]; int discret[2*maxn];int fa[2*maxn][20];int Solve (int l, int r) {int ans = 0;for (int i = n; I >= 0; i--) {int t = Fa[l] [I];//cout << t << endl;if (t = r) return ans+ (1<<i); else if (T < r) L = t, ans + = (1<<i); } return ans; int main () {//freopen ("Input.txt", "R", stdin), while (scanf ("%d%d", &n, &m) = = 2) {MemseT (FA, INF, sizeof (FA)); for (int i = 0; i < n; i++) {scanf ("%d%d", &discret[2*i], &discret[2*i+1]); node[i].l = di Scret[2*i], NODE[I].R = discret[2*i+1];}  Sort (discret, discret+2*n); int cnt = unique (Discret, discret+2*n)-discret;//cout << cnt << endl;for (int i = 0; I < n; i++) {node[i].l = Lower_bound (Discret, discret+cnt, node[i].l)-DISCRET;NODE[I].R = Lower_bound (Discret, discret+cnt, no DE[I].R)-discret;//cout << node[i].r << Endl;} Sort (node, node+n); int cnt2 = n-1;for (int i = cnt-1; I >= 0; i--) {//cout << node[i].l << endl;if (i = = node [CNT2].L] {fa[i][0] = min (NODE[CNT2].R, fa[i+1][0]); while (i = = NODE[CNT2].L) cnt2--;} else fa[i][0] = fa[node[cnt2+1].l][0];} for (int i = 1; i < n; i++) {for (int j = 0; J < CNT; J + +) if (fa[j][i-1]! = INF) Fa[j][i] = fa[fa[j][i-1]][i-1];} for (int i = 0; i < m; i++) {int L, R; scanf ("%d%d", &l, &r); l = Lower_bound (Discret, discret+cnt, L)-discret; R = Upper_bound (Discret, discret+cnt, R)-discret-1;//cout << L << "" << R << endl;printf ("%d\n", Solve (L, R));} cout << fa[0][0] << Endl;} return 0;}

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HDU 4343 Interval Query (multiplication thought + greed)