Codeforces 484E sign on Fence (is a persistent segment tree + dichotomy)

Title Link: Codeforces 484E sign on Fence

Topic: given to a sequence, each position has a value, indicating the height, now there are a number of queries, each query l,r,w, expressed in the interval l,r,

What is the maximum height of the longest continuous length greater than w?

Problem-solving ideas: Can persist segment tree maintenance interval Merge, the front-end time encountered a problem can be persisted dictionary tree, went to check the related papers, probably know is

What is it.

The height is sorted in order from large to small, and then each time a position is inserted, the segment tree maintains the longest continuous interval, as the insertion is in accordance with the large-to-small

Successive maximum lengths in each segment tree are satisfied with height greater than or equal to the current newly inserted height value. For each query, the two-point height,

The height is definitely in the existing height, so just two points below the table can be.

#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace STD;Const intMAXN =1e6+5;typedefpair<int,int> PII;structNode {intLC, RC, LP, RP, L, R, S;intLength () {returnRp-lp +1; }}ND[MAXN <<2];intN, SZ, ROOT[MAXN];p II BLO[MAXN];inline intNewNode () {returnsz++;}inline voidPushup (intu) {intLCID = nd[u].lc, rcid = nd[u].rc; Nd[u]. L = Nd[lcid]. L + (Nd[lcid]. L = = Nd[lcid].length ()? ND[RCID]. L0); Nd[u]. R = Nd[rcid]. R + (Nd[rcid]. R = = Nd[rcid].length ()? ND[LCID]. R:0); Nd[u]. S = Max (Nd[lcid]. R + nd[rcid]. L, Max (Nd[lcid]. S, Nd[rcid]. S));}inlineNode merge (Node A, node B) {node u; U.LP = A.LP;    U.RP = B.RP; U.L = A.l + (A.L = = A.length ()? B.L:0); U.R = B.R + (B.R = = B.length ()? A.R:0); U.S. = max (A.R + B.L, max (A.S, B.S));returnU;}voidBuildint& U,intLintR) {if(U = =0) U = NewNode (); Nd[u] = (Node) {0,0, L, R,0,0,0};if(L = = r)return;intMid = (L + r) >>1;    Build (ND[U].LC, L, mid); Build (nd[u].rc, Mid +1, R); Pushup (u);}intInsertintUintx) {intK = NewNode (); ND[K] = Nd[u];if(ND[K].LP = = x && x = = ND[K].RP) {Nd[k]. S = Nd[k]. L = Nd[k]. R =1;returnK }intMid = (ND[K].LP + nd[k].rp) >>1;if(x <= mid) nd[k].lc = insert (ND[K].LC, x);Elsend[k].rc = insert (nd[k].rc, x); Pushup (k);returnK;} Node Query (intUintLintR) {if(l <= nd[u].lp && nd[u].rp <= R)returnNd[u];intMid = (ND[U].LP + nd[u].rp) >>1;if(R <= Mid)returnQuery (ND[U].LC, L, R);Else if(L > Mid)returnQuery (Nd[u].rc, L, R);Else{Node ll = query (ND[U].LC, L, R); Node rr = query (Nd[u].rc, L, R);returnMerge (ll, RR); }}inline BOOLCMP (Constpii& A,Constpii& b) {returnA.first > B.first;}voidInit () {sz =1;scanf("%d", &n); for(inti =1; I <= N; i++) {scanf("%d", &blo[i].first);    Blo[i].second = i; } sort (Blo +1, Blo +1+ N, CMP); Build (root[0],1, N); for(inti =1; I <= N; i++) Root[i] = insert (root[i-1], blo[i].second);}intMain () {init ();intQ, L, R, W;scanf("%d", &q); while(q--) {scanf("%d%d%d", &l, &r, &w);intL =1, R = N; while(L < R) {intMid = (L + R) >>1;if(Query (Root[mid], L, R). S >= W) R = mid;ElseL = mid +1; }printf("%d\n", Blo[l].first); }return 0;}

Codeforces 484E sign on Fence (is a persistent segment tree + dichotomy)