2015 Multi-school Joint training first field assignment (hdu5289) three kinds of solution

Topic: Give a series, ask how many consecutive sub-sequences, the maximum value of the subsequence-the minimum value

#include <iostream>#include <cstdio>#include <algorithm>#include <string>#include <cmath>using namespace STD;intmaxsum[100000][ -];intminsum[100000][ -];inta[100000];intN,k;voidRmq_init () { for(intj =1; (1&LT;&LT;J) <= N; ++J) for(inti =1; i + (1&LT;&LT;J)-1<= N; ++i) {Maxsum[i][j] = max (maxsum[i][j-1],maxsum[i+ (1<< (J-1))][j-1]); Minsum[i][j] = min (minsum[i][j-1],minsum[i+ (1<< (J-1))][j-1]); }}intQueryintLintR) {intK = log2 (r-l+1);intmax = max (Maxsum[l][k], maxsum[r-(1&LT;&LT;K) +1][k]);intmin = min (minsum[l][k], minsum[r-(1&LT;&LT;K) +1][k]);returnMax-min;}intMain () {intTscanf("%d", &t); while(t--) {scanf("%d%d", &n,&k); for(inti =1; I <= n;++i) {scanf("%d", a+i); maxsum[i][0] = minsum[i][0] = A[i]; } rmq_init ();Long LongAns =0;intL, R; for(inti =1; I <= N; ++i) {L = i, r = N; while(L <= R) {intMid = (l+r)/2;intCha = query (i,mid);if(Cha < k) L = mid+1;ElseR = Mid-1;        } ans + = l-i; }printf("%lld\n", ans); }return 0;}

Two: Monotone queue
Maintain a minimum value with a monotone queue, double pointers, the first second pointer initially points to the first data, the first pointer continuously adds data to the end of the team in order, and when the difference of the maximum minimum value is greater than or equal to K, it means that the newly added one cannot be applied to the position of the current second pointer, and can be calculated The number of successive sub-sequences starting at the second pointer position, and the last statistic sum.

#include <cstdio>#include <cstring>#include <queue>#include <algorithm>using namespace STD;#define LL Long Long deque <LL>Max, Min;//monotone queue, Max Max, min minLL a[100010] ;intMain () {intT, N, I, J; LL k, ans;scanf("%d", &t); while(t--) {scanf("%d%i64d", &n, &k); for(i =0; I < n; i++)scanf("%i64d", &a[i]); while( ! Max.empty ()) Max.pop_back (); while( ! Min.empty ()) Min.pop_back (); for(i =0, j =0, ans =0; I < n; i++)//i in front, J in the back{ while( !            Max.empty () && max.back () < a[i]) Max.pop_back (); Max.push_back (A[i]); while( !            Min.empty () && min.back () > A[i]) min.pop_back (); Min.push_back (A[i]); while( ! Max.empty () &&! Min.empty () && Max.front ()-Min.front () >= k) {ans + = (i-j);if(Max.front () = = A[j]) Max.pop_front ();if(Min.front () = = A[j])                Min.pop_front ();            j + +; }        } while(J < N)            {ans + = (i-j);        j + +; }printf("%lld\n", ans); }return 0;}

Three: rmq+ greedy
This enumeration right endpoint, greedy to select the right endpoint (similar to the ruler)

#include <iostream>#include <cstdio>#include <algorithm>#include <string>#include <cmath>using namespace STD;intmaxsum[100000][ -];intminsum[100000][ -];inta[100000];intN,m;voidRmq_init () { for(intj =1; (1&LT;&LT;J) <= N; ++J) for(inti =1; i + (1&LT;&LT;J)-1<= N; ++i) {Maxsum[i][j] = max (maxsum[i][j-1],maxsum[i+ (1<< (J-1))][j-1]); Minsum[i][j] = min (minsum[i][j-1],minsum[i+ (1<< (J-1))][j-1]); }}intQueryintLintR) {intK = log2 (r-l+1);intmax = max (Maxsum[l][k], maxsum[r-(1&LT;&LT;K) +1][k]);intmin = min (minsum[l][k], minsum[r-(1&LT;&LT;K) +1][k]);returnMax-min;}intMain () {intTscanf("%d", &t); while(t--) {scanf("%d%d", &n,&m); for(inti =1; I <= N; ++i) {scanf("%d", a+i); maxsum[i][0] = minsum[i][0] = A[i]; } rmq_init ();Long LongAns =0;intk=1; for(intI=1; i<=n; i++) { while(Query (k,i) >=m&&k<i) k++; ans+= (i-k+1); }printf("%lld\n", ans); }return 0;}

The following are three algorithms run time (3,2,1)
Visible time Efficiency monotone queue best (405ms), greedy +rmq (733ms) Second, +RMQ (1216ms) worst

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2015 Multi-school Joint training first field assignment (hdu5289) three kinds of solution