"Dynamic Programming DP" RMQ problem (ST algorithm)

"RMQ" range minimum/maximum Query range max problem

"St Algorithm"

An effective algorithm for solving RMQ problems

　　

　　pretreatment pre-treatment to construct D, preprocessing time complexity O (NLOGN)

Using the idea of dynamic programming D (I, j) indicates that the minimum value of the range I ~ i + 2j-1 has a state transition equation

　　D (i, j) = min {d (i, j-1), D (i + 2j-1, j-1)}

　　

Set the original data stored in the array a[], the initial state

D (i, 0) = a[i]

　　Querying for the maximum time complexity of a given interval O (1)

Set the query interval to [L, R]

Interval length = r-l + 1 < = 2k (where k is the largest integer that satisfies the condition)

The query results are min {d (L, K), D (r-2k + 1, k)}

　　

Template

#include <iostream>#include<vector>#include<algorithm>using namespacestd;/*rmq-st algorithm interval min template interval maximum need to change min () to Max () by Chsobin*/Const intMAXN = -;intDP[MAXN][MAXN];voidRmq_init (Constvector<int>&A) {    intn =a.size ();  for(intI=0; i<n;++i) dp[i][0] =A[i];  for(intj=1; (1&LT;&LT;J) <=n; J + +){         for(intI=0; i+ (1&LT;&LT;J)-1<n; i++) {Dp[i][j]= Min (dp[i][j-1], Dp[i + (1<< (J-1))][j-1]); }    } }intRMQ (intLintR) {    intK =0;  while( (1<< (k +1)) <= r-l+1) k++; returnMin (Dp[l][k], dp[r-(1&LT;&LT;K) +1][k]);}intMain () {inta[Ten] = {1,5,8,7,2,6,1,9,3,4}; Vector<int> A (A, A +Ten);    Rmq_init (A);  for(intI=0; I<a.size (); + +i) {cout<< A[i] <<" "; } cout<<Endl;  while(1){        intL, R; cout<<"[L, R]"; CIN>> L >>R; cout<< RMQ (L, R) <<Endl; }    return 0;} 

Topic

hdu3183

/*problem-solving ideas: The use of RMQ, set the original number is n, then after the removal of m number of n-m numbers left. (1) Because there are n-m numbers, then the smallest number in the 1 to m+1 position must be the first number in the result, the smallest number in the number that is recorded as POS (2) and then from the next position of the position pos+1 to the m+2 position must be the second number in the result, Then go back and look backwards. (3) In order to ensure the smallest number, so to ensure that the highest minimum, but also to ensure that the digital length to meet the conditions*/#include<iostream>#include<cstring>using namespacestd;Const intMAXN =1010;CharA[MAXN];intdp[maxn][ One];CharANS[MAXN];intN, M;//Custom Comparison RulesintMIN (intIintj) {    returnA[i] <= a[j]?i:j;}//pretreatmentvoidinit () { for(intI=0; i<n;++i) {dp[i][0] =i; }     for(intj=1;(1&LT;&LT;J) <=n;++j) {         for(intI=0;(i + (1&LT;&LT;J)-1) < n;++i) {Dp[i][j]= MIN (dp[i][j-1], dp[i+ (1<< (J-1))][j-1]); }    }}//EnquiryintQueryintLintR) {    intk=0; inttemp = r-l+1; intsum =1;  while(Sum <=temp) {        ++K; Sum*=2; } k--; returnMIN (Dp[l][k], dp[r-(1&LT;&LT;K) +1][k]); }intMain () {intL =0; intR =0;  while(SCANF ("%s%d", A, &m) = =2){        intindex =0; L=0; R=m; N=strlen (a); M= N-m;                Init (); ////for (int i=0;i<n;++i) {//for (int j=0; (i + (1<<j)-1) < n; ++J) {//printf ("%c", A[dp[i][j]]);//            }//printf ("\ n");//        }                     while(m--){//cout << "L =" << l << "r =" << r << Endl;L =query (L, R); Ans[index++] =A[l]; L++; R++; if(r==n) R = n1;//ensure R does not cross} Ans[index]=' /'; intK =0;  for(; k<index;++k) {//Remove leading 0            if(Ans[k]! ='0') Break; }        if(K==index) printf ("0\n"); Elseprintf"%s\n", ans+k); }    return 0;} 

AC 0ms

Reference

Petition P197

"Dynamic Programming DP" RMQ problem (ST algorithm)