RMQ（st在线算法模板）

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#include<iostream>#include<cmath>#include<algorithm>using namespace std;#define M 100010#define MAXN 500#define MAXM 500int dp[M][18];/**一维RMQ ST算法*构造RMQ数组 makermq(int n,int b[]) O(nlog(n))的算法复杂度*dp[i][j] 表示从i到i+2^j -1中最小的一个值(从i开始持续2^j个数)*dp[i][j]=min{dp[i][j-1],dp[i+2^(j-1)][j-1]}*查询RMQ rmq(int s,int v)*将s-v 分成两个2^k的区间*即 k=(int)log2(s-v+1)*查询结果应该为 min(dp[s][k],dp[v-2^k+1][k])*/void makermq(int n,int b[]){    int i,j;    for(i=1;i<=n;i++)        dp[i][0]=b[i];    for(j=1;(1<<j)<=n;j++)        for(i=1;i+(1<<j)-1<=n;i++)            dp[i][j]=min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);}int rmq(int s,int v){    int k=(int)(log((v-s+1)*1.0)/log(2.0));    return min(dp[s][k],dp[v-(1<<k)+1][k]);}void makeRmqIndex(int n,int b[]) //返回最小值对应的下标{    int i,j;    for(i=1;i<=n;i++)        dp[i][0]=i;    for(j=1;(1<<j)<=n;j++)        for(i=1;i+(1<<j)-1<=n;i++)            dp[i][j]=b[dp[i][j-1]] < b[dp[i+(1<<(j-1))][j-1]]? dp[i][j-1]:dp[i+(1<<(j-1))][j-1];}int rmqIndex(int s,int v,int b[]){    int k=(int)(log((v-s+1)*1.0)/log(2.0));    return b[dp[s][k]]<b[dp[v-(1<<k)+1][k]]? dp[s][k]:dp[v-(1<<k)+1][k];}int main(){    int a[]={3,4,5,7,8,9,0,3,4,5};    //返回下标    makeRmqIndex(sizeof(a)/sizeof(a[0]),a);    cout<<rmqIndex(0,9,a)<<endl;    cout<<rmqIndex(4,9,a)<<endl;    //返回最小值    makermq(sizeof(a)/sizeof(a[0]),a);    cout<<rmq(0,9)<<endl;    cout<<rmq(4,9)<<endl;    return 0;}
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