st表模板，以最小值为例

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#include<iostream>#include<cstring>#include<cstdio>#include<cstdlib>#include<ctype.h>    //tower()#include<set>  #include<map>  #include<iomanip>// cout<<setprecision(1)<<fixed<<a;#include<vector>   #include<time.h>  #include<assert.h>  //assert#include<cmath>#include<algorithm>#include<bitset>#include<limits.h>#include<stack>#include<queue>using namespace std;const int maxn=100010;const int maxm=1001;const int inf=INT_MAX;int n=9,a[]={3,4,6,7,2,0,8,5,1},st[maxn][16],sti[maxn][16];void initrmq(){//记录最小值 for(int i=0;i<n;++i) st[i][0]=a[i];for(int j=1;(1<<j)<=n;++j){for(int i=0;i+(1<<j)-1<n;++i){//i+2^j-1<n,st[i][j]才有意义 st[i][j]=min(st[i][j-1],st[i+(1<<(j-1))][j-1]);}}}int rmq(int u,int v){int k=(int)(log(v-u+1.0)/log(2.0));//向下取整并转换成int，功能同floor()，保证u+1<<k落在[u,v]内。ceil()向上取整 return min(st[u][k],st[v-(1<<k)+1][k]);}void initindex(){//记录最小值下标 dp数组要换个名字！ for(int i=0;i<n;++i) sti[i][0]=i;for(int j=1;(1<<j)<=n;++j){for(int i=0;i+(1<<j)-1<n;++i){sti[i][j]=a[sti[i][j-1]]<a[sti[i+(1<<(j-1))][j-1]]?sti[i][j-1]:sti[i+(1<<(j-1))][j-1];}}} int index(int u,int v){int k=(int)(log(v-u+1.0)/log(2.0));return a[sti[u][k]]<a[sti[v-(1<<k)+1][k]]?sti[u][k]:sti[v-(1<<k)+1][k];}int main(){initrmq();initindex();printf("%d%d\n",rmq(1,7),index(1,7));printf("%d%d\n",rmq(0,4),index(0,4));return 0;}

参：

https://hrbust-acm-team.gitbooks.io/acm-book/content/data_structure/ds_part6.html

http://noalgo.info/489.html

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