RMQ算法 模板

RMQ ：用于求一个区间内的最大最小值

ST算法代码

时间复杂度：预处理O（n * log n）       查询O（1）

#include <stdio.h>  #include <iostream>  #include <string.h>  #include <stack>  #include <algorithm>  #include <queue>  #include <map>  #include <cmath>  #define eps 0.000001  #define pi acos(-1,0)  #define pr 999983  #define LL long long  using namespace std;  int n,m;  int _2[20]={1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192};  int a[10000][100];  void RMQ(){      int z=0;      while (_2[z]<=n) z++;      for (int i = 1; i < z; i++){          for (int j = 0; j <= n-_2[i]; j++){              a[j][i]=min(a[j][i-1],a[j+_2[i-1]][i-1]);          }      }  }   int main(){            while (~scanf("%d",&n)){            for (int i = 0; i < n; i++){              scanf("%d",&a[i][0]);          }          RMQ();          scanf("%d",&m);          while (m--){              int l,r,x;              scanf("%d%d",&l,&r);                  x=(int)(log(r-l+1)/log(2.0));                        printf("%d\n",min(a[l][x],a[r-(1<<x)+1][x]));          }      }      return 0;  }  

不知名大神代码

时间复杂度：预处理O（n * log n）       查询O（log n）

#include <stdio.h>#include <iostream>#include <string.h>#include <stack>#include <algorithm>#include <queue>#include <map>#include <cmath>#define eps 0.000001#define pi acos(-1,0)#define pr 999983#define LL long longusing namespace std;int n,m;int _2[20]={1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192};int a[10000][100];int fd(int l,int x){    int z=0;    while (_2[z]<=x) z++;    z--;    if (_2[z]<x) return min(a[l][z],fd(l+_2[z],x-_2[z]));    return a[l][z];}void RMQ(){int z=0;    while (_2[z]<=n) z++;    for (int i = 1; i < z; i++){        for (int j = 0; j <= n-_2[i]; j++){            a[j][i]=min(a[j][i-1],a[j+_2[i-1]][i-1]);        }    }} int main(){        while (~scanf("%d",&n)){        for (int i = 0; i < n; i++){            scanf("%d",&a[i][0]);        }        RMQ();        scanf("%d",&m);        while (m--){            int l,r,x;            scanf("%d%d",&l,&r);            x=r-l+1;            printf("%d\n",fd(l-1,x));        }    }    return 0;}