第三十九章续：区间最值RMQ问题

问题描述：

找出数组A[]，任意区间[i，j]的最小值

/*RMQ问题区间最值查询*/#include <iostream>using namespace std;#define MAX 100//方法1//M[i][j]表示区间i，j最值的索引//构造M复杂度O(n^2)void RMQ1(int M[][MAX],int A[],int N){int i, j;      for (i =0; i < N; i++)          M[i][i] = i;        for (i = 0; i < N; i++)          for (j = i + 1; j < N; j++)              //若前者小于后者，则把后者的索引值付给M[i][j]              if (A[M[i][j - 1]] < A[j])                  M[i][j] = M[i][j - 1];              //否则前者的索引值付给M[i][j]              else                  M[i][j] = j;  }/*方法2：ST算法M[ i ][ j ] 是以i 开始，长度为 2^j 的子数组的最小值的索引分两个区间，M[i][j]为这两个区间最值的索引则M[i][j]={M[i][j-1],M[i+2^(j-1)+1][j-1]}构造M时间复杂度O(nlogn)如何求区间的最值RMQ[i][j]？设k=log(j-i+1)RMQ[i][j]={M[i][k],M[j-2^k+1][k]}时间复杂度O(1)*/void RMQ2(int M[][MAX], int A[], int N)  {      int i, j;      //initialize M for the intervals with length 1        for (i = 0; i < N; i++)          M[i][0] = i;        //compute values from smaller to bigger intervals      for (j = 1; 1 << j <= N; j++)          for (i = 0; i + (1 << j) - 1 < N; i++)              if (A[M[i][j - 1]] < A[M[i + (1 << (j - 1))][j - 1]])                  M[i][j] = M[i][j - 1];              else                  M[i][j] = M[i + (1 << (j - 1))][j - 1];  }    /*方法3：线段树这里M[i]保存结点i区间最小值的位置。初始时M的所有元素为-1构造M,O(N)查询O(logn)*/void initialize(int node, int b, int e, int M[], int A[])  {      if (b == e){        M[node] = b;  return;}    //compute the values in the left and right subtrees      initialize(2 * node, b, (b + e) / 2, M, A);      initialize(2 * node + 1, (b + e) / 2 + 1, e, M, A);        //search for the minimum value in the first and      //second half of the interval      if (A[M[2 * node]] <= A[M[2 * node + 1]])          M[node] = M[2 * node];      else          M[node] = M[2 * node + 1];  }  int query(int node, int b, int e, int M[], int A[], int i, int j)  {      int p1, p2;      //if the current interval doesn't intersect      //the query interval return -1      if (i > e || j < b)          return -1;        //if the current interval is included in      //the query interval return M[node]      if (b >= i && e <= j)          return M[node];        //compute the minimum position in the      //left and right part of the interval      p1 = query(2 * node, b, (b + e) / 2, M, A, i, j);      p2 = query(2 * node + 1, (b + e) / 2 + 1, e, M, A, i, j);        //return the position where the overall      //minimum is      if (p1 == -1)          return M[node] = p2;      if (p2 == -1)          return M[node] = p1;      if (A[p1] <= A[p2])          return M[node] = p1;      return M[node] = p2;  }  int main(){int A[]={1,2,3,4,5,6,7,8,9,0};int M[MAX][MAX];//ST算法测试RMQ2(M,A,10);int k=log(9+1);cout<<k<<endl;cout<<A[M[0][k]]<<endl;cout<<A[M[9-(1<<k)+1][k]]<<endl;//end//线段树测试int N[MAX];memset(N,-1,sizeof(N));//初始化数组为-1initialize(1,0,10-1,N,A);int index=query(1,0,9,N,A,1,8);cout<<A[index]<<endl;return 0;}