问题及代码:
文件名称:main.cpp
作者:郑孚嘉
问题描述:认真阅读并验证分块查找算法。请用22,4,23,11,20,2,15,13,30,45,26,34,29,35,26,36,55,98,56,
74,61,90,80,96,127,158,116,114,128,113,115,102,184,211,243,188,187,218,195,210,279,307,492,452,408,361,421,399,856,523,704,703,697,535,534,739(共n=56个数据,每块
数据个数s=8)作为数据表,自行构造索引表,分别对查找61、739、200进行测试。
代码:
main.cpp
#include <stdio.h> #define MAXL 100 //数据表的最大长度 #define MAXI 20 //索引表的最大长度 typedef int KeyType; typedef char InfoType[10]; typedef struct { KeyType key; //KeyType为关键字的数据类型 InfoType data; //其他数据 } NodeType; typedef NodeType SeqList[MAXL]; //顺序表类型 typedef struct { KeyType key; //KeyType为关键字的类型 int link; //指向对应块的起始下标 } IdxType; typedef IdxType IDX[MAXI]; //索引表类型 int IdxSearch(IDX I,int m,SeqList R,int n,KeyType k) { int low=0,high=m-1,mid,i; int b=n/m; //b为每块的记录个数 while (low<=high) //在索引表中进行二分查找,找到的位置存放在low中 { mid=(low+high)/2; if (I[mid].key>=k) high=mid-1; else low=mid+1; } //应在索引表的high+1块中,再在线性表中进行顺序查找 i=I[high+1].link; while (i<=I[high+1].link+b-1 && R[i].key!=k) i++; if (i<=I[high+1].link+b-1) return i+1; else return 0; } int main() { int i,n=56,m=7,j; SeqList R; IDX I= {{26,0},{55,8},{102,16},{184,24},{279,32},{523,40},{856,48}}; KeyType a[]= {22,4,23,11,20,2,15,13,30,45,26,34,29,35,26,36,55,98,56,74,61, 90,80,96,127,158,116,114,128,113,115,102,184,211,243,188,187,218,195,210, 279,307,492,452,408,361,421,399,856,523,704,703,697,535,534,739}; KeyType x=200; for (i=0; i<n; i++) R[i].key=a[i]; j=IdxSearch(I,m,R,n,x); if (j!=0) printf("%d是第%d个数据\n",x,j); else printf("未找到%d\n",x); return 0; }
运行结果:
知识点总结:
分块查找要先排序,确定每个块的范围进行分块。在索引表中进行二分查找确定块的位置,再在线性表中进行顺序查找