hdu-4217-线段树-链式存储实现

Data Structure?

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2106    Accepted Submission(s): 671

Problem Description

Data structure is one of the basic skills for Computer Science students, which is a particular way of storing and organizing data in a computer so that it can be used efficiently. Today let me introduce a data-structure-like problem for you.
Original, there are N numbers, namely 1, 2, 3...N. Each round, iSea find out the Ki-th smallest number and take it away, your task is reporting him the total sum of the numbers he has taken away.

 

Input

The first line contains a single integer T, indicating the number of test cases.
Each test case includes two integers N, K, K indicates the round numbers. Then a line with K numbers following, indicating in i (1-based) round, iSea take away the Ki-th smallest away.

Technical Specification
1. 1 <= T <= 128
2. 1 <= K <= N <= 262 144
3. 1 <= Ki <= N - i + 1

 

Output

For each test case, output the case number first, then the sum.

 

Sample Input

23 21 110 33 9 1

 

Sample Output

Case 1: 3Case 2: 14
一开始链式实现超内存，后来加了一个空间释放函数，AC,给大家分享一下吧。。。
#include<stdio.h>#include<stdlib.h>typedef struct Node{    struct Node *lc,*rc;//左右孩子指针    int left,right;//左右区间边界    int len;//区间整点个数 }Tire,*Interval_tire;//定义线段树结构void Build(Interval_tire &T,int left,int right){    int mid;        T=(Interval_tire)malloc(sizeof(Tire));    T->lc=NULL;    T->rc=NULL;    T->left=left;    T->right=right;    T->len=right-left+1;    if(left==right)return;//递归出口    mid=(left+right)>>1;//左移一位，相当于除以2     Build(T->lc,left,mid);    Build(T->rc,mid+1,right);} int Search(Interval_tire T,int key){    T->len--;    if(T->left==T->right)    return T->left;//如果左右端点值相等返回     if(T->lc->len>=key)//左孩子区间长度不小于key表示做区间可容纳至少key个数     return Search(T->lc,key);    return Search(T->rc,key-T->lc->len);}void Ftree(Interval_tire &T){    if(T!=NULL)    {        if(T->lc!=NULL)        Ftree(T->lc);        if(T->rc!=NULL)        Ftree(T->rc);        free(T);    }    return ;}     int main(){    int t,n,k,m=1;    scanf("%d",&t);    while(t--)    {        int i,p;        __int64 sum=0;        Interval_tire T;        scanf("%d %d",&n,&k);         Build(T,1,n);//构建从1到n的线段树         for(i=1;i<=k;i++)        {            scanf("%d",&p);            sum+=Search(T,p);//查找第p小的数并累加到sum        }        printf("Case %d: %I64d\n",m++,sum);        Ftree(T);    }//while   // system("pause");    return 0;}

以下是非链式存储的实现方式，果断AC,
利用结构体数组模拟线段二叉树的构建，
关键要搞明白当前结构体数组的下标乘以2代表的是其对应的左孩子，再加1以后代表的是其又孩子
#include<stdio.h>#include<stdlib.h>#define N 262144struct{    int left,right;//左右区间边界    int len;//区间整点个数 }Tire[3*N];void Build(int p,int left,int right){    int mid,w;    Tire[p].left=left;    Tire[p].right=right;    Tire[p].len=right-left+1;    if(left==right)return;//递归出口    mid=(left+right)>>1;//左移一位，相当于除以2    w=p<<1;//右移一位相当于乘以2，此时代表左孩子节点位置！！！！！！！！     Build(w,left,mid);    Build(w+1,mid+1,right);} int Search(int p,int key){    int w;    Tire[p].len--;    if(Tire[p].left==Tire[p].right)    return Tire[p].left;//如果左右端点值相等返回     w=p<<1;    if(Tire[w].len>=key)//左孩子区间长度不小于key表示做区间可容纳至少key个数     return Search(w,key);    return Search(w+1,key-Tire[w].len);}     int main(){    int t,n,k,m=1;    scanf("%d",&t);    while(t--)    {        int i,p;        __int64 sum=0;        scanf("%d %d",&n,&k);         Build(1,1,n);//构建从1到n的线段树         for(i=1;i<=k;i++)        {            scanf("%d",&p);            sum+=Search(1,p);//查找第p小的数并累加到sum        }        printf("Case %d: %I64d\n",m++,sum);    }//while   // system("pause");    return 0;}