poj 2299 线段树

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Ultra-QuickSort

Time Limit: 7000MS Memory Limit: 65536KTotal Submissions: 54438 Accepted: 20002

Description

In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. For the input sequence
9 1 0 5 4 ,
Ultra-QuickSort produces the output
0 1 4 5 9 .
Your task is to determine how many swap operations Ultra-QuickSort needs to perform in order to sort a given input sequence.

Input

The input contains several test cases. Every test case begins with a line that contains a single integer n < 500,000 -- the length of the input sequence. Each of the the following n lines contains a single integer 0 ≤ a[i] ≤ 999,999,999, the i-th input sequence element. Input is terminated by a sequence of length n = 0. This sequence must not be processed.

Output

For every input sequence, your program prints a single line containing an integer number op, the minimum number of swap operations necessary to sort the given input sequence.

Sample Input

59105431230

Sample Output

转化为线段树：

我们先将原数组每个值附上一个序号index，再将它排序。如题目的例子：

num： 9 1 0 5 4
index： 1 2 3 4 5

排序后：

num： 0 1 4 5 9
index： 3 2 5 4 1

然后由于排序后num为0的点排在原来数组的第3个，所以为了将它排到第一个去，那就至少需要向前移动两次，同时它也等价于最小的数0之前有2个数比它大（所以要移动两次），将0移到它自己的位置后，我们将0删掉（目的是为了不对后面产生影响）。再看第二大的数1，它出现在原数组的第二个，他之前有一个数比它大所以需要移动一次。这样一直循环下去那么着5个数所需要移动的次数就是：

num： 0 1 4 5 9
次数 2 1 2 1 0

将次数全部要加起来就是最后所需要移动的总次数

#include<stdio.h>#include<algorithm>#include<string.h>using namespace std;const int maxx=500010;#define LL long longint sum[maxx<<2];struct record{    int value;    int pos;}a[maxx];bool cmp(record x,record y){    return x.value>y.value;}//------------------------------------//void build(int p,int l,int r){    sum[p]=0;    if(l==r)        return ;    int mid=(l+r)>>1;    build(p<<1,l,mid);    build(p<<1|1,mid+1,r);}//------------------------------------//void update(int p,int l,int r,int m){    sum[p]++;    if(l==r)        return ;    int mid=(l+r)>>1;    if(m<=mid)        update(p<<1,l,mid,m);    else        update(p<<1|1,mid+1,r,m);}//------------------------------------//int query(int p,int l,int r,int left,int right){    if(left==l&&r==right)        return sum[p];    int mid=(l+r)>>1;    if(right<=mid)        return query(p<<1,l,mid,left,right);    else if(left>mid)        return query(p<<1|1,mid+1,r,left,right);    else        return query(p<<1,l,mid,left,mid)+query(p<<1|1,mid+1,r,mid+1,right);}//------------------------------------//int main(){    int n;    //freopen("in.txt","r",stdin);    while(scanf("%d",&n),n)    {        build(1,1,n);        for(int i=0;i<n;i++){            scanf("%d",&a[i]);            a[i].pos=i+1;        }        sort(a,a+n,cmp);        LL ans=0;        for(int i=0;i<n;i++){            update(1,1,n,a[i].pos);            if(a[i].pos==1)                continue;            ans+=query(1,1,n,1,a[i].pos-1);        }        printf("%I64d\n",ans);    }    return 0;}

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