POJ训练计划2299_Ultra-QuickSort(线段树/单点更新)

解题报告

题意：

求逆序数。

思路：

线段树离散化处理。

#include <algorithm>#include <iostream>#include <cstring>#include <cstdio>#define LL long longusing namespace std;LL sum[2001000],num[501000],_hash[501000];void push_up(int rt){    sum[rt]=sum[rt*2]+sum[rt*2+1];}void update(int rt,int l,int r,int p,LL v){    if(l==r)    {        sum[rt]=+v;        return;    }    int mid=(l+r)/2;    if(p<=mid)update(rt*2,l,mid,p,v);    else update(rt*2+1,mid+1,r,p,v);    push_up(rt);}LL q_sum(int rt,int l,int r,int ql,int qr){    if(ql>r||qr<l)return 0;    if(ql<=l&&r<=qr)return sum[rt];    int mid=(l+r)/2;    return q_sum(rt*2,l,mid,ql,qr)+q_sum(rt*2+1,mid+1,r,ql,qr);}int main(){    int n,i,j;    while(~scanf("%d",&n))    {        LL ans=0;        if(!n)break;        memset(_hash,0,sizeof(_hash));        memset(num,0,sizeof(num));        memset(sum,0,sizeof(sum));        for(i=0; i<n; i++)        {            scanf("%LLd",&num[i]);            _hash[i]=num[i];        }        sort(_hash,_hash+n);        int m=unique(_hash,_hash+n)-_hash;        for(i=0; i<n; i++)        {            int t=lower_bound(_hash,_hash+m,num[i])-_hash+1;            ans+=q_sum(1,1,m,t+1,m);            update(1,1,m,t,1);        }        printf("%lld\n",ans);    }    return 0;}

Ultra-QuickSort

Time Limit: 7000MS Memory Limit: 65536KTotal Submissions: 41278 Accepted: 14952

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

60

Source

Waterloo local 2005.02.05