poj2299--B - Ultra-QuickSort（线段树，离散化）

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Ultra-QuickSort

Time Limit: 7000MS Memory Limit: 65536KTotal Submissions: 41215 Accepted: 14915

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

题意：求出最小交换次数，使得数组变得有序（由小到大）。也就是求出逆序数，这个题归并排序，树状数组，线段树又可以解，可以当做练手用

使用线段树解的话，因为给出的数值很大，所以需要离散化，一开始直接使用map果断的超时了，只能想一个更简便的离散的方法，使用结构体存下一开始的值k和它初始的序号（id1），sort对k进行排序，得到新的序号（id2），通过id1直接改变给出的数组，变为id2，这样只用n + logn的时间就可以离散完成，（注意要判断重复的值，重复的值共享一个id2）。至于线段树部分就是一个模板

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define LL __int64#define maxn 600000#define lmin 1#define rmax n#define lson l,(l+r)/2,rt<<1#define rson (l+r)/2+1,r,rt<<1|1#define root lmin,rmax,1#define now l,r,rt#define int_now LL l,LL r,LL rtstruct node{    LL id1 , id2 ;    LL k ;}p[maxn] ;LL cl[maxn<<2] , a[maxn] ;bool cmp(node a,node b){    return a.k < b.k ;}void push_up(int_now){    cl[rt] = cl[rt<<1] + cl[rt<<1|1] ;}void update(LL i,int_now){    if( i < l || i > r )        return ;    if( i == l && i==r )    {        cl[rt]++ ;        return ;    }    update(i,lson);    update(i,rson);    push_up(now);    return ;}LL query(int ll,int rr,int_now){    if( ll > r || rr < l )        return 0;    if( ll <= l && r <= rr )        return cl[rt] ;    return query(ll,rr,lson) + query(ll,rr,rson);}int main(){    LL i , n , m , l , r , x , num ;    while(scanf("%I64d", &m) && m)    {        for(i = 0 ; i < m ; i++)        {            scanf("%I64d", &a[i]);            p[i].k = a[i] ;            p[i].id1 = i ;        }        sort(p,p+m,cmp);        int temp = -1 ;        n = 0 ;        for(i = 0 ; i < m ; i++)        {            if( p[i].k == temp )                p[i].id2 = n ;            else            {                p[i].id2 = ++n ;                temp = p[i].k ;            }        }        for(i = 0 ; i < m ; i++)            a[ p[i].id1 ] = p[i].id2 ;        memset(cl,0,sizeof(cl));        num = 0 ;        for(i = 0 ; i < m ; i++)        {            num += (i - query(1,a[i],root));            update(a[i],root);        }        printf("%I64d\n", num);    }    return 0;}

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