POJ 2299 离散化线段树

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Ultra-QuickSort

Time Limit: 7000MS Memory Limit: 65536KTotal Submissions: 40827 Accepted: 14752

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

求冒泡排序交换的次数。

由于这些数可能太大，且差距很大，所以离散化一下，然后求一下逆序数，边查询边插入边即可。

//32684K1579MS#include<stdio.h>#include<string.h>#include<algorithm>#define M 500007#define ll __int64using namespace std;int s[M],n;struct Tree{    int l,r,mid;    ll val;}tree[M<<1];struct sa{    int id;    ll val;}p[M*2];int cmp(sa a,sa b){    return a.val>b.val;}void build(int left,int right,int i){    tree[i].l=left;tree[i].r=right;tree[i].mid=(left+right)>>1;tree[i].val=0;    if(left==right){return;}    build(left,tree[i].mid,i*2);    build(tree[i].mid+1,right,i*2+1);}int query(int x,int i){    if(tree[i].l==tree[i].r)return tree[i].val;    if(x<=tree[i].mid)return query(x,i*2)+tree[i].val;    else return query(x,i*2+1)+tree[i].val;}void insert(int left,int right,int i){    if(tree[i].l==left&&tree[i].r==right){tree[i].val++;return;}    if(right<=tree[i].mid)insert(left,right,2*i);    else if(left>tree[i].mid)insert(left,right,2*i+1);    else {insert(left,tree[i].mid,i*2);insert(tree[i].mid+1,right,i*2+1);}}void discretization(){    int tmp=p[1].val,pos=1;    for(int i=1;i<=n;i++)        if(p[i].val!=tmp)p[i].val=++pos,tmp=p[i].val;        else p[i].val=pos;    for(int i=1;i<=n;i++)        s[p[i].id]=p[i].val;}int main(){    while(scanf("%d",&n)&&n)    {        ll ans=0;        build(0,M,1);        memset(s,0,sizeof(s));        for(int i=1;i<=n;i++)        {            scanf("%I64d",&p[i].val);            p[i].id=i;        }        sort(p+1,p+n+1,cmp);        discretization();        for(int i=1;i<=n;i++)            printf("%d ",s[i]);        printf("\n");        for(int i=1;i<=n;i++)        {                ans+=query(s[i],1);                insert(s[i],M,1);        }        printf("%I64d\n",ans);    }    return 0;}