[POJ2299] Ultra-QuickSort - 逆序对

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题目描述

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.

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输入格式

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.

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输出格式

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.

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样例数据

样例输入

5
9
1
0
5
4
3
1
2
3
0

样例输出

6
0

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题目分析

求逆序对。
一个一个插入，统计比他大的。
我为了方便，倒序插入统计小的。
a[]太大，需要离散化。
注意开long long

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源代码

#include<algorithm>#include<iostream>#include<iomanip>#include<cstring>#include<cstdlib>#include<vector>#include<cstdio>#include<cmath>#include<queue>#include<map>using namespace std;inline const int Get_Int() {    int num=0,bj=1;    char x=getchar();    while(x<'0'||x>'9') {        if(x=='-')bj=-1;        x=getchar();    }    while(x>='0'&&x<='9') {        num=num*10+x-'0';        x=getchar();    }    return num*bj;}const int maxn=500005;struct BIT { //树状数组    long long n,c[maxn];    inline int Lowbit(int x) { //低位操作        return x&(-x);    }    void init(int n) {        this->n=n;        memset(c,0,sizeof(c));    }    void add(int x,int v) {        for(int i=x; i<=n; i+=Lowbit(i))c[i]+=v;    }    long long sum(int x) { //求出1~s的区间和        long long s=0;        for(int i=x; i; i-=Lowbit(i))s+=c[i];        return s;    }} bit;map<long long,long long>M;map<long long,long long>::iterator it;long long n,a[500005],b[500005];void Discretization() { //a是待离散数组 b是离散后数组    M.clear();    memset(b,0,sizeof(b));    for(int i=1; i<=n; i++)M[a[i]]=1;     int i=1;    for(it=M.begin(); it!=M.end(); it++,i++)it->second=i;    for(int i=1; i<=n; i++)b[i]=M[a[i]];}long long ans=0;int main() {    while(true) {        n=Get_Int();        if(n==0)break;        ans=0;        bit.init(n);        for(int i=1; i<=n; i++)a[i]=Get_Int();        Discretization();        for(int i=n; i>=1; i--) {            bit.add(b[i],1);            ans+=bit.sum(b[i]-1);        }        printf("%lld\n",ans);    }    return 0;}

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