poj 2299(离散化+树状数组)

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 53777		Accepted: 19766

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

5
9
1
0
5
4
3
1
2
3
0

Sample Output

6
0

Source

题意:求解一个串的逆序数的个数是多少？？

题解:离散化数组变成下标，然后每次将离散化的下标放进树状数组，放进去之后统计小于他的数的个数是多少。用 i - getsum(a[i])即为大于它的数的个数，其中 i 为当前已经插入的数的个数。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <vector>
#include <algorithm>
using namespace std;
const int N = ;

struct Node{
    int v,id;
}node[N];
int a[N],c[N],n;
int lowbit(int x){
    return x&(-x);
}
void update(int idx,int v){
    for(int i=idx;i<=N;i+=lowbit(i)){
        c[i]+=v;
    }
}
int getsum(int idx){
    int sum = ;
    for(int i=idx;i>=;i-=lowbit(i)){
        sum+=c[i];
    }
    return sum;
}
int cmp(Node a,Node b){
    return a.v<b.v;
}
int main()
{
    while(scanf("%d",&n)!=EOF,n){
        memset(c,,sizeof(c));
        for(int i=;i<=n;i++){
            scanf("%d",&node[i].v);
            node[i].id = i;
        }
        sort(node+,node+n+,cmp);
        for(int i=;i<=n;i++){
            a[node[i].id] = i;
        }
        long long  cnt = ;
        for(int i=;i<=n;i++){
            update(a[i],);
            cnt=cnt+ i - getsum(a[i]);
        }
        printf("%lld\n",cnt);
    }
    return ;
}