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 = 500005;

struct Node{
int v,id;
}node
;
int a
,c
,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 = 0;
for(int i=idx;i>=1;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,0,sizeof(c));
for(int i=1;i<=n;i++){
scanf("%d",&node[i].v);
node[i].id = i;
}
sort(node+1,node+n+1,cmp);
for(int i=1;i<=n;i++){
a[node[i].id] = i;
}
long long  cnt = 0;
for(int i=1;i<=n;i++){
update(a[i],1);
cnt=cnt+ i - getsum(a[i]);
}
printf("%lld\n",cnt);
}
return 0;
}