poj 2299 线段树

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 54438		Accepted: 20002

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

转化为线段树：

我们先将原数组每个值附上一个序号index，再将它排序。如题目的例子：

num：   9  1  0  5  4

index：  1  2  3  4  5

排序后：

num：   0  1  4  5  9

index：  3  2  5  4  1

然后由于排序后num为0的点排在原来数组的第3个，所以为了将它排到第一个去，那就至少需要向前移动两次，同时它也等价于最小的数0之前有2个数比它大（所以要移动两次），将0移到它自己的位置后，我们将0删掉（目的是为了不对后面产生影响）。再看第二大的数1，它出现在原数组的第二个，他之前有一个数比它大所以需要移动一次。这样一直循环下去那么着5个数所需要移动的次数就是：

num：  0  1  4  5  9

次数      2  1  2  1  0

将次数全部要加起来就是最后所需要移动的总次数

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;

const int maxx=500010;
#define LL long long
int sum[maxx<<2];

struct record
{
int value;
int pos;
}a[maxx];

bool cmp(record x,record y)
{
return x.value>y.value;
}

//------------------------------------//
void build(int p,int l,int r)
{
sum[p]=0;
if(l==r)
return ;

int mid=(l+r)>>1;
build(p<<1,l,mid);
build(p<<1|1,mid+1,r);
}
//------------------------------------//

void update(int p,int l,int r,int m)
{
sum[p]++;
if(l==r)
return ;

int mid=(l+r)>>1;
if(m<=mid)
update(p<<1,l,mid,m);
else
update(p<<1|1,mid+1,r,m);
}

//------------------------------------//

int query(int p,int l,int r,int left,int right)
{
if(left==l&&r==right)
return sum[p];

int mid=(l+r)>>1;
if(right<=mid)
return query(p<<1,l,mid,left,right);
else if(left>mid)
return query(p<<1|1,mid+1,r,left,right);
else
return query(p<<1,l,mid,left,mid)+query(p<<1|1,mid+1,r,mid+1,right);
}

//------------------------------------//
int main()
{
int n;
//freopen("in.txt","r",stdin);
while(scanf("%d",&n),n)
{
build(1,1,n);
for(int i=0;i<n;i++){
scanf("%d",&a[i]);
a[i].pos=i+1;
}
sort(a,a+n,cmp);

LL ans=0;
for(int i=0;i<n;i++){
update(1,1,n,a[i].pos);
if(a[i].pos==1)
continue;
ans+=query(1,1,n,1,a[i].pos-1);
}
printf("%I64d\n",ans);
}
return 0;
}