poj 2299 Ultra-QuickSort（求逆序对）

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 52778		Accepted: 19348

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

题意：给一组数，每次只能交换相邻两个数的位置，问最少交换多少次，使得这个序列从小到大排序；
题解：求逆序数的对数

线段树：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define LL long long
#define MAX 500500
using namespace std;
int n,m;
struct node
{
int val,pos;
}num[MAX];
int sum[MAX<<2];
void pushup(int o)
{
sum[o]=sum[o<<1]+sum[o<<1|1];
}
void gettree(int o,int l,int r)
{
sum[o]=0;
if(l==r)
return ;
int mid=(l+r)>>1;
gettree(o<<1,l,mid);
gettree(o<<1|1,mid+1,r);
pushup(o);
}

void update(int o,int l,int r,int pos)
{
if(l==r)
{
sum[o]+=1;
return ;
}
int mid=(l+r)>>1;
if(pos<=mid) update(o<<1,l,mid,pos);
else update(o<<1|1,mid+1,r,pos);
pushup(o);
}
int find(int o,int l,int r,int L,int R)
{
if(L<=l&&R>=r)
return sum[o];
int mid=(l+r)>>1;
int ans=0;
if(L<=mid) ans+=find(o<<1,l,mid,L,R);
if(R>mid) ans+=find(o<<1|1,mid+1,r,L,R);
return ans;
}
bool cmp(node a,node b)
{
if(a.val!=b.val) return a.val<b.val;
else return a.pos<b.pos;
}
int main()
{
int j,i,t,k;
while(scanf("%d",&n),n)
{
gettree(1,1,n);
for(i=1;i<=n;i++)
{
scanf("%d",&num[i].val);
num[i].pos=i;
}
sort(num+1,num+1+n,cmp);
LL ant=0;
for(i=1;i<=n;i++)
{
update(1,1,n,num[i].pos);
ant+=i-find(1,1,n,1,num[i].pos);
}
printf("%lld\n",ant);
}
return 0;
}

　　树状数组：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define LL long long
#define MAX 500500
using namespace std;
int n,m;
int c[MAX<<1];
struct node
{
int val,pos;
}num[MAX];
bool cmp(node a,node b)
{
if(a.val!=b.val)
return a.val<b.val;
else
return a.pos<b.pos;
}
void update(int x)
{
while(x<=n)
{
c[x]+=1;
x+=x&-x;
}
}
LL sum(int x)
{
LL ans=0;
while(x>=1)
{
ans+=c[x];
x-=x&-x;
}
return ans;
}
int main()
{
int j,i,t,k;
while(scanf("%d",&n),n)
{
for(i=1;i<=n;i++)
{
scanf("%d",&num[i].val);
num[i].pos=i;
}
sort(num+1,num+n+1,cmp);
memset(c,0,sizeof(c));
LL ant=0;
for(i=1;i<=n;i++)
{
update(num[i].pos);
ant+=i-sum(num[i].pos);
}
printf("%lld\n",ant);
}
return 0;
}