poj 2299 Ultra-QuickSort 【线段树求和（点更新）】

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 50810		Accepted: 18618

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

Waterloo local 2005.02.05

思路：

先将需要排序的数输入，然后用sort 进行排序（从大到小），然后将第一个数对应的位置标记为1，然后在这个数的前面的位置找看有没有为1的数，如果有的话，就加到ans 上；然后这个数前面的区间对应的数组的值（也就是使包含这个数的区间都加1），

遍历完之后，ans保存的就是你需要找的值了！

代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define MAX 500000+10
typedef long long ll;
using namespace std;
struct node
{
int val,pos;
}a[MAX];
int sum[MAX<<2];
int cmp(node a,node b)
{
return a.val>b.val;
}
void build(int o,int l,int r)
{
sum[o]=0;
if(l==r)
return;
int mid=(l+r)>>1;
build(o<<1,l,mid);
build(o<<1|1,mid+1,r);
}
void updata(int o,int l,int r,int L)
{
sum[o]+=1;
if(l==r)
return;
int mid=(l+r)>>1;
if(L<=mid)
updata(o<<1,l,mid,L);
else
updata(o<<1|1,mid+1,r,L);
}
int query(int o,int l,int r,int L,int R)
{
if(L == l && R == r)
{
return sum[o];
}
int mid = (l+r) >> 1;
if(R <= mid)
{
return query(o<<1,l,mid,L,R);
}
else if(L > mid)
{
return query(o<<1|1,mid+1,r,L,R);
}
else
{
return query(o<<1,l,mid,L,mid)+query(o<<1|1,mid+1,r,mid+1,R);
}
}
int main()
{
int n;
while(scanf("%d",&n),n)
{
for(int i=0;i<n;i++)
{
scanf("%d",&a[i].val);
a[i].pos=i+1;
}
build(1,1,n);
sort(a,a+n,cmp);
ll ans=0;
for(int i=0;i<n;i++)
{
updata(1,1,n,a[i].pos);
if(a[i].pos==1)
continue;
ans+=query(1,1,n,1,a[i].pos-1);
}
printf("%lld\n",ans);
}
return 0;
}