[樹狀數組]POJ 2299 Ultra-QuickSort

傳送門：Ultra-QuickSort

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 37112		Accepted: 13350

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

解題報告：

這題用樹狀數組，需要離散化纔行。。
離散化的步驟就是， 先用一個結構體，用num 保存 原來的數，id保存原來數組所在的位置。。然後按num排序，那麼得到一個從小到大有序的序列。 然後通過id找到原數組所在的位置，並用另一個數組存放 通過映射得到的新的類似原來序列的數組。

代碼如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 500010
using namespace std;
struct node{
    int num,id;
}s[maxn];
int ans[maxn],result[maxn];
bool cmp(const node &a,const node &b){
    return a.num<b.num;
}
int lowbit(int n){   //lowbit
    return n & (-n);
}
void modify(int x){  //修改modify
    for(int i = x; i<maxn; i+=lowbit(i))
        ans[i]++;
}
int getSum(int x){  //求和sum
    long long sum= 0;
    for (int i = x; i > 0; i -= lowbit(i))
        sum += ans[i];
    return sum;
}
int main(){
    int n;
    while(scanf("%d",&n)==1&&n){
        memset(ans,0,sizeof(ans));
        for(int i=1;i<=n;i++){
            scanf("%d",&s[i].num);
            s[i].id=i;
        }
        sort(s+1,s+n+1,cmp);
        result[s[1].id]=1;
        for(int i=2;i<=n;i++){
            if(s[i].num==s[i-1].num)
                result[s[i].id]=result[s[i-1].id];
            else
                result[s[i].id]=i;
        }
        long long sum=0;
        for(int i=n;i>=1;i--){
            sum+=getSum(result[i]-1);
            modify(result[i]);
        }
        printf("%lld\n",sum);
    }
    return 0;
}