Ultra-QuickSort
Time Limit: 7000MS |
|
Memory Limit: 65536K |
Total Submissions: 22147 |
|
Accepted: 7897 |
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
此题大意:找逆序数有多少个,逆序数就是那一个序列中,右边的数比左边的数大的个数之和。其实就是和找冒泡的次数一样,不过用冒泡来做,50万个数据直接死亡,这道题可以用树状数组来做,之前不会做,看到解题报告中有说树状数组可以,就开始尝试写,后来发现竟然可以这样想:每次找前面(左边)的数比自己大的个数的总和!!!这样子想的话树状数组可以轻松解决。
不过还有一个问题,就是他输入的数据的值很大0 ≤ a[i] ≤ 999,999,999,但是数据的数量不是很大n < 500,000 ,所以用到离散化解决!!!
链接:http://poj.org/problem?id=2299
代码:
#include <iostream>
#include <stdio.h>
#include <cmath>
#include <algorithm>
using namespace std;
int b[500005], c[500005];
int n;
struct node
{
int num, id;
}a[500005];
bool cmp(node a, node b)
{
return a.num < b.num;
}
void update(int i, int x)
{
while(i <= n)
{
c[i] += x;
i += i&(-i);
}
}
int sum(int i)
{
int sum = 0;
while(i > 0)
{
sum += c[i];
i -= i&(-i);
}
return sum;
}
int main()
{
int i;
long long ans;
while(scanf("%d", &n), n)
{
memset(b, 0, sizeof(b));
memset(c, 0, sizeof(c));
for(i = 1; i <= n; i++)
{
scanf("%d", &a[i].num); //输入数值num
a[i].id = i; //记录序号id
}
///开始离散化
sort(a+1, a+n+1, cmp); //先排序
/*因为a[1].num是最小的,id是它的位置,所以b[a[1].id]=1最小,
最小的数变成1,第二小的变成2,如此类推从而达到离散化*/
b[a[1].id] = 1;
for(i = 2; i <= n; i++)
{
if(a[i].num != a[i-1].num)
b[a[i].id] = i;
else b[a[i].id] = b[a[i-1].id];
}
///离散化完毕
ans = 0;
for(i = 1; i <= n; i++)
{
update(b[i], 1);
//这里很巧妙,每一次更新后,判断此数比左边的数小的数有多少
ans += (sum(n)-sum(b[i]));
}
//从而求到:右边的数比左边的数大的个数的总和
printf("%I64d\n", ans);
}
return 0;
}
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