Problem 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
很有意思的一道题,首先看图片就像一个通马桶的工具。
这里只提及树状数组。
这道题考的考点是数据的处理,逆序数。
所以思路来了,逆序数不就比大小吗,直接就标上序号,来一个排序加上数据处理,OK!
数据处理的方法网上一般称之为离散化,我对离散化的理解就是简化问题,使一个连续(不可解)的问题变得离散(可解)。
本题考的就是数据,直接加和会爆炸。于是用123......n,表示这些数的价值就变得可解,这个过程算是离散化。
#include<iostream> #include<cstdio> #include<algorithm> #include<cstring> using namespace std; #define MAX 500050 typedef long long ll; int a[MAX]; int c[MAX]; struct node { int num; ll v; bool operator < (const node &b ) const //重载一下运算符,这里的const可加可不加,对于不同编译器是有区别的 { return v<b.v; } }b[MAX]; int lowbit(int i) { return i&(-i); } void add(int x,int v) { while(x<=MAX) { c[x]+=v; x+=lowbit(x); } } int sum(int x) { int res=0; while(x>0) { res+=c[x]; x-=lowbit(x); } return res; } int main() { int n; while(scanf("%d",&n),n) { for(int i=1;i<=n;i++) { scanf("%d",&b[i].v); b[i].num=i; } sort(b+1,b+n+1); //值排序 memset(a,0,sizeof(a)); a[b[1].num]=1; //对于最小值当然标最小啦 ll ans=0; for(int i=2;i<=n;i++) { if(b[i].v==b[i-1].v) a[b[i].num]=a[b[i-1].num]; else a[b[i].num]=i; // 记录前面比他小的数。 } memset(c,0,sizeof(c)); for(int i=1;i<=n;i++) { add(a[i],1); ans+=sum(n)-sum(a[i]); } printf("%lld\n",ans); } }