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.
Ultra-QuickSort produces the output
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
思路:求逆序数;用树状数组?开不了c[999,999,999]这么大的数组;离散化即可;
AC代码:
#include <iostream> #include<cstdio> #include<cstring> #include<algorithm> #define lowbit(x) x&(-x) typedef long long ll; using namespace std; int n; int b[500010]; int c[500010]; struct A{ int val,ind; }a[500010]; bool cmp(A a,A b){ return a.val<b.val; } void add(int x,int val){ for(int i=x;i<=n;i+=lowbit(i)){ c[i]+=val; } } int getsum(int x){ int ret=0; for(int i=x;i>0;i-=lowbit(i)){ ret+=c[i]; } return ret; } int main() { while(scanf("%d",&n)!=EOF&&n){for(int i=1;i<=n;i++) {scanf("%d",&a[i].val); a[i].ind=i;} sort(a+1,a+1+n,cmp); for(int i=1;i<=n;i++) b[a[i].ind]=i;//离散化 ll ans=0; memset(c,0,sizeof(c)); for(int i=1;i<=n;i++){ add(b[i],1); ans+=(i-getsum(b[i])); } cout<<ans<<endl; } return 0; }