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 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
题意:有一种排序,规则为如果相邻两数左比右大就交换他们,求最小交换次数?
题解:显然最小次数为逆序对数,至于逆序对,可以归并排序求,也可以树状数组/线段树求,自然是选择简单的喽!
代码如下:
#include<queue> #include<string> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define lson root<<1 #define rson root<<1|1 #define hi puts("hi!"); using namespace std; struct node { int kd,val; }a[500005]; int n,m,cnt[500050]; long long tr[2000050]; bool cmp(node a,node b) { return a.val<b.val; } void push_up(int root) { tr[root]=tr[lson]+tr[rson]; } void build(int root,int l,int r) { if(l==r) { tr[root]=0; return; } int mid=(l+r)>>1; build(lson,l,mid); build(rson,mid+1,r); push_up(root); } void add(int root,int l,int r,int x,int p) { if(l==r) { tr[root]=1; return; } int mid=(l+r)>>1; if(p<=mid) { add(lson,l,mid,x,p); } if(p>mid) { add(rson,mid+1,r,x,p); } push_up(root); } long long query(int root,int l,int r,int x,int y) { long long ans=0; if(x<=l&&y>=r) { return tr[root]; } int mid=(l+r)>>1; if(x<=mid) { ans+=query(lson,l,mid,x,y); } if(y>mid) { ans+=query(rson,mid+1,r,x,y); } return ans; } int main() { while(scanf("%d",&n)==1&&n) { long long ans1=0; memset(tr,0,sizeof(tr)); build(1,1,n); for(int i=1;i<=n;i++) { scanf("%d",&a[i].val); a[i].kd=i; } sort(a+1,a+n+1,cmp); for(int i=1;i<=n;i++) { cnt[a[i].kd]=i; } for(int i=n;i>=1;i--) { ans1+=query(1,1,n,1,cnt[i]); add(1,1,n,1,cnt[i]); } printf("%lld\n",ans1); } }
