poj 2299 Ultra-QuickSort（求逆序对）

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 52778		Accepted: 19348

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

题意：给一组数，每次只能交换相邻两个数的位置，问最少交换多少次，使得这个序列从小到大排序；
题解：求逆序数的对数

线段树：

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#include<stdio.h> #include<string.h> #include<algorithm> #define LL long long #define MAX 500500 using namespace std; int n,m; struct node {     int val,pos; }num[MAX]; int sum[MAX<<2]; void pushup(int o) {     sum[o]=sum[o<<1]+sum[o<<1|1]; } void gettree(int o,int l,int r) {     sum[o]=0;     if(l==r)         return ;     int mid=(l+r)>>1;     gettree(o<<1,l,mid);     gettree(o<<1|1,mid+1,r);     pushup(o); }   void update(int o,int l,int r,int pos) {     if(l==r)     {         sum[o]+=1;         return ;     }     int mid=(l+r)>>1;     if(pos<=mid) update(o<<1,l,mid,pos);     else update(o<<1|1,mid+1,r,pos);     pushup(o); } int find(int o,int l,int r,int L,int R) {     if(L<=l&&R>=r)     return sum[o];     int mid=(l+r)>>1;     int ans=0;     if(L<=mid) ans+=find(o<<1,l,mid,L,R);     if(R>mid) ans+=find(o<<1|1,mid+1,r,L,R);     return ans; } bool cmp(node a,node b) {     if(a.val!=b.val) return a.val<b.val;     else return a.pos<b.pos; } int main() {     int j,i,t,k;     while(scanf("%d",&n),n)     {         gettree(1,1,n);         for(i=1;i<=n;i++)         {             scanf("%d",&num[i].val);             num[i].pos=i;         }         sort(num+1,num+1+n,cmp);         LL ant=0;         for(i=1;i<=n;i++)         {             update(1,1,n,num[i].pos);             ant+=i-find(1,1,n,1,num[i].pos);         }         printf("%lld\n",ant);     }     return 0; }

　　树状数组：

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#include<stdio.h> #include<string.h> #include<algorithm> #define LL long long #define MAX 500500 using namespace std; int n,m; int c[MAX<<1]; struct node {     int val,pos; }num[MAX]; bool cmp(node a,node b) {     if(a.val!=b.val)         return a.val<b.val;     else         return a.pos<b.pos; } void update(int x) {     while(x<=n)     {         c[x]+=1;         x+=x&-x;     } } LL sum(int x) {     LL ans=0;     while(x>=1)     {         ans+=c[x];         x-=x&-x;     }     return ans; } int main() {     int j,i,t,k;     while(scanf("%d",&n),n)     {         for(i=1;i<=n;i++)         {             scanf("%d",&num[i].val);             num[i].pos=i;         }         sort(num+1,num+n+1,cmp);         memset(c,0,sizeof(c));         LL ant=0;         for(i=1;i<=n;i++)         {             update(num[i].pos);             ant+=i-sum(num[i].pos);         }         printf("%lld\n",ant);     }     return 0; }