HDU 1394：Minimum Inversion Number（线段树区间求和单点更新）

http://acm.hdu.edu.cn/showproblem.php?pid=1394

Minimum Inversion Number

Problem Description

 

The inversion number of a given number sequence a1, a2, ..., an is the number of pairs (ai, aj) that satisfy i < j and ai > aj.

For a given sequence of numbers a1, a2, ..., an, if we move the first m >= 0 numbers to the end of the seqence, we will obtain another sequence. There are totally n such sequences as the following:

a1, a2, ..., an-1, an (where m = 0 - the initial seqence)
a2, a3, ..., an, a1 (where m = 1)
a3, a4, ..., an, a1, a2 (where m = 2)
...
an, a1, a2, ..., an-1 (where m = n-1)

You are asked to write a program to find the minimum inversion number out of the above sequences.

 

Input

 

The input consists of a number of test cases. Each case consists of two lines: the first line contains a positive integer n (n <= 5000); the next line contains a permutation of the n integers from 0 to n-1.

 

Output

 

For each case, output the minimum inversion number on a single line.

 

Sample Input

 

10
1 3 6 9 0 8 5 7 4 2

 

Sample Output

 

16

 

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 using namespace std;
 #define N 5005
 #define lson rt<<1,l,m
 #define rson rt<<1|1,m+1,r
 struct node
 {
     int l,r,value;
 }tree[N<<];
 int a[N];
 /*
 求移位后的最小逆序数
 Update单点增减
 Query区间求和
 */
 void Build(int rt,int l,int r)
 {
     tree[rt].l=l;
     tree[rt].r=r;
     tree[rt].value=;
     if(l==r) return ;
     int m=(l+r)>>;
     Build(rt<<,l,m);
     Build(rt<<|,m+,r);
 }

 void Update(int rt,int num)
 {
     if(tree[rt].l==num&&tree[rt].r==num){
         tree[rt].value=;
         return ;
     }
     int m=(tree[rt].l+tree[rt].r)>>;
     if(num<=m) Update(rt<<,num);
     else Update(rt<<|,num);
     tree[rt].value=tree[rt<<].value+tree[rt<<|].value;
 }

 int Query(int rt,int l,int r)
 {
 /*
 画个图就明白了，搜寻的区间为[l,r]，m代表当前节点的左儿子和右儿子的分支
 如果左儿子有包含该区间的元素，即l<=m，那么就要继续递归搜寻左儿子
 如果右儿子有包含该区间的元素，即m<r,那么就要继续递归搜寻右儿子
 当当前的节点的区间被搜寻的区间[l,r]覆盖的话，就说明该段区间完全在[l,r]里面
 那么返回该节点的值
 */
     if(l<=tree[rt].l&&tree[rt].r<=r){
         return tree[rt].value;
     }
     else{
         int m=(tree[rt].l+tree[rt].r)>>;
         int s1=,s2=;
         if(l<=m) s1=Query(rt<<,l,r);
         if(r>m) s2=Query(rt<<|,l,r);
         return s1+s2;
     }
 }

 int main()
 {
     int n;
     while(~scanf("%d",&n)){
         Build(,,n);
         int sum=;
         for(int i=;i<n;i++){
             scanf("%d",&a[i]);
             a[i]++;
             sum+=Query(,a[i]+,n);
             Update(,a[i]);
         }
         int res = sum;
         for(int i=n-;i>=;i--){
             sum=sum-(n-a[i])+(a[i]-);
             if(sum<res) res=sum;
         }
         printf("%d\n",res);
     }
     return ;
 }