ACM Minimum Inversion Number 解题报告 -线段树

C - Minimum Inversion Number

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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

 //线段树专题解题。
 //其实一开始没想通，后来手动模拟了一下整个树建立的过程就全部清楚了。
 //详细AC代码在下面： 

 #include"iostream"
 #include"algorithm"
 #include"cstdio"
 #include"cstring"
 #include"cmath"
 #define max(a,b) a>b?a:b
 #define min(a,b) a<b?a:b
 #define lson l,m,rt<<1
 #define rson m+1,r,rt<<1|1
 using namespace std;
 const int MX = +;
 int sum[MX<<];
 void PushUp(int rt) {
     sum[rt]=sum[rt<<]+sum[rt<<|];  //更新节点  父节点为子节点之和
 }

 void Build(int l,int r,int rt) {
     sum[rt]=;            //建立一棵空树  【这里之前放在了判断里面，叶节点确实清空了，枝节点漏掉了】
     if(r==l) return ;
     int    m=(r+l)>>;
     Build(lson);//建立左节点
     Build(rson);//建立右节点
 }

 void UpData(int p,int l,int r,int rt) {
     if(r==l) {            //找到并更新目标点
         sum[rt]++;
         return ;
     }
     int m=(r+l)>>;
     if(p<=m) UpData(p,lson);  //如果不是目标点向左右寻找
     if(p >m) UpData(p,rson);
     PushUp(rt);//将更新过的每个点的子节点的和更新。
 }

 int Query(int L,int R,int l,int r,int rt) {
     if(L<=l&&R>=r)         //大小超过整个范围
         return sum[rt];  //返回总数
     int m=(r+l)>>;
     int ret=;
     if(L<= m) ret += Query(L,R,lson);  //比x[i]大的树的左值和
     if(R > m) ret += Query(L,R,rson);  //比x[i]大的树的右值和
     return ret;
 }
 int x[MX];
 int main() {
     int n;
     int sums;
     char s[];
     while(~scanf("%d",&n)) {
         sums=;
         Build(,n-,);         //【这里应该从0~n-1比较好，从1~n的话0的位置不好放在哪里了。后面也就一样了。】
         for(int i=; i<n; i++) {
             scanf("%d",&x[i]);
             sums+=Query(x[i],n-,,n-,);
             UpData(x[i],,n-,);
         }
         int ret=sums;
         for(int i=; i<n; i++) {
             sums=sums+n-*x[i]-;
             ret=min(ret,sums);
         }
         printf("%d\n",ret);
     }
     return ;
 }