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 = 200000+10;
int sum[MX<<2];
void PushUp(int rt) {
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];  //更新节点  父节点为子节点之和
}

void Build(int l,int r,int rt) {
    sum[rt]=0;            //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)>>1;
    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)>>1;
    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)>>1;
    int ret=0;
    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[2];
    while(~scanf("%d",&n)) {
        sums=0;            
        Build(0,n-1,1);         //【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=0; i<n; i++) {
            scanf("%d",&x[i]);
            sums+=Query(x[i],n-1,0,n-1,1);
            UpData(x[i],0,n-1,1);
        }
        int ret=sums;
        for(int i=0; i<n; i++) {
            sums=sums+n-2*x[i]-1;
            ret=min(ret,sums);
        }
        printf("%d
",ret);
    }
    return 0;
}