线段树（5）逆序对数。

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14885    Accepted Submission(s): 9087

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

 

 

Author

CHEN, Gaoli

 

 

Source

ZOJ Monthly, January 2003

 

 每输出一个数字，查找比ta大的数字有多少个，加和即可。用线段树单点更新，区间求和，时间复杂度O(nlog(n))。

#include<cstdio>
#include<algorithm>
#include<cstring>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const int maxn = 5005;
using namespace std;
int sum[maxn<<2];
int a[maxn];
void pushup(int rt){
    sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}
void build(int l,int r,int rt){
    sum[rt] = 0;
    if(l==r)return;
    int m = (l+r)>>1;
    build(lson);
    build(rson);
}
void update(int x,int l,int r,int rt){
    if(l==r){
        sum[rt]++;
        return;
    }
    int m = (l+r)>>1;
    if(x<=m)update(x,lson);
    else update(x,rson);
    pushup(rt);
}
int query(int L,int R,int l,int r,int rt){
    if(L<=l && R>=r){
        return sum[rt];
    }
    int cnt = 0;
    int m = (l+r)>>1;
    if(L<=m)cnt += query(L,R,lson);
    if(R>m) cnt += query(L,R,rson);
    return cnt;
}
int main(){
    int n;
    while(~scanf("%d",&n)){
        build(1,n,1);
        int ans = 0;
        for(int i = 1; i <= n; i++){
            scanf("%d",&a[i]);
            a[i]++;
            ans += query(a[i],n,1,n,1);//区间求和;
            update(a[i],1,n,1);
        }
        int temp = ans;
        for(int i = 1; i < n; i++){
            temp = temp + (n-a[i]) - (a[i]-1);
            ans = min(ans,temp);
        }
        printf("%d\n",ans);
    }
    return 0;
}