HDU-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

题意：
题意：0~n-1且不一定有序排列的数字串，不断的将第一个数字放到数字串的最后，可以得到n个不同的数字串。每一个数字串都有一个逆序数，求n个数字串中最小的逆序数。
思路：
很简单的线段树，树状数组也可以做，把数拿下来每次往线段树里面查，或者往一个新数组里面插，就xjb些就行了

做这道题的时候初识线段树，竟然敲了一大段代码手写了一个for循环，坑爹的是竟然 842Ms 水过了
后来for循环写了一下 102Ms就过了 —– 菜鸡如我
仔细学习了一下线段树之后理解了线段树的区间查询再写了一次 42 Ms 过了 在这里写一下心得体会，
线段树的区间查询操作之所以是O(lg(n))的复杂度，完全是因为分治，每次都查询一半，如果每一个都遍历一次的话，
跟for循环就一模一样了，而且百度学习线段树的时候还学习到了懒标记的使用，也是线段树的一大重点，能够很好的降低时间复杂度，值得学习

———————————————————-AC code—————————————

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
using namespace std;
const int maxn = 6000+5;
struct segtree{
    int l,r;
    int sum;
}ss[maxn<<2];

int v[maxn];

void build(int l,int r,int rt ) {
    ss[rt].l = l,ss[rt].r = r;
    if ( ss[rt].l == ss[rt].r ) { ss[rt].sum = 0; return;}
    int mid = (l+r)>>1;
    build(lson); build(rson);
    ss[rt].sum = 0;
}

void update(int l,int r,int rt){
    if ( ss[rt].l == ss[rt].r ) { ss[rt].sum++; return; }
    int mid = (ss[rt].l+ss[rt].r)>>1;
    if ( mid >= r ) update(l,r,rt<<1);
    else update(l,r,rt<<1|1);
    ss[rt].sum = ss[rt<<1].sum + ss[rt<<1|1].sum;
}

int query(int l,int r,int rt){
    if ( ss[rt].l == l && ss[rt].r == r ){
        return ss[rt].sum;
    }
    int mid = (ss[rt].l+ss[rt].r)>>1; int s = 0;
    if ( mid >= r ) {
        s += query(l,r,rt<<1);
    } else if ( mid < l ) {
        s += query(l,r,rt<<1|1);
    } else {
        s += query(l,mid,rt<<1);
        s += query(mid+1,r,rt<<1|1);
    }
    return s;
}

int main(){
    int t;
    while(~scanf("%d",&t)){
        build(1,t,1);
        int sum = 0;
        for (int i = 1;i<=t;i++) {
            scanf("%d",&v[i]); v[i]++;
            if ( v[i] != t )
                sum += query(v[i]+1,t,1);
            update(v[i],v[i],1);
        }
        int minn = sum;
        for (int i = 1;i<=t;i++){
            sum = sum + t - 2*v[i] + 1;
            minn = min(minn,sum);
        }
        printf("%d
",minn);
    }
    return 0;
}

其实树状数组也可以实现

——————————————- AC code ————————————————-

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define debug puts("ok!");
using namespace std;
//typedef long long ll;
const int maxn = 50000+5;

int bit[maxn],t;

int lowbit(int pos){
    return (pos & (-pos));
}
int query(int pos){
    int ret = 0;
    while(pos > 0){
        ret += bit[pos];
        pos -= lowbit(pos);
    }
    return ret;
}
void update(int pos)
{
    while(pos<=t){
        bit[pos] ++;
        pos += lowbit(pos);
    }
} 

int v[maxn];

int main(){
    while(cin>>t){
        int sum = 0;
        memset(bit,0,sizeof(bit));
        for (int i = 1;i<=t;i++) {
            scanf("%d",&v[i]);    v[i]++;
            sum += query(t) - query(v[i]);
            //printf("%d
",query(t));
            update(v[i]);
        }
        //printf("%d
",sum);
        int minn = sum;
        for (int i = 1;i<=t;i++){
            sum = sum + t - 2*v[i] + 1;
            minn = min(minn,sum);
        }
        printf("%d
",minn);
    }
    return 0;
}