HDOJ1394 求逆序对

最新推荐文章于 2021-12-13 18:32:51 发布

原创最新推荐文章于 2021-12-13 18:32:51 发布 · 347 阅读

1 ·

CC 4.0 BY-SA版权

文章标签：

#线段树

线段树专栏收录该内容

4 篇文章

订阅专栏

本文介绍了一种算法，用于解决给定序列通过不同轮转方式得到的所有序列中逆序对数量最小的问题。采用树状数组优化逆序对的计算过程。

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21799 Accepted Submission(s): 13029

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

Author

CHEN, Gaoli

Source

ZOJ Monthly, January 2003

Recommend

Ignatius.L | We have carefully selected several similar problems for you: 1540 1542 1255 1828 2871

本题不难，先求一遍逆序对，然后逐个枚举情况就好了。

逆序对求法很多，归并排序，树状数组，线段树都可以。

代码如下：

#include <cstdio>
#include <cstring>
using namespace std;

const int maxn = 5000;
struct node{
   int l,r,val;
}tree[maxn*4+10];
int num[maxn+10];
int a[maxn+10];
int n;

void build(int rt, int l, int r){
    tree[rt].l = l;
    tree[rt].r = r;
    tree[rt].val = 0;

    if (l==r) return;
    int mid = (l+r)/2;
    build(rt<<1,l,mid);
    build(rt<<1|1,mid+1,r);
}

void update(int rt, int pos){
    int l = tree[rt].l;
    int r = tree[rt].r;
    if (l==r) {num[rt]++; return;}

    int mid = (l+r)>>1;
    if (pos<=mid) update(rt<<1,pos);
    else update(rt<<1|1,pos);

    num[rt] = num[rt<<1] + num[rt<<1|1];
}

int query(int rt, int L, int R){
    int l = tree[rt].l;
    int r = tree[rt].r;
    if (L<=l && r<=R) return num[rt];

    int mid = (l+r)>>1,ans=0;
    if (R>mid) ans += query(rt<<1|1,L,R);
    if (L<=mid) ans += query(rt<<1,L,R);
    return ans;
}

inline int MIN(int x, int y){return x<y?x:y;}
int main(){
    int sum;
    while (scanf("%d",&n)!=EOF){
        sum = 0;
        memset(num,0,sizeof(num));
        build(1,0,n-1);
        for (int i=0; i<n; i++) {
            scanf("%d",&a[i]);
            update(1,a[i]);
            sum += query(1,a[i]+1,n-1);
        }
        int mn = sum;
        for (int i=0; i<n; i++) {
            sum = sum + n-1-a[i]-a[i];
            mn = MIN(mn,sum);
        }
        printf("%d\n",mn);
    }
    return 0;
}