HDOJ4911 归并排序求逆序数

最新推荐文章于 2020-05-10 20:32:16 发布

罗侯

最新推荐文章于 2020-05-10 20:32:16 发布

阅读量258

点赞数

CC 4.0 BY-SA版权

分类专栏：逆序对-归并排序文章标签：归并排序

本文链接：https://blog.youkuaiyun.com/computer_user/article/details/78217503

逆序对-归并排序专栏收录该内容

1 篇文章

订阅专栏

本文介绍了一种通过归并排序算法来计算给定序列在允许交换相邻元素一定次数后的最小逆序对数量的方法。使用C++实现，适用于算法竞赛等场景。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Inversion

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 5034 Accepted Submission(s): 1773

Problem Description
bobo has a sequence a1,a2,…,an. He is allowed to swap two adjacent numbers for no more than k times.

Find the minimum number of inversions after his swaps.

Note: The number of inversions is the number of pair (i,j) where 1≤i

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

typedef long long ll;
const int maxn = 1e6+10;
int n;
int a[maxn],tmp[maxn];
ll ans;

void init(){
    scanf("%d",&n);
    for (int i=1; i<=n; i++) scanf("%d",&a[i]);
    ans = 0;
}

void Merge(int left, int mid, int right){
    int i = left, j = mid+1, k = left;
    while (i<=mid && j<=right){
        if (a[i]>a[j]){
            tmp[k++] = a[j++];
            ans += mid+1-i;
        }
        else tmp[k++] = a[i++];
    }
    while (i<=mid) tmp[k++] = a[i++];
    while (j<=right) tmp[k++] = a[j++];
    for (int i=left; i<=right; i++ ) a[i] = tmp[i];
}

void merge_sort(int left, int right){
    if (left<right){
        int mid = (left+right)>>1;
        merge_sort(left,mid);
        merge_sort(mid+1,right);
        Merge(left,mid,right);
    }
}

int main(){
    int T;
    scanf("%d",&T);
    while (T--){
        init();
        merge_sort(1,n);
        printf("%lld\n",ans);
    }
    return 0;
}