hdu5497 Inversion 树状数组 待补完！！！

Inversion

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1174    Accepted Submission(s): 351

Problem Description

You have a sequence {a1,a2,...,an} and you can delete a contiguous subsequence of length m. So what is the minimum number of inversions after the deletion.

 

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first line contains two integers n,m(1≤n≤105,1≤m<n) - the length of the seuqence. The second line contains n integers a1,a2,...,an(1≤ai≤n).

The sum of n in the test cases will not exceed 2×106.

 

Output

For each test case, output the minimum number of inversions.

 

Sample Input

23 11 2 34 24 1 3 2

 

Sample Output

01

 

Source

BestCoder Round #58 (div.2)

 

半年前的问题一直困然至今。。

题意：一个数组，你可以从中删除连续的k个数，求一个最优的方案使得操作后的数组内逆序数对最少。

学长讲过这个，但是暑假的我还是在无限wa。。。枚举删除连续k个数的起点，每次

啊啊啊回头在写。。。。

待补完

/* ━━━━━┒ ┓┏┓┏┓┃μ'sic foever!! ┛┗┛┗┛┃＼○／ ┓┏┓┏┓┃ / ┛┗┛┗┛┃ノ) ┓┏┓┏┓┃ ┛┗┛┗┛┃ ┓┏┓┏┓┃ ┛┗┛┗┛┃ ┓┏┓┏┓┃ ┛┗┛┗┛┃ ┓┏┓┏┓┃ ┃┃┃┃┃┃ ┻┻┻┻┻┻ */#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <map>#include <stack>#include <vector>using namespace std;const int maxn=1e5;int save[maxn+5];int before[maxn+5];int after[maxn+5];int n;int lowbit(int x){    return x&(-x);}void update(int x,int val,bool type){    if(type){        while(x<=maxn){            before[x] += val;            x += lowbit(x);        }    }else{        while(x<=maxn){            after[x] += val;            x += lowbit(x);        }    }}int getsum(int x,bool type){    int sum=0;    if(type){        while(x){            sum += before[x];            x -=lowbit(x);        }    }else{        while(x){            sum += after[x];            x -=lowbit(x);        }    }    return sum;}int main(){    int t;    scanf("%d",&t);    int m,i;    while(t--){        memset(before, 0, sizeof(before));        memset(after, 0, sizeof(after));        scanf("%d%d",&n,&m);        long long ans=0;        for(i=1;i<=n;i++){            scanf("%d",&save[i]);        }        for(i=m+1;i<=n;i++){            ans += i-m-1-getsum(save[i],false);            update(save[i], 1, false);        }        long long now=ans;        for(i=1;i<=n-m;i++){            now -= getsum(save[i+m]-1,false);            now -= getsum(save[n], true) - getsum(save[i+m], true);            update(save[i+m], -1, false);            now += getsum(save[i]-1, false);            now += getsum(save[n], true) - getsum(save[i],true);            update(save[i], 1, true);            ans = min(ans,now);        }        printf("%lld\n",ans);    }    return 0;}