Codeforces 785E 分块+树状数组

Anton and Permutation

time limit per test
4 seconds

memory limit per test
512 megabytes

input
standard input

output
standard output

Anton likes permutations, especially he likes to permute their elements. Note that a permutation of n elements is a sequence of numbers{a1, a2, ..., an},
in which every number from 1 to n appears
exactly once.

One day Anton got a new permutation and started to play with it. He does the following operation q times: he takes two elements of
the permutation and swaps these elements. After each operation he asks his friend Vanya, how many inversions there are in the new permutation. The number of inversions in a permutation is the number of distinct pairs (i, j) such
that 1 ≤ i < j ≤ n and ai > aj.

Vanya is tired of answering Anton's silly questions. So he asked you to write a program that would answer these questions instead of him.

Initially Anton's permutation was {1, 2, ..., n}, that is ai = i for
all i such that 1 ≤ i ≤ n.

Input

The first line of the input contains two integers n and q (1 ≤ n ≤ 200 000, 1 ≤ q ≤ 50 000) —
the length of the permutation and the number of operations that Anton does.

Each of the following q lines of the input contains two integers li and ri (1 ≤ li, ri ≤ n) —
the indices of elements that Anton swaps during the i-th operation. Note that indices of elements that Anton swaps during the i-th
operation can coincide. Elements in the permutation are numbered starting with one.

Output

Output q lines. The i-th
line of the output is the number of inversions in the Anton's permutation after the i-th operation.

Examples

input

5 4
4 5
2 4
2 5
2 2

output

1
4
3
3

input

2 1
2 1

output

1

input

6 7
1 4
3 5
2 3
3 3
3 6
2 15 1

output

5
6
7
7
10
118

Note

Consider the first sample.

After the first Anton's operation the permutation will be {1, 2, 3, 5, 4}. There is only one inversion in it: (4, 5).

After the second Anton's operation the permutation will be {1, 5, 3, 2, 4}. There are four inversions: (2, 3), (2, 4), (2, 5) and (3, 4).

After the third Anton's operation the permutation will be {1, 4, 3, 2, 5}. There are three inversions: (2, 3), (2, 4) and (3, 4).

After the fourth Anton's operation the permutation doesn't change, so there are still three inversions.

题意：1-n的序列  q个操作  每个操作后输出有多少个逆序数

题解：分块暴力  每块建个树状数组搞一搞就行了

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;
int a[200001],L[450],R[450],C[449][200001],belong[200001],n;
void change(int k,int x,int v){
for(;x<=n;x+=x&-x)
C[k][x]+=v;
}
int sum(int k,int x){
int ret=0;
for(;x;x-=x&-x)
ret+=C[k][x];
return ret;
}

int main(){
int i,j,m;
scanf("%d%d",&n,&m);
int len=sqrt(n);
for(i=1;i<=449;i++)L[i]=999999;
if(len*len<n)len++;
int kuai=(n+len-1)/len;
for(i=1;i<=n;i++){
a[i]=i;
belong[i]=(i+kuai-1)/kuai;
L[belong[i]]=min(L[belong[i]],i);
R[belong[i]]=max(R[belong[i]],i);
change(belong[i],i,1);
}
ll now=0;
int l,r;
for(i=1;i<=m;i++){
scanf("%d%d",&l,&r);
if(l>r)swap(l,r);
if(l==r){
printf("%lld\n",now);
continue;
}
int bl=belong[l],br=belong[r];
int sum1=0,sum2=0,len=r-l-1;
if(bl==br){
for(j=l+1;j<r;j++){
if(a[j]>a[l])sum1++;
if(a[j]<a[r])sum2++;
}
swap(a[l],a[r]);
now=now+sum1-(len-sum1);
now=now+sum2-(len-sum2);
if(a[l]>a[r])now++;
else now--;
printf("%lld\n",now);
continue;
}
else{
for(j=l+1;j<=R[bl];j++){
if(a[j]>a[l])sum1++;
if(a[j]<a[r])sum2++;
}
for(j=L[br];j<=r-1;j++){
if(a[j]>a[l])sum1++;
if(a[j]<a[r])sum2++;
}
for(j=bl+1;j<br;j++){
sum1+=sum(j,n)-sum(j,a[l]);
sum2+=sum(j,a[r]);
}
}
change(bl,a[l],-1);
change(bl,a[r],1);
change(br,a[r],-1);
change(br,a[l],1);
swap(a[l],a[r]);
now=now+sum1-(len-sum1);
now=now+sum2-(len-sum2);
if(a[l]>a[r])now++;
else now--;
printf("%lld\n",now);
}
return 0;
}