codechef Chef and Swaps

Problem Description

This time, Chef has given you an array A containing N elements.
He had also asked you to answer M of his questions. Each question sounds like: "How many inversions will the array A contain, if we swap the elements at the i-th and the j-th positions?".
The inversion is such a pair of integers (i, j) that i < j and Ai > Aj.

Input

The first line contains two integers N and M - the number of integers in the array A and the number of questions respectively.
The second line contains N space-separated integers - A1, A2, ..., AN, respectively.
Each of next M lines describes a question by two integers i and j - the 1-based indices of the numbers we'd like to swap in this question.

Output

Output M lines. Output the answer to the i-th question of the i-th line.

Constraints

1 ≤ N, M ≤ 2 * 105
1 ≤ i, j ≤ N
1 ≤ Ai ≤ 109
Mind that we don't actually swap the elements, we only answer "what if" questions, so the array doesn't change after the question.

Example

Input:

6 3
1 4 3 3 2 5
1 1
1 3
2 5

Output:

5
6
0

Explanation

Inversions for the first case: (2, 3), (2, 4), (2, 5), (3, 5), (4, 5).
Inversions for the second case: (1, 3), (1, 5), (2, 3), (2, 4), (2,5), (4, 5).
In the third case the array looks like 1 2 3 3 4 5 and there are no inversions.

题解

tyvj《逆序对加强版》的加强版。同样的我们也考虑离线的做法。在那道题中，我们可以得到“修改一个点之后的逆序对”，这道题其实相当于同时改两个点。但要知道，每次交换算答案时要分情况讨论：当a[x]==a[y]和当a[x]!=a[y]时答案的计算是不一样的.

#include<cstdio>#include<cstring>#include<cstdlib>#include<iostream>#include<cmath>#include<algorithm>#define ll long long#define MAXN 200002using namespace std;int n,m,a[MAXN],ha[MAXN],hs[MAXN],maxs,zz;struct qus{int l,r;} b[MAXN];int lastl[MAXN],prel[MAXN],lastr[MAXN],prer[MAXN];ll v[MAXN],cgl[MAXN],cgr[MAXN],tot,tr[MAXN];void init(){scanf("%d%d",&n,&m);int i;for(i=1;i<=n;i++)   {scanf("%d",&a[i]);    ha[i]=a[i];   }sort(ha+1,ha+n+1);for(i=1;i<=n;i++)   {if(ha[i]!=ha[i-1])       hs[++zz]=ha[i];   }for(i=1;i<=m;i++)   {scanf("%d%d",&b[i].l,&b[i].r);    prel[i]=lastl[b[i].l]; lastl[b[i].l]=i;    prer[i]=lastr[b[i].r]; lastr[b[i].r]=i;   }}int getw(int x){int s=1,t=zz,mid;while(s<=t)   {mid=(s+t)>>1;if(hs[mid]<x) s=mid+1;else t=mid-1;   }return s;}int lowbit(int x) {return x&(-x);}ll find(int x){ll sum=0;for(;x>0;x-=lowbit(x)) sum+=tr[x];return sum;}void insert(int x){for(;x<=zz;x+=lowbit(x)) tr[x]++;}void work(){int i,j,k,x,y;maxs=zz;for(i=1;i<=n;i++)   {j=lastl[i]; k=lastr[i];    x=getw(a[i]);    v[i]+=find(maxs)-find(x);    tot+=v[i];    while(j)       {y=getw(a[b[j].r]);    cgl[j]+=find(maxs)-find(y);    j=prel[j];   }while(k)       {y=getw(a[b[k].l]);    cgr[k]+=find(maxs)-find(y);    k=prer[k];   }insert(x);   }memset(tr,0,sizeof(tr));for(i=n;i>0;i--)   {j=lastl[i]; k=lastr[i];    x=getw(a[i]);    v[i]+=find(x-1);    while(k)       {y=getw(a[b[k].l]);    cgr[k]+=find(y-1);    k=prer[k];   }    while(j)       {y=getw(a[b[j].r]);    cgl[j]+=find(y-1);    j=prel[j];   }insert(x);   }for(i=1;i<=m;i++)   {x=b[i].l; y=b[i].r;    if(a[x]==a[y]) printf("%lld\n",tot);    else printf("%lld\n",tot-v[x]-v[y]+cgl[i]+cgr[i]+1);   }}int main(){init(); work();return 0;}