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 200002
using 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;
}