【bzoj3295】[Cqoi2011]动态逆序对

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Description
对于序列A，它的逆序对数定义为满足i
#include<cstdio>#include<iostream>#include<cstring>#include<cmath>#include<algorithm>#define inf 0x7fffffff#define N 100005#define M 5000005#define ll long longll ans;int n,m,size;int A[30],B[30];int num[N],pos[N],a1[N],a2[N];int t[N],root[N];int ls[M],rs[M],sum[M];using namespace std;inline int read(){    int x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}    return x*f;}inline int lowbit(int x){    return x&(-x);}int getans(int x){    int ans=0;    while (x)    {        ans+=t[x];        x-=lowbit(x);    }    return ans;}void update(int &k,int l,int r,int x){    if (!k) k=++size;    sum[k]++;    if (l==r) return;    int mid=(l+r)>>1;    if (x<=mid)update(ls[k],l,mid,x);    else update(rs[k],mid+1,r,x);}int querymore(int x,int y,int num){    A[0]=B[0]=0;int tmp=0;x--;    for (int i=x;i;i-=lowbit(i))A[++A[0]]=root[i];    for (int i=y;i;i-=lowbit(i))B[++B[0]]=root[i];    int l=1,r=n;    while (l!=r)    {        int mid=(l+r)>>1;        if (num<=mid)        {            for (int i=1;i<=A[0];i++)tmp-=sum[rs[A[i]]];            for (int i=1;i<=B[0];i++)tmp+=sum[rs[B[i]]];            for (int i=1;i<=A[0];i++)A[i]=ls[A[i]];            for (int i=1;i<=B[0];i++)B[i]=ls[B[i]];            r=mid;        }        else        {            for (int i=1;i<=A[0];i++)A[i]=rs[A[i]];            for (int i=1;i<=B[0];i++)B[i]=rs[B[i]];            l=mid+1;        }    }    return tmp;}int queryless(int x,int y,int num){    A[0]=B[0]=0;int tmp=0;x--;    for (int i=x;i;i-=lowbit(i))A[++A[0]]=root[i];    for (int i=y;i;i-=lowbit(i))B[++B[0]]=root[i];    int l=1,r=n;    while (l!=r)    {        int mid=(l+r)>>1;        if (num>mid)        {            for (int i=1;i<=A[0];i++)tmp-=sum[ls[A[i]]];            for (int i=1;i<=B[0];i++)tmp+=sum[ls[B[i]]];            for (int i=1;i<=A[0];i++)A[i]=rs[A[i]];            for (int i=1;i<=B[0];i++)B[i]=rs[B[i]];            l=mid+1;        }        else        {            for (int i=1;i<=A[0];i++)A[i]=ls[A[i]];            for (int i=1;i<=B[0];i++)B[i]=ls[B[i]];            r=mid;        }    }    return tmp;}int main(){    n=read();m=read();    for (int i=1;i<=n;i++)    {        num[i]=read();pos[num[i]]=i;        a1[i]=getans(n)-getans(num[i]);//前面比它大的个数         ans+=a1[i];        for (int j=num[i];j<=n;j+=lowbit(j))            t[j]++;    }    for (int i=1;i<=n;i++)        t[i]=0;    for (int i=n;i>=1;i--)    {        a2[i]=getans(num[i]-1);//后面比它小的个数        for (int j=num[i];j<=n;j+=lowbit(j))            t[j]++;     }    for (int i=1;i<=m;i++)    {        printf("%lld\n",ans);        int x=read();x=pos[x];        ans-=(a1[x]+a2[x]-querymore(1,x-1,num[x])-queryless(x+1,n,num[x]));        for (int j=x;j<=n;j+=lowbit(j))        {            update(root[j],1,n,num[x]);        }    }    return 0;}
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