题目链接：BZOJ3295

【分析】如果是静态的逆序对，很简单，只要用树状数组或者线段树，并归等nlogn处理出来；而这题是会删除序列的，可以通过树套树N*logn*logn来解决。这里我用了树状数组套BST，观察发现树状数组的每个点表示的范围i-lowbit(i)+1 ~ i的总点数是n*logn个，不会爆内存。所以用BST表示当前树状数组结点表示的范围还没被删除的点，则可以先预处理出初始序列的逆序对，然后每删除一个数只要计算出在他前面有几个比他大和在他后面有几个比他小，减去即可。

或者可以用分块写代码更短，由于m比较大，所以比上面的方法慢，在UVA超时了，BZOJ 4248ms

【AC CODE(树套树)】 3744ms

#include <cstdio>
#include <cstring>
#include <cctype>
#include <cmath>
#include <map>
//#include <unordered_set>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <algorithm>
#include <ctime>
using namespace std;
typedef long long LL;
#define rep(i,a,n) for(int i = a; i < n; i++)
#define repe(i,a,n) for(int i = a; i <= n; i++)
#define per(i,n,a) for(int i = n; i >= a; i--)
#define clc(a,b) memset(a,b,sizeof(a))
const int INF = 0x3f3f3f3f, MAXN = 200000+10;
struct NODE{
    bool in;
    int sum, v;
    NODE *lc,*rc;
    NODE(int a = 0){
        in = sum = 1;
        v = a;
    }
};
NODE *rt[MAXN];
NODE *null = new NODE(-1);
int a[MAXN], num[MAXN], id[MAXN], n;

inline int lowbit(int x){return x&-x;}
inline bool cmp(int x, int y){
    return a[x] < a[y];
}
void bulid(NODE *&u, int x, int y)
{
    if(x > y) return;
    int m = (x+y)>>1;
    u = new NODE(a[num[m]]);
    u->lc = u->rc = null;
    bulid(u->lc,x,m-1);
    bulid(u->rc,m+1,y);
    u->sum = u->lc->sum+u->rc->sum+1;
}
bool fd;
int find_rk(NODE *u, int k, bool b)
{
    if(u == null)
    {
        fd = false;
        return 0;
    }
    if(k == u->v)
    {
        fd = u->in;
        return u->lc->sum+fd;
    }
    if(k < u->v) return find_rk(u->lc,k,b);
    return find_rk(u->rc,k,b)+u->lc->sum+u->in;//注意已经点已经被删除的情况
}
LL query(int x, bool b, int v)//a[1～x]有多少个大于(a = true,否则是小于)v的数
{
    LL sum = 0;
    while(x > 0)
    {
        fd = true;
        int k = find_rk(rt[x],v,b);
        if(b)
        {
            sum += rt[x]->sum-k;
            //if(fd) sum++;
        }
        else
        {
            sum += k-1;
            if(!fd) sum++;
        }
        x -= lowbit(x);
    }
    return sum;
}
void del(NODE *u, int v)//在u树删除v
{
    if(u->v == v)
    {
        u->sum--;
        u->in = false;
        return;
    }
    if(v < u->v) del(u->lc,v);
    else del(u->rc,v);
    u->sum = u->lc->sum + u->rc->sum+u->in;
}
void update(int x)//删除a[x]
{
    int v = a[x];
    while(x <= n)
    {
        del(rt[x], v);
        x += lowbit(x);
    }
}
void pt(NODE *u)
{
    if(null == u) return;
    pt(u->lc);
    if(u->in)
        printf("=%d=\n", u->v);
    pt(u->rc);
}
int main()
{
#ifdef SHY
    freopen("d:\\1.txt", "r", stdin);
#endif
    int m;
    null->sum = 0;
    while(~scanf("%d %d%*c", &n, &m))
    {
        LL sum = 0;
        repe(i,1,n) scanf("%d%*c", &a[i]), id[a[i]] = i;
        repe(i,1,n)
        {
            int st = i-lowbit(i)+1;
            repe(j,st,i) num[j-st] = j;
            sort(num,num+lowbit(i),cmp);
            bulid(rt[i], 0,lowbit(i)-1);
            sum += query(i,1,a[i]);
        }
        while(m--)
        {
            printf("%lld\n", sum);
            int v;
            scanf("%d\n", &v);
            update(id[v]);
            LL b = query(id[v], 1,v);
            //printf("b = %d\n", b);
            LL c = query(n,0,v);
            LL d = query(id[v],0,v);
            sum -= b+(c-d);
        }
    }
    return 0;
}

 

【分块】4248ms

#include <cstdio>
#include <cstring>
#include <cctype>
#include <cmath>
#include <set>
//#include <unordered_set>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;
typedef long long LL;
#define rep(i,a,n) for(int i = a; i < n; i++)
#define repe(i,a,n) for(int i = a; i <= n; i++)
#define per(i,n,a) for(int i = n; i >= a; i--)
#define clc(a,b) memset(a,b,sizeof(a))
const int INF = 0x3f3f3f3f, MAXN = 200000+10, SIZE = 500;
int c[MAXN],n,a[MAXN],pos[MAXN];

inline int lowbit(int x){
    return x&-x;
}
int get_sum(int x)
{
    int sum = 0;
    while(x > 0)
    {
        sum += c[x];
        x -= lowbit(x);
    }
    return sum;
}
void add(int x)
{
    while(x <= n)
    {
        c[x]++;
        x += lowbit(x);
    }
}
int cnt, block[MAXN/SIZE+1][SIZE], bc[MAXN/SIZE+1];
int query(int x, int y, int v, bool biger)//a[x~y]有几个比v(biger大，否则小)
{
    int bx = x/SIZE,by = y/SIZE, sum = 0;
    if(bx == by)
    {
        if(biger)
        {
            repe(i,x,y)
                if(~a[i] && a[i] > v) sum++;
        }
        else
        {
            repe(i,x,y)
                if(~a[i] && a[i] < v) sum++;
        }
    }
    else
    {
        if(biger)
        {
            per(i,(bx+1)*SIZE-1,x)
                if(~a[i] && a[i] > v) sum++;
            repe(i,by*SIZE,y)
                if(~a[i] && a[i] > v) sum++;
            rep(i,bx+1,by)
                sum += bc[i]-(lower_bound(block[i],block[i]+bc[i],v)-block[i]);
        }
        else
        {
            per(i,(bx+1)*SIZE-1,x)
                if(~a[i] && a[i] < v) sum++;
            repe(i,by*SIZE,y)
                if(~a[i] && a[i] < v) sum++;
            rep(i,bx+1,by)
                sum += lower_bound(block[i],block[i]+bc[i],v)-block[i];
        }
    }
    return sum;
}
void del(int x)//删除a[x]
{
    int *b = block[x/SIZE], k = 0, mx = bc[x/SIZE];
    while(b[k] != a[x]) k++;
    b[k] = a[x] = -1;
    for(int i = k; i < mx-1; i++)
        swap(b[i],b[i+1]);
    bc[x/SIZE]--;
}
int main()
{
#ifdef SHY
    freopen("d:\\1.txt", "r", stdin);
#endif
    int m;
    while(~scanf("%d %d%*c", &n, &m))
    {
        clc(c,0);
        int j = 0;
        LL sum = 0;
        cnt = 0;
        clc(block,-1);
        rep(i,0,n)
        {
            scanf("%d%*c", &a[i]);
            pos[a[i]] = i;
            sum += get_sum(n)-get_sum(a[i]);
            add(a[i]);
            block[cnt][j++] = a[i];
            if(j == SIZE) cnt++,j = 0;
        }
        rep(i,0,cnt) sort(block[i],block[i]+SIZE),bc[i] = SIZE;
        sort(block[cnt],block[cnt]+j);
        bc[cnt] = j;
        while(m--)
        {
            printf("%lld\n", sum);
            int v;
            scanf("%d%*c", &v);
            del(pos[v]);
            sum -= query(0,pos[v],v,1)+query(pos[v],n-1,v,0);
        }
    }
    return 0;
}