题面

洛谷

和动态逆序对那道题没有什么区别
把一个交换换成两个删除和两个插入

#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;
const int N = 1e5 + 5;
const double eps = 1e-9;
const int inf = 0x3f3f3f3f;
typedef pair<int, int> PII;
typedef pair<double, int> PDI;
struct Node{
    int x, y, z, d, cnt;    
}node[N];
bool rule_yzx(Node x, Node y){
    if(x.y != y.y) return x.y < y.y;    
    if(x.z != y.z) return x.z < y.z;
    return x.x < y.x;
}
int n, m, nsize, a[N], b[N];
int tim, dfn[N];
long long ans[N];
struct BIT{
    int w[N];
    void ins(int x, int d){while(x<=n){w[x] += d; x += x & -x;}}    
    int qry(int x){int res = 0; while(x){res += w[x]; x -= x & -x;}return res;}
    void print(){for(int i = 1; i <= n; ++i) printf("%d", w[i]); printf("\n");}
}bit;
    
inline void add(int x1, int x2, int x3, int x4){
    ++nsize;
    node[nsize].x = x1, node[nsize].y = x2, node[nsize].z = x3, node[nsize].d = x4;
}

void cdq(int L, int R){
    if(L == R) return ;
    int mid = L + ((R - L) >> 1);
    cdq(L, mid); cdq(mid + 1, R);   
    sort(node + L, node + mid + 1, rule_yzx);
    sort(node + mid + 1, node + R + 1, rule_yzx);
    int j = L; 
    for(int i = mid + 1; i <= R; ++i){
        while(j <= mid && node[j].y <= node[i].y){
            bit.ins(node[j].z, node[j].d); ++j;
        }
        node[i].cnt += node[i].d * (bit.qry(n) - bit.qry(node[i].z));
    }
    //printf("L %d R %d\n", L, R);
    //bit.print();
    while(j > L){--j; bit.ins(node[j].z, -node[j].d);}
    j = mid;
    for(int i = R; i >= mid + 1; --i){
        while(j >= L && node[j].y >= node[i].y){
            bit.ins(node[j].z, node[j].d); --j; 
        }   
        node[i].cnt += node[i].d * bit.qry(node[i].z - 1);
    }
    while(j < mid){++j; bit.ins(node[j].z, -node[j].d);} 
}

int main() {
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i){
        scanf("%d", &a[i]); b[i] = a[i];
    }
    sort(b + 1, b + n + 1);
    for(int i = 1; i <= n; ++i){
        a[i] = (lower_bound(b + 1, b + n + 1, a[i]) - b);
        add(0, i, a[i], 1);
        //printf("%d %d %d\n", node[i].x, node[i].y, node[i].z);
    }
    scanf("%d", &m);
    for(int i = 1, x, y; i <= m; ++i){
        scanf("%d%d", &x, &y);
        add(++tim, x, a[x], -1);
        add(++tim, y, a[y], -1);
        swap(a[x], a[y]);
        add(++tim, x, a[x], 1);
        add(++tim, y, a[y], 1);
        dfn[i] = tim;
    }
    //for(int i = 1; i <= nsize; ++i) printf("%d %d %d\n", node[i].x, node[i].y, node[i].z);
    //已经是升序了
    cdq(1, nsize);
    for(int i = 1; i <= nsize; ++i){
        //printf("%d %d %d %d\n", node[i].x, node[i].y, node[i].z, node[i].cnt);    
        ans[node[i].x] += node[i].cnt;
    } 
    for(int i = 1; i <= tim; ++i) ans[i] += ans[i - 1];
    for(int i = 0; i <= m; ++i) printf("%d\n", ans[dfn[i]]); 
    system("PAUSE");
    return 0;
}