bzoj3744: Gty的妹子序列 （BIT && 分块）

强制在线的区间询问逆序对数

如果不是强制在线

就是可以用莫队乱搞啦

强制在线的话

用f[i][j]记录第i块到第j个点之间的逆序对数

用s[i][j]记录前i块中小于等于j的数字个数

离散化一下

BIT用来处理需要暴力的地方即可

下面是代码

 #include <cstdio>
 #include <algorithm>
 #include <cmath>
 #include <cstring>
 using namespace std;
 #define isdigit(x) (x <= '9' && x >= '0')
 #define lowbit(x) (x & (-x))
 const int N = 5e4 + ;
 const int M = ;

 struct s {
     int u, v;
     inline bool operator < (const s &o) const {
         return u < o.u;
     }
 } a[N];

 inline void read(int &ans) {
     ans = ;
     static char buf = getchar();
     for (; !isdigit(buf); buf = getchar());
     for (; isdigit(buf); buf = getchar())
         ans = ans *  + buf - '';
 }

 int n, cnt, maxn, sz;
 int s[M][N], f[M][N], c[N], d[N], b[N];

 inline void add(int x, int a) {
     while (x <= maxn) {
         c[x] += a;
         x += lowbit(x);
     }
 }

 inline int query(int x) {
     int ans = ;
     while (x > ) {
         ans += c[x];
         x -= lowbit(x);
     }
     return ans;
 }

 inline void work(int x) {
     int h = (x - ) * sz + ;
     int t = x * sz;
     for (int i = h; i <= n; i++)
         add(d[i], ), f[x][i] = f[x][i - ] + i - h +  - query(d[i]);
     memset(c, , sizeof(c));
     for (int i = h; i <= t; i++)    s[x][d[i]]++;
     for (int i = ; i <= maxn; i++)    s[x][i] += s[x][i - ];
     for (int i = ; i <= maxn; i++)    s[x][i] += s[x - ][i];
 }

 int main() {
     read(n);
     sz = sqrt(n);
     for (int i = ; i <= n; i++) {
         read(a[i].u); a[i].v = i;
         b[i] = (i - ) / sz + ;
     }
     cnt = b[n];
     sort(a + , a + n + );
     int last = ; d[a[].v] = ;
     for (int i = ; i <= n; i++) {
         if (a[i].u == a[i - ].u)    d[a[i].v] = last;
         else    d[a[i].v] = ++last;
     }
     maxn = last;
     for (int i = ; i <= cnt; i++)
         work(i);
     int m; read(m);
     int ans = ;
     while (m--) {
         int l, r;
         read(l); read(r);
         l = l ^ ans; r = r ^ ans;
         ans = ;
         if (l > r)    swap(l, r);
         if (b[l] == b[r]) {
             for (int i = l; i <= r; i++)
                 add(d[i], ), ans += i - l +  - query(d[i]);
             for (int i = l; i <= r; i++)    add(d[i], -);
         }
         else {
             ans = f[b[l] + ][r];
             for (int i = (b[r] - ) * sz + ; i <= r; i++)    add(d[i], );
             for (int i = b[l] * sz; i >= l; i--)
                 add(d[i], ), ans += query(d[i] - ) + s[b[r] - ][d[i] - ] - s[b[l]][d[i] - ];
             for (int i = (b[r] - ) * sz + ; i <= r; i++)    add(d[i], -);
             for (int i = l; i <= b[l] * sz; i++)    add(d[i], -);
         }
         printf("%d\n", ans);
     }
 }