【bzoj3295】[Cqoi2011]动态逆序对

题目描述：

对于序列A，它的逆序对数定义为满足i<j，且Ai>Aj的数对(i,j)的个数。给1到n的一个排列，按照某种顺序依次删除m个元素，你的任务是在每次删除一个元素之前统计整个序列的逆序对数。

输入：

输入第一行包含两个整数n和m，即初始元素的个数和删除的元素个数。以下n行每行包含一个1到n之间的正整数，即初始排列。以下m行每行一个正整数，依次为每次删除的元素。

输出：
输出包含m行，依次为删除每个元素之前，逆序对的个数。

样例输入：
5 4
1
5
3
4
2
5
1
4
2

样例输出：
5
2
2
1

题解：
动态逆序对，如果把逆序对当做二维偏序，动态的相当于多一维时间。于是就变成了三维偏序问题，然后就可以用cdq来做了。

代码：

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#ifdef WIN32#define LL "%I64d"#else#define LL "%lld"#endif#ifdef CT#define debug(...) printf(__VA_ARGS__)#define setfile() #else#define debug(...)#define filename ""#define setfile() freopen(filename".in", "r", stdin); freopen(filename".out", "w", stdout);#endif#define R register#define getc() (S == T && (T = (S = B) + fread(B, 1, 1 << 15, stdin), S == T) ? EOF : *S++)#define dmax(_a, _b) ((_a) > (_b) ? (_a) : (_b))#define dmin(_a, _b) ((_a) < (_b) ? (_a) : (_b))#define cmax(_a, _b) (_a < (_b) ? _a = (_b) : 0)#define cmin(_a, _b) (_a > (_b) ? _a = (_b) : 0)char B[1 << 15], *S = B, *T = B;inline int FastIn(){R char ch; R int cnt = 0; R bool minus = 0;while (ch = getc(), (ch < '0' || ch > '9') && ch != '-') ;ch == '-' ? minus = 1 : cnt = ch - '0';while (ch = getc(), ch >= '0' && ch <= '9') cnt = cnt * 10 + ch - '0';return minus ? -cnt : cnt;}#define maxn 100010#define maxm 50010int pos[maxn], bit[maxn], last[maxn], now, n, m;struct Event{int pos, t, val;inline bool operator < (const Event &that) const{return t < that.t || (t == that.t && (pos < that.pos || (pos == that.pos && val < that.val)));}}p[maxn], t[maxn];int ans[maxn];#define lowbit(_x) ((_x) & -(_x))inline void add(R int x, R int val){for (; x <= n; x += lowbit(x)){if (last[x] != now)bit[x] = 0;last[x] = now;bit[x] += val;}}inline int query(R int x){R int ret = 0;for (; x ; x -= lowbit(x))if (last[x] == now)ret += bit[x];return ret;}void cdq(R int left, R int right){if (left == right) return ;R int mid = left + right >> 1;R int i, j, k;for (i = k = left, j = mid + 1; k <= right; ++k)t[p[k].t <= mid ? i++ : j++] = p[k];for (R int i = left; i <= right; ++i)p[i] = t[i];++now;for (R int i = left, j = mid + 1; j <= right; ++j){for (; i <= mid && p[i].pos <= p[j].pos; ++i)add(p[i].val, 1);ans[p[j].t] += i - left - query(p[j].val);}++now;for (R int i = mid, j = right; j > mid; --j){for (; i >= left && p[i].pos >= p[j].pos; --i)add(p[i].val, 1);ans[p[j].t] += query(p[j].val);}cdq(left, mid); cdq(mid + 1, right);}long long sum[maxn];int main(){//setfile();n = FastIn(); m = FastIn();for (R int i = 1; i <= n; ++i){R int val = FastIn();p[i] = (Event) {i, 0, val}, pos[val] = i;}for (R int i = 1; i <= m; ++i)p[pos[FastIn()]].t = n - i + 1;R int mcnt = 0;for (R int i = 1; i <= n; ++i)if (!p[i].t) p[i].t = ++mcnt;cdq(1, n);for (R int i = 1; i <= n; ++i) sum[i] = sum[i - 1] + ans[p[i].t];for (R int i = n; i > n - m; --i) printf("%lld\n",sum[i] );return 0;}/*5 41 5 3 4 25142*/