[BZOJ 2141] 排队
[题目链接]
https://www.lydsy.com/JudgeOnline/problem.php?id=2141
[算法]
首先求出原序列中的逆序对个数
考虑交换两个数的位置 ， 对答案产生的影响
显然 , (x , y)对答案的贡献为 : -(x对答案的贡献) - (y对答案的贡献) + x的新贡献 + y的新贡献
我们需要快速计算一个形如" 区间[l , r]中大于 / 小于某个数的数有多少个"的问题
树状数组套主席树即可 ， 时间复杂度 : O(NlogN ^ 2)
[代码]
#include<bits/stdc++.h> using namespace std; #define MAXN 100010 #define MAXP 10000010 #define MAXLOG 20 typedef long long ll; typedef long double ld; typedef unsigned long long ull; int n , m , len , L , R; int a[MAXN] , tmp[MAXN] , fen[MAXN] , root[MAXN] , ql[MAXLOG] , qr[MAXLOG]; struct Presitent_Segment_Tree { int sz; int ls[MAXP] , rs[MAXP] , cnt[MAXP]; Presitent_Segment_Tree() { sz = 0; memset(ls , 0 , sizeof(ls)); memset(rs , 0 , sizeof(rs)); memset(cnt , 0 , sizeof(cnt)); } inline void modify(int &k , int old , int l , int r , int pos , int value) { if (!k) k = ++sz; ls[k] = ls[old] , rs[k] = rs[old]; cnt[k] = cnt[old] + value; if (l == r) return; int mid = (l + r) >> 1; if (mid >= pos) modify(ls[k] , ls[k] , l , mid , pos , value); else modify(rs[k] , rs[k] , mid + 1 , r , pos , value); } inline int queryA(int l , int r , int k) { if (r <= k) return 0; if (l > k) { int ret = 0; for (int i = 1; i <= L; i++) ret -= cnt[ql[i]]; for (int i = 1; i <= R; i++) ret += cnt[qr[i]]; return ret; } int mid = (l + r) >> 1; if (mid >= k) { int ret = 0; for (int i = 1; i <= L; i++) ret -= cnt[rs[ql[i]]]; for (int i = 1; i <= R; i++) ret += cnt[rs[qr[i]]]; for (int i = 1; i <= L; i++) ql[i] = ls[ql[i]]; for (int i = 1; i <= R; i++) qr[i] = ls[qr[i]]; ret += queryA(l , mid , k); return ret; } else { for (int i = 1; i <= L; i++) ql[i] = rs[ql[i]]; for (int i = 1; i <= R; i++) qr[i] = rs[qr[i]]; return queryA(mid + 1 , r , k); } } inline int queryB(int l , int r , int k) { if (l >= k) return 0; if (r < k) { int ret = 0; for (int i = 1; i <= L; i++) ret -= cnt[ql[i]]; for (int i = 1; i <= R; i++) ret += cnt[qr[i]]; return ret; } int mid = (l + r) >> 1; if (mid < k) { int ret = 0; for (int i = 1; i <= L; i++) ret -= cnt[ls[ql[i]]]; for (int i = 1; i <= R; i++) ret += cnt[ls[qr[i]]]; for (int i = 1; i <= L; i++) ql[i] = rs[ql[i]]; for (int i = 1; i <= R; i++) qr[i] = rs[qr[i]]; ret += queryB(mid + 1 , r , k); return ret; } else { for (int i = 1; i <= L; i++) ql[i] = ls[ql[i]]; for (int i = 1; i <= R; i++) qr[i] = ls[qr[i]]; return queryB(l , mid , k); } } } PST; struct Binary_Indexed_Tree { inline int lowbit(int x) { return x & (-x); } inline void modify(int pos , int x , int value) { for (int i = pos; i <= n; i += lowbit(i)) PST.modify(root[i] , root[i] , 1 , len , x , value); } inline int queryA(int l , int r , int k) { if (l > r) return 0; L = 0 , R = 0; for (int i = l - 1; i >= 1; i -= lowbit(i)) ql[++L] = root[i]; for (int i = r; i >= 1; i -= lowbit(i)) qr[++R] = root[i]; return PST.queryA(1 , len , k); } inline int queryB(int l , int r , int k) { if (l > r) return 0; L = 0 , R = 0; for (int i = l - 1; i >= 1; i -= lowbit(i)) ql[++L] = root[i]; for (int i = r; i >= 1; i -= lowbit(i)) qr[++R] = root[i]; return PST.queryB(1 , len , k); } } BIT; template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); } template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); } template <typename T> inline void read(T &x) { T f = 1; x = 0; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -f; for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0'; x *= f; } inline int lowbit(int x) { return x & (-x); } inline void change(int x , int w) { while (x <= n) { fen[x] += w; x += lowbit(x); } } inline int ask(int x) { int ret = 0; while (x >= 1) { ret += fen[x]; x -= lowbit(x); } return ret; } inline int ask(int l , int r) { return ask(r) - ask(l - 1); } int main() { read(n); for (int i = 1; i <= n; i++) { read(a[i]); tmp[i] = a[i]; } sort(tmp + 1 , tmp + n + 1); len = unique(tmp + 1 , tmp + n + 1) - tmp - 1; for (int i = 1; i <= n; i++) a[i] = lower_bound(tmp + 1 , tmp + len + 1 , a[i]) - tmp; ll ans = 0; for (int i = 1; i <= n; i++) { ans += (ll)ask(a[i] + 1 , len); change(a[i] , 1); } printf("%d\n" , ans); for (int i = 1; i <= n; i++) BIT.modify(i , a[i] , 1); read(m); for (int i = 1; i <= m; i++) { int x , y; read(x); read(y); if (x > y) swap(x , y); ans += BIT.queryA(x + 1 , y - 1 , a[x]); ans -= BIT.queryB(x + 1 , y - 1 , a[x]); ans += BIT.queryB(x + 1 , y - 1 , a[y]); ans -= BIT.queryA(x + 1 , y - 1 , a[y]); if (a[x] > a[y]) --ans; if (a[x] < a[y]) ++ans; printf("%lld\n" , ans); BIT.modify(x , a[x] , -1); BIT.modify(y , a[y] , -1); swap(a[x] , a[y]); BIT.modify(x , a[x] , 1); BIT.modify(y , a[y] , 1); } return 0; }
