Codefores 785E Anton and Permutation(分块)

E. Anton and Permutation
time limit per test 4 seconds
memory limit per test 512 megabytes
input standard input
output standard output
Anton likes permutations, especially he likes to permute their elements. Note that a permutation of n elements is a sequence of numbers{a1, a2, …, an}, in which every number from 1 to n appears exactly once.

One day Anton got a new permutation and started to play with it. He does the following operation q times: he takes two elements of the permutation and swaps these elements. After each operation he asks his friend Vanya, how many inversions there are in the new permutation. The number of inversions in a permutation is the number of distinct pairs (i, j) such that 1 ≤ i < j ≤ n and ai > aj.
Vanya is tired of answering Anton’s silly questions. So he asked you to write a program that would answer these questions instead of him.
Initially Anton’s permutation was {1, 2, …, n}, that is ai = i for all i such that 1 ≤ i ≤ n.
Input
The first line of the input contains two integers n and q (1 ≤ n ≤ 200 000, 1 ≤ q ≤ 50 000) — the length of the permutation and the number of operations that Anton does.
Each of the following q lines of the input contains two integers li and ri (1 ≤ li, ri ≤ n) — the indices of elements that Anton swaps during the i-th operation. Note that indices of elements that Anton swaps during the i-th operation can coincide. Elements in the permutation are numbered starting with one.
Output
Output q lines. The i-th line of the output is the number of inversions in the Anton’s permutation after the i-th operation.
Examples
input
5 4
4 5
2 4
2 5
2 2
output
1
4
3
3
input
2 1
2 1
output
1
input
6 7
1 4
3 5
2 3
3 3
3 6
2 1
5 1
output
5
6
7
7
10
11
8
Note
Consider the first sample.
After the first Anton’s operation the permutation will be {1, 2, 3, 5, 4}. There is only one inversion in it: (4, 5).
After the second Anton’s operation the permutation will be {1, 5, 3, 2, 4}. There are four inversions: (2, 3), (2, 4), (2, 5) and (3, 4).
After the third Anton’s operation the permutation will be {1, 4, 3, 2, 5}. There are three inversions: (2, 3), (2, 4) and (3, 4).
After the fourth Anton’s operation the permutation doesn’t change, so there are still three inversions.
题意：给你一个初始升序序列1~n，q次交换，求每次交换后的逆序对数有多少。

分块去做，维护一个原数组，和一些保持升序的块，每次交换之后，找u-v位置的数与比a[u],a[v]大或者小的数的个数，更新答案，然后找的时候块内二分，两端暴力，询问完之后，交换a[u],a[v],然后更新两个块，使得块仍然有序。

区间操作貌似用分块挺有用的。

参考了别人的代码~~

#include<iostream>    #include<cmath>    #include<queue>    #include<cstdio>    #include<queue>    #include<algorithm>    #include<cstring>    #include<string>    #include<utility>  #include<map>  #include<vector>  #define maxn 200005   #define inf 0x3f3f3f3f    using namespace std;  typedef long long LL;  const double eps = 1e-8;  vector<int>v[maxn];  int Left[maxn], Right[maxn], belong[maxn], a[maxn];  void build(int n){      int num = sqrt(n);      int block = n / num;      if (n%num)          block++;      for (int i = 1; i <= block; i++){          Left[i] = (i - 1)*num + 1;          Right[i] = i*num;      }      for (int i = 1; i <= n; i++){          belong[i] = (i - 1) / num + 1;      }      for (int i = 1; i <= block; i++){          for (int j = Left[i]; j <= Right[i]; j++){              v[i].push_back(j);          }      }  }  LL query(int l, int r, int now){      if (l > r)          return 0;      LL ans = 0;      if (belong[l] == belong[r]){          for (int i = l; i <= r; i++){              if (a[i] < now)                  ans++;          }      }      else{          for (int i = l; i <= Right[belong[l]]; i++){              if (a[i] < now)                  ans++;          }          for (int i = belong[l] + 1; i < belong[r]; i++){              int pos = upper_bound(v[i].begin(), v[i].end(), now) - v[i].begin();              ans += pos;          }          for (int i = Left[belong[r]]; i <= r; i++){              if (a[i] < now)                  ans++;          }      }      return ans;  }  void update(int x, int y){      int id = belong[x];      v[id].erase(lower_bound(v[id].begin(), v[id].end(), a[x]));      v[id].insert(upper_bound(v[id].begin(), v[id].end(), a[y]), a[y]);      id = belong[y];      v[id].erase(lower_bound(v[id].begin(), v[id].end(), a[y]));      v[id].insert(upper_bound(v[id].begin(), v[id].end(), a[x]), a[x]);      swap(a[x], a[y]);  }  int main(){      int n, q;      scanf("%d%d", &n, &q);      for (int i = 1; i <= n; i++){          a[i] = i;      }      build(n);      LL ans = 0;      while (q--){          int x, y;          scanf("%d%d", &x, &y);          if (x > y)              swap(x, y);          if (x == y)              printf("%lld\n", ans);          else{              LL sub = query(x + 1, y - 1, a[x]);              LL add = y - 1 - x - sub;              ans -= sub;              ans += add;              add = query(x + 1, y - 1, a[y]);              sub = y - 1 - x - add;              ans += add;              ans -= sub;              if (a[x] < a[y])                  ans++;              else                  ans--;              printf("%lld\n", ans);              update(x, y);          }      }  }