题面
The Smart Beaver has recently designed and built an innovative nanotechnologic all-purpose beaver mass shaving machine, "Beavershave 5000". Beavershave 5000 can shave beavers by families! How does it work? Very easily!
题意
在一个数组里面,取出值从a到b的序列,但是每次取出数的必须是连续的.
求每次最少取多少次.
询问中存在交换操作
思路
线段数维护区间和.
点\(x\)表示\(x+1\)是否在\(x\)后面.

#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>#define fuck(x) cerr<<#x<<" = "<<x<<endl;
#define debug(a, x) cerr<<#a<<"["<<x<<"] = "<<a[x]<<endl;
#define lson l,mid,ls
#define rson mid+1,r,rs
#define ls (rt<<1)
#define rs ((rt<<1)|1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int loveisblue = 486;
const int maxn = 300086;
const int maxm = 100086;
const int inf = 0x3f3f3f3f;
const ll Inf = 999999999999999999;
const int mod = 1000000007;
const double eps = 1e-6;
const double pi = acos(-1);int num[maxn], pos[maxn];int bit[maxn];int lowbit(int x) {return x & -x;
}int query(int pos) {int ans = 0;while (pos) {ans += bit[pos];pos -= lowbit(pos);}return ans;
}void update(int pos, int val) {while (pos < maxn) {bit[pos] += val;pos += lowbit(pos);}
}int main() {ios::sync_with_stdio(true);
#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);
#endifint n;scanf("%d", &n);for (int i = 1; i <= n; i++) {scanf("%d", &num[i]);pos[num[i]] = i;}for (int i = 1; i < n; i++) {if (pos[i] > pos[i + 1]) {update(i, 1);}}int m;scanf("%d", &m);while (m--) {int type;scanf("%d", &type);if (type == 1) {int l, r;scanf("%d%d", &l, &r);printf("%d\n", query(r-1) - query(l - 1) + 1);} else {int x,y;scanf("%d%d",&x,&y);swap(num[x],num[y]);x=num[x];y=num[y];int t1 = pos[x];int t2 = pos[y];if(t1>t2){swap(x,y);swap(t1,t2);}if(pos[x+1]>=pos[x]&&pos[x+1]<=pos[y]){update(x,1);}if(x>1&&x-1!=y&&pos[x-1]>=pos[x]&&pos[x-1]<=pos[y]){update(x-1,-1);}if(pos[y+1]>=pos[x]&&pos[y+1]<=pos[y]){update(y,-1);}if(y>1&&y-1!=x&&pos[y-1]>=pos[x]&&pos[y-1]<=pos[y]){update(y-1,1);}swap(pos[x],pos[y]);}}return 0;
}