北京师范大学第十四届ACM决赛- BSquared Permutation 树状数组

a[a[l]] a[a[r]]是最后的值。

要是l和r交换，首先改变的是a[a[l]] a[a[r]]

要是id[a[i]]也就是a[i]对应的位置，不为当前这个要换的位置，那么也要变

所以最大可能变4次

///                 .-~~~~~~~~~-._       _.-~~~~~~~~~-.///             __.'              ~.   .~              `.__///           .'//                  \./                  \\`.///        .'//                     |                     \\`.///       .'// .-~"""""""~~~~-._     |     _,-~~~~"""""""~-. \\`.///     .'//.-"                 `-.  |  .-'                 "-.\\`.///   .'//______.============-..   \ | /   ..-============.______\\`./// .'______________________________\|/______________________________`.#pragma GCC optimize(2)#pragma comment(linker, "/STACK:102400000,102400000")#include <vector>#include <iostream>#include <string>#include <map>#include <stack>#include <cstring>#include <queue>#include <list>#include <stdio.h>#include <set>#include <algorithm>#include <cstdlib>#include <cmath>#include <iomanip>#include <cctype>#include <sstream>#include <functional>#include <stdlib.h>#include <time.h>#include <bitset>using namespace std;#define pi acos(-1)#define s_1(x) scanf("%d",&x)#define s_2(x,y) scanf("%d%d",&x,&y)#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)#define S_1(x) scan_d(x)#define S_2(x,y) scan_d(x),scan_d(y)#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)#define PI acos(-1)#define endl '\n'#define srand() srand(time(0));#define me(x,y) memset(x,y,sizeof(x));#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)#define close() ios::sync_with_stdio(0); cin.tie(0);#define FOR(x,n,i) for(int i=x;i<=n;i++)#define FOr(x,n,i) for(int i=x;i<n;i++)#define fOR(n,x,i) for(int i=n;i>=x;i--)#define fOr(n,x,i) for(int i=n;i>x;i--)#define W while#define sgn(x) ((x) < 0 ? -1 : (x) > 0)#define bug printf("***********\n");#define db double#define ll long long#define mp make_pair#define pb push_backtypedef long long LL;typedef pair <int, int> ii;const int INF=~0U>>1;const LL LINF=0x3f3f3f3f3f3f3f3fLL;const int dx[]={-1,0,1,0,1,-1,-1,1};const int dy[]={0,1,0,-1,-1,1,-1,1};const int maxn=1e5+2e4;const int maxx=1e3+10;const double EPS=1e-8;const double eps=1e-8;const int mod=1e9+7;template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}template <class T>inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}if(IsN) num=-num;return true;}void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}void print(LL a){ Out(a),puts("");}//freopen( "in.txt" , "r" , stdin );//freopen( "data.txt" , "w" , stdout );//cerr << "run time is " << clock() << endl;//void readString(string &s)//{//static char str[maxn];//scanf("%s", str);//s = str;//}int n;int a[maxn],id[maxn];LL c[maxn];int f[6],f1[6];int lowbit(int x)  {      return x&-x;  }  void update(int i, int x)//update(int i,int x) 把a[i]加上x  {      while(i<=n)      {          c[i]+=x;          i+=lowbit(i);      }  }  ll getsum(int x)//sum[1,x]  求这一段的和  {      ll ans = 0;      while(x>0)      {          ans+=c[x];          x-=lowbit(x);      }      return ans;  }  void solve() {me(id,0);s_1(n);FOR(1,n,i){s_1(a[i]);c[i]=0;id[a[i]]=i;}FOR(1,n,i)update(i,a[a[i]]);int q;s_1(q);W(q--){int op,l,r;s_3(op,l,r);if(op==1){f[1]=l;f1[1]=a[a[l]];f[2]=r;f1[2]=a[a[r]];int tot=2;if(id[l]!=l&&id[l]!=r)f[++tot]=id[l],f1[tot]=a[a[id[l]]];if(id[r]!=l&&id[r]!=r)f[++tot]=id[r],f1[tot]=a[a[id[r]]];swap(a[l],a[r]);id[a[l]]=l;id[a[r]]=r;FOR(1,tot,i)update(f[i],a[a[f[i]]]-f1[i]);}else print(getsum(r)-getsum(l-1));}}int main(){    //freopen( "1.in" , "r" , stdin );    //freopen( "1.out" , "w" , stdout );    int t=1;    //init();    s_1(t);    for(int cas=1;cas<=t;cas++)    {        //printf("Case #%d: ",cas);        solve();    }}