标签：cond sample 代码 amp += with struct 输入输出 after
You are given two permutations aa and bb , both consisting of nn elements. Permutation of nn elements is such a integer sequence that each value from 11 to nn appears exactly once in it.
You are asked to perform two types of queries with them:
Print the answer for each query of the first type.
It is guaranteed that there will be at least one query of the first type in the input.
输入格式：
The first line contains two integers nn and mm ( 2 \le n \le 2 \cdot 10^52≤n≤2⋅105 , 1 \le m \le 2 \cdot 10^51≤m≤2⋅105 ) — the number of elements in both permutations and the number of queries.
The second line contains nn integers a_1, a_2, \dots, a_na1?,a2?,…,an? ( 1 \le a_i \le n1≤ai?≤n ) — permutation aa . It is guaranteed that each value from 11 to nn appears in aa exactly once.
The third line contains nn integers b_1, b_2, \dots, b_nb1?,b2?,…,bn? ( 1 \le b_i \le n1≤bi?≤n ) — permutation bb . It is guaranteed that each value from 11 to nn appears in bb exactly once.
Each of the next mm lines contains the description of a certain query. These are either:
输出格式：
Print the answers for the queries of the first type, each answer in the new line — the number of values which appear in both segment [l_a; r_a][la?;ra?] of positions in permutation aa and segment [l_b; r_b][lb?;rb?] of positions in permutation bb .
6 7 5 1 4 2 3 6 2 5 3 1 4 6 1 1 2 4 5 2 2 4 1 1 2 4 5 1 2 3 3 5 1 1 6 1 2 2 4 1 1 4 4 1 3
Consider the first query of the first example. Values on positions [1; 2][1;2] of aa are [5, 1][5,1] and values on positions [4; 5][4;5] of bb are [1, 4][1,4] . Only value 11 appears in both segments.
After the first swap (the second query) permutation bb becomes [2, 1, 3, 5, 4, 6][2,1,3,5,4,6] .
After the second swap (the sixth query) permutation bb becomes [5, 1, 3, 2, 4, 6][5,1,3,2,4,6] .
1 #include <cstdio> 2 #include <iostream> 3 #include <cstring> 4 using namespace std; 5 const int N=2e5+10,M=1e6*40; 6 int n,m,cnt,top,a[N],b[N],d[N],st[N],root[N]; 7 struct node { int num,ch[2]; }e[M]; 8 int lowbit(int x) { return (x)&(-x); } 9 void add(int &d,int x,int y,int l,int r) 10 { 11 if (!d) d=top?st[top--]:++cnt; e[d].num+=y; 12 if (l==r) return; 13 int mid=l+r>>1; 14 if (x<=mid) add(e[d].ch[0],x,y,l,mid); else add(e[d].ch[1],x,y,mid+1,r); 15 if (!e[d].num) st[++top]=d,d=0; 16 } 17 void add(int x,int y,int v) { for (;x<=n;x+=lowbit(x)) add(root[x],y,v,1,n); } 18 int query(int d,int l,int r,int L,int R) 19 { 20 if (!d) return 0; 21 if (L<=l&&r<=R) return e[d].num; 22 int mid=l+r>>1,tot=0; 23 if (L<=mid) tot+=query(e[d].ch[0],l,mid,L,R); 24 if (R>mid) tot+=query(e[d].ch[1],mid+1,r,L,R); 25 return tot; 26 } 27 int query(int L,int R,int l,int r) 28 { 29 int ans=0; 30 for (int i=L-1;i;i-=lowbit(i)) ans-=query(root[i],1,n,l,r); 31 for (int i=R;i;i-=lowbit(i)) ans+=query(root[i],1,n,l,r); 32 return ans; 33 } 34 int main() 35 { 36 scanf("%d%d",&n,&m); 37 for (int i=1;i<=n;i++) scanf("%d",&a[i]),d[a[i]]=i; 38 for (int i=1,x;i<=n;i++) scanf("%d",&x),b[i]=d[x]; 39 for (int i=1;i<=n;i++) add(i,b[i],1); 40 for (int l,r,L,R,op;m;m--) 41 { 42 scanf("%d",&op); 43 if (op==1) scanf("%d%d%d%d",&L,&R,&l,&r),printf("%d\n",query(l,r,L,R)); 44 else scanf("%d%d",&l,&r),add(l,b[l],-1),add(r,b[r],-1),add(l,b[r],1),add(r,b[l],1),swap(b[l],b[r]); 45 } 46 }
[树状数组][权值线段树] Codeforces 1093E Intersection of Permutations
标签：cond sample 代码 amp += with struct 输入输出 after
原文地址：https://www.cnblogs.com/Comfortable/p/11206582.html
