题意翻译
给出一个数列,有q个操作,每种操作是把区间[l,r]中等于x的数改成y.输出q步操作完的数列.
输入输出格式
输入格式：
The first line contains one integer n n n ( 1<=n<=200000 1<=n<=200000 1<=n<=200000 ) — the size of array a a a .
The second line contains n n n integers a1 a_{1} a1 , a2 a_{2} a2 , ..., an a_{n} an ( 1<=ai<=100 1<=a_{i}<=100 1<=ai<=100 ) — the elements of array a a a .
The third line contains one integer q q q ( 1<=q<=200000 1<=q<=200000 1<=q<=200000 ) — the number of queries you have to process.
Then q q q lines follow. i i i -th line contains four integers l l l , r r r , x x x and y y y denoting i i i -th query ( 1<=l<=r<=n 1<=l<=r<=n 1<=l<=r<=n , 1<=x,y<=100 1<=x,y<=100 1<=x,y<=100 ).
输出格式：
Print n n n integers — elements of array a a a after all changes are made.
输入输出样例
输入样例#1： 复制
输出样例#1： 复制
$c[L(rt)][i]=c[rt][c[L(rt)][i]]$
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<algorithm> 6 using namespace std; 7 int c[800001][101]; 8 bool flag[200001][101]; 9 int n,a[200001],ans[200001]; 10 void build(int rt,int l,int r) 11 {int i; 12 for (i=1;i<=100;i++) 13 c[rt][i]=i; 14 if (l==r) return; 15 int mid=(l+r)/2; 16 build(rt<<1,l,mid); 17 build(rt<<1|1,mid+1,r); 18 } 19 void pushdown(int rt) 20 {int i; 21 for (i=1;i<=100;i++) 22 { 23 c[rt<<1][i]=c[rt][c[rt<<1][i]]; 24 c[rt<<1|1][i]=c[rt][c[rt<<1|1][i]]; 25 } 26 for (i=1;i<=100;i++) 27 c[rt][i]=i; 28 } 29 void update(int rt,int l,int r,int L,int R,int x,int y) 30 {int i; 31 if (l>=L&&r<=R) 32 { 33 for (i=1;i<=100;i++) 34 if (c[rt][i]==x) c[rt][i]=y; 35 return; 36 } 37 pushdown(rt); 38 int mid=(l+r)/2; 39 if (L<=mid) update(rt<<1,l,mid,L,R,x,y); 40 if (R>mid) update(rt<<1|1,mid+1,r,L,R,x,y); 41 } 42 void query(int rt,int l,int r) 43 { 44 if (l==r) 45 { 46 ans[l]=c[rt][a[l]]; 47 return; 48 } 49 int mid=(l+r)/2; 50 pushdown(rt); 51 query(rt<<1,l,mid); 52 query(rt<<1|1,mid+1,r); 53 } 54 int main() 55 {int i,q,l,r,x,y; 56 cin>>n; 57 for (i=1;i<=n;i++) 58 { 59 scanf("%d",&a[i]); 60 } 61 build(1,1,n); 62 cin>>q; 63 for (i=1;i<=q;i++) 64 { 65 scanf("%d%d%d%d",&l,&r,&x,&y); 66 update(1,1,n,l,r,x,y); 67 } 68 query(1,1,n); 69 for (i=1;i<n;i++) 70 printf("%d ",ans[i]); 71 cout<<ans[n]<<endl; 72 }