codefroces 911G Mass Change Queries

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

5

1 2 3 4 5

3

3 5 3 5

1 5 5 1

1 5 1 5

输出样例#1： 复制

5 2 5 4 5

维护100颗线段树$c[rt][i]=j$，表示在这个区间里i颜色变成j

区间修改，将修改下传

初始$c[rt][i]=i$

至于下传状态：

$c[L(rt)][i]=c[rt][c[L(rt)][i]]$

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 using namespace std;
 int c[][];
 bool flag[][];
 int n,a[],ans[];
 void build(int rt,int l,int r)
 {int i;
   for (i=;i<=;i++)
     c[rt][i]=i;
   if (l==r) return;
   int mid=(l+r)/;
   build(rt<<,l,mid);
   build(rt<<|,mid+,r);
 }
 void pushdown(int rt)
 {int i;
   for (i=;i<=;i++)
     {
       c[rt<<][i]=c[rt][c[rt<<][i]];
       c[rt<<|][i]=c[rt][c[rt<<|][i]];
     }
   for (i=;i<=;i++)
     c[rt][i]=i;
 }
 void update(int rt,int l,int r,int L,int R,int x,int y)
 {int i;
   if (l>=L&&r<=R)
     {
       for (i=;i<=;i++)
     if (c[rt][i]==x) c[rt][i]=y;
       return;
     }
   pushdown(rt);
   int mid=(l+r)/;
   if (L<=mid) update(rt<<,l,mid,L,R,x,y);
   if (R>mid) update(rt<<|,mid+,r,L,R,x,y);
 }
 void query(int rt,int l,int r)
 {
   if (l==r)
     {
       ans[l]=c[rt][a[l]];
       return;
     }
   int mid=(l+r)/;
   pushdown(rt);
   query(rt<<,l,mid);
   query(rt<<|,mid+,r);
 }
 int main()
 {int i,q,l,r,x,y;
   cin>>n;
   for (i=;i<=n;i++)
     {
       scanf("%d",&a[i]);
     }
   build(,,n);
   cin>>q;
   for (i=;i<=q;i++)
     {
       scanf("%d%d%d%d",&l,&r,&x,&y);
       update(,,n,l,r,x,y);
     }
   query(,,n);
   for (i=;i<n;i++)
     printf("%d ",ans[i]);
   cout<<ans[n]<<endl;
 }