Description
您需要写一种数据结构（可参考题目标题），来维护一个有序数列，其中需要提供以下操作：翻转一个区间，例如原有序序列是5 4 3 2 1，翻转区间是[2,4]的话，结果是5 2 3 4 1
Input
第一行为n,m n表示初始序列有n个数，这个序列依次是(1,2……n-1,n) m表示翻转操作次数
接下来m行每行两个数[l,r] 数据保证 1<=l<=r<=n
Output
输出一行n个数字，表示原始序列经过m次变换后的结果
Sample Input
5 3
1 3
1 3
1 4
1 3
1 3
1 4
Sample Output
4 3 2 1 5
HINT
N,M<=100000
Source
做完了维修数列之后，这就是个大水题
按题意反转序列，最后更新所有节点，输出中序遍历即可。。。。。。
1 // It is made by XZZ 2 #include<cstdio> 3 #include<algorithm> 4 #define Fname "sph" 5 using namespace std; 6 #define rep(a,b,c) for(rg int a=b;a<=c;a++) 7 #define drep(a,b,c) for(rg int a=b;a>=c;a--) 8 #define erep(a,b) for(rg int a=fir[b];a;a=nxt[a]) 9 #define il inline 10 #define rg register 11 #define vd void 12 #define Now t[now] 13 #define pr pair<int,int> 14 #define mp make_pair 15 typedef long long ll; 16 il int gi(){ 17 rg int x=0,f=1;rg char ch=getchar(); 18 while(ch<'0'||ch>'9')f=ch=='-'?-1:f,ch=getchar(); 19 while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar(); 20 return x*f; 21 } 22 ll seed=19260817; 23 int root; 24 const int maxn=101010; 25 il int Rand(){return (int)(seed=seed*48271%2147483647);} 26 struct node{int data,size,ls,rs,rand;bool rev;}t[maxn]; 27 int newnode(int i){t[i]=(node){i,1,0,0,Rand(),0};return i;} 28 il vd down(int now){ 29 if(now&&Now.rev)swap(Now.ls,Now.rs),Now.rev=0,t[Now.ls].rev^=1,t[Now.rs].rev^=1; 30 } 31 il vd reset(int now){ 32 down(Now.ls),down(Now.rs); 33 Now.size=t[Now.ls].size+t[Now.rs].size+1; 34 } 35 il int build(int n){ 36 int stack[maxn],top=0,last; 37 rep(i,1,n){ 38 int now=newnode(i);last=0; 39 while(top&&t[stack[top]].rand>Now.rand)reset(stack[top]),last=stack[top--]; 40 if(top)t[stack[top]].rs=now;Now.ls=last,stack[++top]=now; 41 } 42 while(top)reset(stack[top--]); 43 return stack[1]; 44 } 45 il int merge(int a,int b){ 46 if(!a||!b)return a|b; 47 if(t[a].rand<t[b].rand){down(a),t[a].rs=merge(t[a].rs,b),reset(a);return a;} 48 else{down(b),t[b].ls=merge(a,t[b].ls),reset(b);return b;} 49 } 50 il pr split(int now,int num){ 51 if(!now)return mp(0,0); 52 down(now); 53 int ls=Now.ls,rs=Now.rs; 54 if(num==t[Now.ls].size){Now.ls=0,reset(now);return mp(ls,now);} 55 if(num==t[Now.ls].size+1){Now.rs=0,reset(now);return mp(now,rs);} 56 if(num<t[Now.ls].size){ 57 pr T=split(Now.ls,num); 58 Now.ls=T.second,reset(now); 59 return mp(T.first,now); 60 }else{ 61 pr T=split(Now.rs,num-t[Now.ls].size-1); 62 Now.rs=T.first,reset(now); 63 return mp(now,T.second); 64 } 65 } 66 il vd dfs(int now){ 67 if(now)down(Now.ls),down(Now.rs),dfs(Now.ls),printf("%d ",Now.data),dfs(Now.rs); 68 } 69 int main(){ 70 freopen(Fname".in","r",stdin); 71 freopen(Fname".out","w",stdout); 72 int n=gi(),m=gi(); 73 root=build(n); 74 while(m--){ 75 int pos=gi(),tot=gi()-pos+1; 76 pr T=split(root,pos-1),TT=split(T.second,tot); 77 t[TT.first].rev^=1; 78 root=merge(T.first,merge(TT.first,TT.second)); 79 } 80 down(root),dfs(root); 81 return 0; 82 }