題面如下:

Tyvj 1729 文藝平衡樹

Time Limit:?1 Sec ?Memory Limit:?128 MB

Description

您需要寫一種數(shù)據(jù)結(jié)構(gòu)（可參考題目標(biāo)題），來維護一個有序數(shù)列，其中需要提供以下操作：翻轉(zhuǎn)一個區(qū)間，例如原有序序列是5?4?3?2?1，翻轉(zhuǎn)區(qū)間是[2,4]的話，結(jié)果是5?2?3?4?1?

Input

第一行為n,m?n表示初始序列有n個數(shù)，這個序列依次是(1,2……n-1,n)??m表示翻轉(zhuǎn)操作次數(shù)
接下來m行每行兩個數(shù)[l,r]?數(shù)據(jù)保證?1<=l<=r<=n?

Output

輸出一行n個數(shù)字，表示原始序列經(jīng)過m次變換后的結(jié)果?

Sample Input

5 3 1 3 1 3 1 4

Sample Output

4 3 2 1 5

啊,又是一道板子題

題目要求支持的操作只有區(qū)間翻轉(zhuǎn),其中定位需要維護$size$,翻轉(zhuǎn)要維護翻轉(zhuǎn)標(biāo)記,所以總體來說還是比較好寫的.Splay寫這題思路也很明確,每次Splay區(qū)間左/右端點的左/右邊結(jié)點到根/根的右孩子,然后對根的右子結(jié)點的左子樹打翻轉(zhuǎn)標(biāo)記即可w

直接貼代碼w

GitHub

1 /********************************* 2 Judge Result:Accepted 3 4 *********************************/ 5 #include <cstdio> 6 #include <vector> 7 #include <cstring> 8 #include <cstring> 9 #include <iostream> 10 #include <algorithm> 11 12 #define lch chd[0] 13 #define rch chd[1] 14 #define kch chd[k] 15 #define xch chd[k^1] 16 17 const int INF=0x2FFFFFFF; 18 19 class SplayTree{ 20 private: 21 struct Node{ 22 int k; 23 int sz; 24 bool rev; 25 Node* prt; 26 Node* chd[2]; 27 Node(const int& key){ 28 this->k=key; 29 this->sz=1; 30 this->prt=NULL; 31 this->lch=NULL; 32 this->rch=NULL; 33 this->rev=false; 34 } 35 ~Node(){ 36 if(this->lch!=NULL) 37 delete this->lch; 38 if(this->rch!=NULL) 39 delete this->rch; 40 } 41 inline void Maintain(){ 42 if(this!=NULL) 43 this->sz=this->lch->size()+this->rch->size()+1; 44 } 45 inline void Swap(){ 46 if(this!=NULL){ 47 this->rev=!this->rev; 48 std::swap(this->lch,this->rch); 49 } 50 } 51 inline void PushDown(){ 52 if(this->rev){ 53 this->rev=false; 54 this->lch->Swap(); 55 this->rch->Swap(); 56 } 57 } 58 inline int key(){ 59 return this==NULL?0:this->k; 60 } 61 inline int size(){ 62 return this==NULL?0:this->sz; 63 } 64 }*root; 65 inline void Rotate(Node* root,int k){ 66 Node* tmp=root->xch; 67 root->PushDown(); 68 tmp->PushDown(); 69 tmp->prt=root->prt; 70 if(root->prt==NULL) 71 this->root=tmp; 72 else if(root->prt->lch==root) 73 root->prt->lch=tmp; 74 else 75 root->prt->rch=tmp; 76 root->xch=tmp->kch; 77 if(tmp->kch!=NULL) 78 tmp->kch->prt=root; 79 tmp->kch=root; 80 root->prt=tmp; 81 root->Maintain(); 82 tmp->Maintain(); 83 } 84 void Splay(Node* root,Node* prt=NULL){ 85 while(root->prt!=prt){ 86 int k=root->prt->lch==root; 87 if(root->prt->prt==prt){ 88 Rotate(root->prt,k); 89 } 90 else{ 91 int d=root->prt->prt->lch==root->prt; 92 Rotate(k==d?root->prt->prt:root->prt,k); 93 Rotate(root->prt,d); 94 } 95 } 96 } 97 Node* Build(const std::vector<int>& v,int l,int r){ 98 if(l>r) 99 return NULL; 100 int mid=(l+r)>>1; 101 Node* tmp=new Node(v[mid]); 102 tmp->lch=Build(v,l,mid-1); 103 tmp->rch=Build(v,mid+1,r); 104 if(tmp->lch!=NULL) 105 tmp->lch->prt=tmp; 106 if(tmp->rch!=NULL) 107 tmp->rch->prt=tmp; 108 tmp->Maintain(); 109 return tmp; 110 } 111 void PrintTree(Node* root,int deep){ 112 for(int i=0;i<deep;i++) 113 fputc(' ',stderr); 114 fprintf(stderr, "(root=0x%X,key=%dsize=%d)\n", root,root->key(),root->size()); 115 if(root==NULL) 116 return; 117 PrintTree(root->lch,deep+1); 118 PrintTree(root->rch,deep+1); 119 } 120 public: 121 SplayTree(){ 122 this->root=new Node(-INF); 123 this->root->rch=new Node(-INF); 124 this->root->rch->prt=this->root; 125 } 126 SplayTree(const std::vector<int>& v){ 127 this->root=Build(v,0,v.size()-1); 128 } 129 ~SplayTree(){ 130 delete this->root; 131 } 132 Node* Kth(int pos){ 133 ++pos; 134 Node* root=this->root; 135 while(root!=NULL){ 136 root->PushDown(); 137 int k=root->lch->size()+1; 138 if(pos<k) 139 root=root->lch; 140 else if(pos==k) 141 return root; 142 else{ 143 pos-=k; 144 root=root->rch; 145 } 146 } 147 return NULL; 148 } 149 inline void Reverse(const int& l,const int& r){ 150 this->Splay(this->Kth(l-1)); 151 this->Splay(this->Kth(r+1),this->root); 152 this->root->rch->lch->Swap(); 153 this->root->rch->Maintain(); 154 this->root->Maintain(); 155 } 156 inline void Insert(const int& pos,SplayTree* data){ 157 this->Splay(this->Kth(pos)); 158 this->Splay(this->Kth(pos+1),this->root); 159 Node* tmp=data->root; 160 data->root=NULL; 161 this->root->rch->lch=tmp; 162 tmp->prt=this->root->rch; 163 this->root->rch->Maintain(); 164 this->root->Maintain(); 165 } 166 void Print(){ 167 this->PrintTree(this->root,0); 168 } 169 }; 170 171 int FastRead(); 172 173 int main(){ 174 freopen("sph.in","r",stdin); 175 freopen("sph.out","w",stdout); 176 SplayTree* tree=new SplayTree(); 177 std::vector<int> v; 178 int n=FastRead(); 179 int m=FastRead(); 180 int a,b; 181 for(int i=1;i<=n;i++){ 182 v.push_back(i); 183 } 184 tree->Insert(0,new SplayTree(v)); 185 for(int i=0;i<m;i++){ 186 a=FastRead(); 187 b=FastRead(); 188 tree->Reverse(a,b); 189 } 190 for(int i=1;i<=n;i++){ 191 printf("%d ",tree->Kth(i)->key()); 192 } 193 putchar('\n'); 194 return 0; 195 } 196 197 int FastRead(){ 198 int ans=0; 199 bool neg=false; 200 register char ch=getchar(); 201 while(!isdigit(ch)){ 202 if(ch=='-') 203 neg=true; 204 ch=getchar(); 205 } 206 while(isdigit(ch)){ 207 ans=ans*10+ch-'0'; 208 ch=getchar(); 209 } 210 if(neg) 211 ans=-ans; 212 return ans; 213 } Backup

然后放圖~

?

轉(zhuǎn)載于:https://www.cnblogs.com/rvalue/p/7230060.html

總結(jié)

以上是生活随笔為你收集整理的[Tyvj 1729] 文艺平衡树的全部內(nèi)容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網(wǎng)站內(nèi)容還不錯，歡迎將生活随笔推薦給好友。