HDU 3487 Play with Chain | Splay

Play with Chain

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

【Problem Description】

YaoYao is fond of playing his chains. He has a chain containing n diamonds on it. Diamonds are numbered from 1 to n. At first, the diamonds on the chain is a sequence: 1, 2, 3, …, n. He will perform two types of operations:

CUT a b c: He will first cut down the chain from the ath diamond to the bth diamond. And then insert it after the cth diamond on the remaining chain. For example, if n=8, the chain is: 1 2 3 4 5 6 7 8; We perform “CUT 3 5 4”, Then we first cut down 3 4 5, and the remaining chain would be: 1 2 6 7 8. Then we insert “3 4 5” into the chain before 5th diamond, the chain turns out to be: 1 2 6 7 3 4 5 8.
FLIP a b: We first cut down the chain from the ath diamond to the bth diamond. Then reverse the chain and put them back to the original position. For example, if we perform “FLIP 2 6” on the chain: 1 2 6 7 3 4 5 8. The chain will turn out to be: 1 4 3 7 6 2 5 8
He wants to know what the chain looks like after perform m operations. Could you help him?

 

【Input】

There will be multiple test cases in a test data. For each test case, the first line contains two numbers: n and m (1≤n, m≤3*100000), indicating the total number of diamonds on the chain and the number of operations respectively. Then m lines follow, each line contains one operation. The command is like this: CUT a b c   // Means a CUT operation, 1 ≤ a ≤ b ≤ n, 0≤ c ≤ n-(b-a+1). FLIP a b    // Means a FLIP operation, 1 ≤ a < b ≤ n. The input ends up with two negative numbers, which should not be processed as a case.

 

【Output】

For each test case, you should print a line with n numbers. The ith number is the number of the ith diamond on the chain.

 

【Sample Input】

8 2 
CUT 3 5 4 
FLIP 2 6 
-1 -1

【Sample Output】

1 4 3 7 6 2 5 8

 

【题意】

给出一列数，而后对整个数列执行两种操做：切下一段插入到另外的位置，或者把其中的一整段整个翻转一下。

求通过一系列操做以后，数列最后的样子。

 

【分析】

数据范围最高可以到达3e5那么大，所以算法至少要是O(nlogn)复杂度如下才可能达到要求。

考虑采用Splay解决（这样的题目只能用这种动态维护的树结构不是么？）

 

初始先建树，把1~n加入Splay树。因为数列在后面是要被打乱顺序的，Splay二叉平衡树的性质只有在初始的时候是被保持的，以后是靠size，即每一个点在中序遍历中的位置来维护。最后输出数列则只须要中序遍历一遍便可。

切割操做：若要切下a~b段，则把第a-1个结点移到根，把第b+1个结点移到根如下（即跟的右子树），则整个a~b段就落在b+1的左子树上，切出来。插入到c的时候，将c移到根，c+1移到根的右子树，则切出来的插入到c+1的左子树便可

翻转操做：用上面相同的方法把a~b整合到一棵子树上，而后能够参考线段树标记的方法，经过标记来完成访问结点的翻转等操做。

具体能够在纸上模拟一下......

 

【教训】

教训仍是比较惨痛的...卡在这道题上很久了。

首先是输入输出之后要特别注意结尾方式，两个负数结尾仍是两个-1结尾

把各类可能出现的不一样状况考虑完整

 

  1 /* ***********************************************
  2 MYID    : Chen Fan
  3 LANG    : G++
  4 PROG    : HDU3487
  5 ************************************************ */
  6 
  7 #include <iostream>
  8 #include <cstdio>
  9 #include <cstring>
 10 #include <algorithm>
 11 
 12 using namespace std;
 13 
 14 #define MAXN 300010
 15 
 16 int sons[MAXN][2];
 17 int father[MAXN],size[MAXN],data[MAXN],list[MAXN];
 18 bool flag[MAXN];
 19 int spt=0,spttail=0;
 20 
 21 void down(int x)
 22 {
 23     if (flag[x])
 24     {
 25         flag[x]=0;
 26         swap(sons[x][0],sons[x][1]);
 27         flag[sons[x][0]]^=1;
 28         flag[sons[x][1]]^=1;
 29     }
 30 }
 31 
 32 void rotate(int x,int w) //rotate(node,0/1)
 33 {
 34     int y=father[x];
 35     down(y);down(x);
 36     sons[y][!w]=sons[x][w];
 37     if (sons[x][w]) father[sons[x][w]]=y;
 38 
 39     father[x]=father[y];
 40     if (father[y]) sons[father[y]][y==sons[father[y]][1]]=x;
 41 
 42     sons[x][w]=y;
 43     father[y]=x;
 44 
 45     size[x]=size[y];
 46     size[y]=size[sons[y][0]]+size[sons[y][1]]+1;
 47 }
 48 
 49 void splay(int x,int y) //splay(node,position)
 50 {
 51     down(x);
 52     while(father[x]!=y)
 53     {
 54         if (father[father[x]]==y) rotate(x,x==sons[father[x]][0]);
 55         else 
 56         {
 57             int t=father[x];
 58             int w=(sons[father[t]][0]==t);
 59             if (sons[t][w]==x)
 60             {
 61                 rotate(x,!w);
 62                 rotate(x,w);
 63             } else 
 64             {
 65                 rotate(t,w);
 66                 rotate(x,w);
 67             }
 68         }
 69     }
 70     if (!y) spt=x;
 71 }
 72 
 73 void select(int x,int v,int p) //select(root,k,position)
 74 {
 75     down(x);
 76     while(v!=size[sons[x][0]]+1)
 77     {
 78         if (v<=size[sons[x][0]]) 
 79         {
 80             x=sons[x][0];
 81             down(x);
 82         }
 83         else 
 84         {
 85             v-=size[sons[x][0]]+1;
 86             x=sons[x][1];
 87             down(x);
 88         }
 89     }
 90     splay(x,p);
 91 }
 92 
 93 bool done=false;
 94 
 95 void outp(int x)
 96 {
 97     down(x);
 98     if (sons[x][0]) outp(sons[x][0]);
 99     if (done) printf(" ");
100     done=true;
101     printf("%d",data[x]);
102     if (sons[x][1]) outp(sons[x][1]);
103 }
104 
105 void maketree(int l,int r)
106 {
107     spttail++;
108     int now=spttail,w=(l+r)/2,ls=0,rs=0;
109     data[now]=w;
110     flag[now]=false;
111     sons[now][0]=0;
112     sons[now][1]=0;
113     
114     if (l<=w-1)
115     {
116         ls=spttail+1;
117         sons[now][0]=ls;
118         father[ls]=now;
119         maketree(l,w-1);
120     }
121     if (w+1<=r)
122     {
123         rs=spttail+1;
124         sons[now][1]=rs;
125         father[rs]=now;
126         maketree(w+1,r);
127     }
128 
129     size[now]=size[ls]+size[rs]+1;
130 }
131 
132 int main()
133 {
134     freopen("3487.txt","r",stdin);
135     
136     int n,m;
137     scanf("%d%d",&n,&m);
138     while(!(n<0&&m<0))
139     {
140         spt=1;
141         spttail=0;
142         father[1]=0;
143         maketree(1,n);
144 
145         for (int i=1;i<=m;i++)
146         {
147             char s[10];
148             scanf("%s",&s);
149             if (s[0]=='C')
150             {
151                 int a,b,c,temp;
152                 scanf("%d%d%d",&a,&b,&c);
153 
154                 if (a>1)
155                 {
156                     select(spt,a-1,0);
157                     if (b<n)
158                     {
159                         select(spt,b+1,spt);
160                         temp=sons[sons[spt][1]][0];
161                         sons[sons[spt][1]][0]=0;
162                         size[spt]-=size[temp];
163                         size[sons[spt][1]]-=size[temp];
164                     } else 
165                     {
166                         temp=sons[spt][1];
167                         sons[spt][1]=0;
168                         size[spt]-=size[temp];
169                     }
170                 } else 
171                 {
172                     if (b<n)
173                     {
174                         select(spt,b+1,0);
175                         temp=sons[spt][0];
176                         sons[spt][0]=0;
177                         size[spt]-=size[temp];
178                     } else temp=spt;
179                 }
180 
181                 if (c>0)
182                 {
183                     select(spt,c,0);
184                     if (c==size[spt])
185                     {
186                         sons[spt][1]=temp;
187                         father[temp]=spt;
188                         size[spt]+=size[temp];
189                     } else
190                     {
191                         select(spt,c+1,spt);
192                         sons[sons[spt][1]][0]=temp;
193                         father[temp]=sons[spt][1];
194                         size[spt]+=size[temp];
195                         size[sons[spt][1]]+=size[temp];
196                     }
197                 } else 
198                 {
199                     if (spt!=temp)
200                     {
201                         select(spt,1,0);
202                         sons[spt][0]=temp;
203                         father[temp]=spt;
204                         size[spt]+=size[temp];
205                     }
206                 }  
207             } else 
208             {
209                 int a,b,temp;
210                 scanf("%d%d",&a,&b);
211                 if (a>1)
212                 {
213                     select(spt,a-1,0);
214                     if (b<n)
215                     {
216                         select(spt,b+1,spt);
217                         temp=sons[sons[spt][1]][0];
218                     } else 
219                     {
220                         temp=sons[spt][1];
221                     }
222                 } else 
223                 {
224                     if (b<n)
225                     {
226                         select(spt,b+1,0);
227                         temp=sons[spt][0];
228                     } else temp=spt;
229                 }
230                 flag[temp]^=1;
231             }
232         }
233         done=false;
234         outp(spt);
235         printf("\n");
236         scanf("%d%d",&n,&m);
237     }
238 
239     return 0;
240 }

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