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  • Data Handler 大模拟 + 双端链表 hdu 4268

    E - Data Handler
    Time Limit:10000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u
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    Description

      You are in charge of data in a company, so you are called "Data Handler". Different from the data in computer, the data you have are really in huge volume, and each data contains only one integer. All the data are placed in a line from left to right. There are two "hand" to handle the data, call hand "L" and hand "R". Every hand is between two adjacent data or at the end of the data line.
      In one day, the company gives you many commands to handle these data, so you should finish them one by one. At the beginning, there are N data, and hand "L" and "R" are in some positions. Each command is one the following formats:
      (1)MoveLeft L/R: it means that you should move the hand "L"/"R" left one data unit;


      (2)MoveRight L/R: it means that you should move the hand "L"/"R" right one data unit;



      (3)Insert L X: it means that you should insert the data that contains X at the right of the hand "L";



      (4)Insert R X: it means that you should insert the data that contains X at the left of the hand "R";



      (5)Delete L: it means that you should delete the one data at the right of the hand "L";



      (6)Delete R: it means that you should delete the one data at the left of the hand "R";



      (7)Reverse: it means that you should reverse all the data between hand "L" and hand "R".


      After finish all the commands, you should record all the data from left to right. So please do it.
     

    Input

      The first line contains an integer T(1<=T<=10), the number of test cases.
      Then T test cases follow. For each test case, the first line contains an integer N(1<=N<=500000), the number of data at the beginning. The second line contains N integers, means the integer in each data, from left to right. The third line contains two integers L and R (1<=L<=R<=N), the positions of hand "L" and hand "R". It means that hand "L" is at the left of the L-th data and hand "R" is at the right of the R-th data. The fourth line contains one integer M(1<=M<=500000), the number of commands. Then M lines follow, each line contains a command in the above format. All the integers in the data will in range [-10000,10000].
      It is guaranteed that there are always some data between hand "L" and "R", and if the hand is at the left/right end of the data line, it will not receive the command MoveLeft/MoveRight.
      Because of large input, please use scanf instead of cin.
     

    Output

      For each test case, output the integers in the data from left to right in one line, separated in a single space.
      Because of large output, please use printf instead of cout.
      
     

    Sample Input

    2
    5
    1 2 3 4 5
    1 5
    5
    MoveLeft R
    Insert R 6
    Reverse
    Delete R
    Insert L 7
    5
    6536 5207 2609 6604 -4046
    1 3
    5
    Delete L
    Insert R -9221
    Reverse
    Delete L
    MoveRight L
     

    Sample Output

    7 6 4 3 2 5
    2609 5207 6604 -4046
     
    题意: 给出一串数字,左端L 指向左边某个数字下标,右端R 指向右边某个数字下标,
           现在给出一些可行操作 MoveLeft Insert Reverse Delete (详见题意配图)
        要求在L和R中间对这串数据进行操作 最后输出完成操作的这串数字
    分析:
       建立一个双端链表 将每个数字连接起来 每次进行操作只需重新连接某个数字的端点
       插入和移动操作 单向链表就可以进行 然而双向链表最大的优势在于 翻转
     
       给出的数据最大为500000
       考虑纯暴力(每一次都将L与R间数据全部翻转)的最坏情况 500000*500000 很明显超时
        这里的暴力需要一点技巧
        
      比如 一串数字为
            1 2 3 4 5
             L指向1 的下标   R指向 5的下标
      现在我们把这串数字分为三段 L及其左边一段 L与R中间一段 R及其右边一段
      数字之间 左端接左数字下标 右端接右数字下标 中间段就只需标记读取方向 而不改变端点的连接方式
      
      那么
      如果 L与R中间的数 是从左到右读取 那么MoveLeft Insert操作只需按照题意进行
      如果 是从右向左读取  MoveLeft L 就相当于是 将1 连到 4 与 5 之间 再将L指向 虚拟下标 0 这样做是为了保证L与R之间的读取方向不变
       其他操作同理
     
    来分析一下第一组数据吧
       0 | 1 2 3 4 5 | 6
       L = 0  R = 6
    MoveLeft R
    0 | 1 2 3 4 5 | 6
    L           R    L与R间读取方向 :左→右
    Insert R 6
    0 | 1 2 3 4 6 5 | 6
    L             R    L与R间读取方向 :左→右
    Reverse
    0 | 1 2 3 4 6 5 | 6
    L             R    L与R间读取方向 :右→左
    Delete R
    0 | 2 3 4 6 5 | 6
    L           R    L与R间读取方向 :右→左
    Insert L 7
    0 | 2 3 4 6 7 5 | 6
    L             R    L与R间读取方向 :右→左
    那么此时输出为 7 6 4 3 2 5 L与R间数据就按照了从右向左输出 而连接顺序不变
     
    附上AC代码
      1 #include<cstring>
      2 #include<cstdio>
      3 #include<iostream>
      4 #include<algorithm>
      5 #include<cmath>
      6 
      7 using namespace std;
      8 
      9 #define AA struct ss
     10 
     11 AA
     12 {
     13     int r,l;
     14     int num;
     15     int ans;
     16 }T[1500006];
     17 
     18 int L,R;
     19 int vis;
     20 int flag;
     21 int sum;
     22 int n;
     23 
     24 bool MoveLeft(char *s)
     25 {
     26     if(strcmp(s,"MoveLeft")!=0) return false;
     27 
     28     char ss[5];
     29     scanf("%s",ss);
     30 
     31     if(ss[0]=='R')
     32     {
     33         int p=T[R].l;
     34         if(!flag) R=T[p].num;
     35         else
     36         {
     37             int q=T[L].r;
     38             int q1=T[q].r;
     39             T[R].l=T[q].num;
     40             T[p].r=T[q].num;
     41 
     42             T[q].l=T[p].num;
     43             T[q].r=T[R].num;
     44 
     45             T[q1].l=T[L].num;
     46             T[L].r=T[q1].num;
     47 
     48             R= T[q].num;
     49         }
     50         sum--;
     51     }
     52     else
     53     {
     54         int p=T[L].l;
     55         if(!flag) L=T[p].num;
     56         else
     57         {
     58             int q=T[L].r;
     59             int q1=T[R].l;
     60             T[p].r=T[q].num;
     61             T[q].l=T[p].num;
     62 
     63             T[q1].r=T[L].num;
     64             T[L].r=T[R].num;
     65 
     66             T[R].l=T[L].num;
     67             T[L].l=T[q1].num;
     68 
     69             L = T[p].num;
     70         }
     71         sum++;
     72     }
     73     return true;
     74 }
     75 
     76 bool MoveRight(char *s)
     77 {
     78     if(strcmp(s,"MoveRight")!=0) return false;
     79 
     80     char ss[5];
     81     scanf("%s",ss);
     82 
     83     if(ss[0]=='R')
     84     {
     85         int p=T[R].r;
     86         if(!flag) R=T[p].num;
     87         else
     88         {
     89             int q=T[L].r;
     90             int q1=T[R].l;
     91             T[L].r= T[R].num;
     92             T[R].r= T[q].num;
     93 
     94             T[q].l= T[R].num;
     95             T[R].l= T[L].num;
     96 
     97             T[q1].r= T[p].num;
     98             T[p].l= T[q1].num;
     99 
    100             R= T[p].num;
    101         }
    102         sum++;
    103     }
    104     else
    105     {
    106         int p=T[L].r;
    107         if(!flag) L=T[p].num;
    108         else
    109         {
    110             int q= T[R].l;
    111             int q1= T[q].l;
    112 
    113             T[L].r= T[q].num;
    114             T[q].l= T[L].num;
    115 
    116             T[q].r= T[p].num;
    117             T[p].l= T[q].num;
    118 
    119             T[q1].r=T[R].num;
    120             T[R].l=T[q1].num;
    121 
    122             L= T[q].num;
    123         }
    124         sum--;
    125     }
    126     return true;
    127 }
    128 
    129 bool Insert(char *s)
    130 {
    131     if(strcmp(s,"Insert")!=0 ) return  false;
    132 
    133     char p[6];
    134     scanf("%s",p);
    135     scanf("%d",&T[vis].ans);
    136     T[vis].num=vis;
    137 
    138     if(p[0]=='R')
    139     {
    140         int q = T[R].l;
    141 
    142         if(!flag)
    143         {
    144             T[q].r=T[vis].num;
    145             T[vis].l=T[q].num;
    146             T[vis].r=T[R].num;
    147             T[R].l=T[vis].num;
    148         }
    149         else
    150         {
    151             q=T[L].r;
    152             T[L].r=T[vis].num;
    153             T[vis].l=T[L].num;
    154             T[vis].r=T[q].num;
    155             T[q].l=T[vis].num;
    156         }
    157     }
    158     else
    159     {
    160         int q = T[L].r;
    161 
    162         if(!flag)
    163         {
    164             T[L].r=T[vis].num;
    165             T[vis].l=T[L].num;
    166             T[vis].r=T[q].num;
    167             T[q].l=T[vis].num;
    168         }
    169         else
    170         {
    171             q = T[R].l;
    172             T[q].r=T[vis].num;
    173             T[vis].l=T[q].num;
    174             T[vis].r=T[R].num;
    175             T[R].l=T[vis].num;
    176         }
    177 
    178     }
    179     vis++;
    180     sum++;
    181 
    182     return true;
    183 }
    184 
    185 bool Delete(char *s)
    186 {
    187     if(strcmp(s,"Delete")!=0 ) return  false;
    188 
    189     char p[5];
    190     scanf("%s",p);
    191 
    192     if(sum==0) return true;
    193     if(p[0]=='R')
    194     {
    195         if(flag){
    196         int q1 = T[L].r;
    197         int q2 = T[q1].r;
    198 
    199         T[L].r=T[q2].num;
    200         T[q2].l=T[L].num;
    201 
    202         T[q1].r=q1;
    203         T[q1].l=q1;
    204         }
    205         else
    206         {
    207             int q1=T[R].l;
    208             int q2=T[q1].l;
    209 
    210             T[q2].r=T[R].num;
    211             T[R].l=T[q2].num;
    212 
    213             T[q1].r=q1;
    214             T[q1].l=q1;
    215         }
    216     }
    217     else
    218     {
    219         if(!flag){
    220         int q1 = T[L].r;
    221         int q2 = T[q1].r;
    222 
    223         T[q2].l=T[L].num;
    224         T[L].r=q2;
    225 
    226         T[q1].l=q1;
    227         T[q1].r=q1;
    228         }
    229         else
    230         {
    231             int q1=T[R].l;
    232             int q2=T[q1].l;
    233 
    234             T[q2].r=T[R].num;
    235             T[R].l=T[q2].num;
    236 
    237             T[q1].r=q1;
    238             T[q1].l=q1;
    239         }
    240     }
    241     sum--;
    242     return true;
    243 }
    244 
    245 bool Reverse(char *s)
    246 {
    247     if(strcmp(s,"Reverse")!=0) return false;
    248 
    249     if(!flag) flag=1;
    250     else flag=0;
    251     return true;
    252 }
    253 
    254 void change()
    255 {
        //这里是将L 与 R 端点 链表节点重新连接一下 方便输出
    256 if(!sum) return ; 257 int q1=T[R].l,q2=T[L].r; 258 259 T[L].r = T[q1].num; 260 T[q1].r = T[L].num; 261 T[q2].l = T[R].num; 262 T[R].l = T[q2].num; 263 } 264 265 void pr() 266 { 267 if(flag) change(); 268 269 int q=0; 270 for(int i=T[0].r ; T[i].num!=n+1 ; ) 271 { 272 printf("%d",T[i].ans); 273 int s=i; 274 275 if(T[i].l==q) i=T[i].r; 276 else if(T[i].r==q) i=T[i].l; //这里的输出用了一点小技巧 :判断当前数据在上一数据的位置l 或者 r 然后改变输出 i 的方向 277 278 q=T[s].num; 279 if(T[i].num!=n+1) printf(" "); 280 } 281 } 282 283 int main() 284 { 285 int t; 286 scanf("%d",&t); 287 while(t--) 288 { 289 scanf("%d",&n); 290 291 T[0].l=0; 292 T[0].r=1; 293 T[0].num=0; 294 T[0].ans=0; 295 296 T[n+1].l=n; 297 T[n+1].r=n+1; 298 T[n+1].num=n+1; 299 T[n+1].ans=0; 300 vis = n+2; 301 flag = 0; 302 303 for(int i=1;i<=n;i++) 304 { 305 scanf("%d",&T[i].ans); 306 T[i].num=i; 307 T[i].r=i+1; 308 T[i].l=i-1; 309 } 310 311 scanf("%d%d",&L,&R); 312 L--,R++; 313 sum = R-L-1; 314 315 int time; 316 scanf("%d",&time); 317 318 while(time--) 319 { 320 char s[50]; 321 scanf("%s",s); 322 if( MoveLeft(s) ) ; 323 else if( MoveRight(s) ) ; 324 else if( Insert(s) ) ; 325 else if( Delete(s) ) ; 326 else if( Reverse(s) ) ; 327 if(sum<=1) flag=0; 328 } 329 330 pr(); 331 puts(""); 332 333 } 334 return 0; 335 }

    还有一种是 双端队列做法 利用已有的函数简化了链表操作

     
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  • 原文地址:https://www.cnblogs.com/wsaaaaa/p/4427947.html
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