UVa101 - The Blocks Problem

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桌上放有若干“blocks”

$\begin{figure}\centering\setlength{\unitlength}{0.0125in} %\begin{picture}(2......raisebox{0pt}[0pt][0pt]{$\bullet\bullet \bullet$ }}}\end{picture}\end{figure}$ Figure: Initial Blocks World

初始位置如图所示；有五种操作：

（1）move a onto b ：先将a和b之上的所有block都复位（恢复为initial状态），然后把a放在b上；

（2）move a over b ：先将a之上的所有block都复位，然后把a放在b所在的堆顶端；

（3）pile a onto b ：先将b之上的所有block都复位，然后把a以及a之上的所有block放在b之上；

（4）pile a over b ：将a以及a之上所有的block放在b所在的堆顶端

（5）quit ：退出

Sample Input 10move 9 onto 1move 8 over 1move 7 over 1move 6 over 1pile 8 over 6pile 8 over 5move 2 over 1move 4 over 9quitSample Output  0: 0 1: 1 9 2 4 2: 3: 3 4: 5: 5 8 7 6 6: 7: 8: 9:

解题思路：

应该用链表来做，我用数组模拟了链表，因为觉得指针容易出错。

代码：

#include <string.h>#include <stdio.h>#define MAXNUM 25typedef struct{int adress ;int start ;int pre ;int next ;} BlockType ;BlockType arr[MAXNUM];void resetinit(int from){int q ;int p = arr[from].next ;arr[from].next = -1 ;while(p != -1){arr[p].adress = p ;arr[p].start = p ;arr[p].pre = -1 ;q = arr[p].next ;arr[p].next = -1 ;p = q ;}}int main(){int T ;scanf("%d" , &T) ;for(int i = 0 ; i < T ; i ++){arr[i].start = i ;arr[i].adress = i ;arr[i].next = -1 ;arr[i].pre = -1 ;}char cmd1[10],cmd2[10];int pmt1,pmt2;while(scanf("%s" , cmd1)){if(! strcmp(cmd1,"quit"))break ;scanf("%d%s%d" , &pmt1 , cmd2 , &pmt2) ;int cmd ;if(! strcmp(cmd1,"move")){if(! strcmp(cmd2,"onto"))cmd = 1 ;elsecmd = 2 ;}else{if(! strcmp(cmd2,"onto"))cmd = 3 ;elsecmd = 4 ;}int p,q ;switch(cmd){case 1 : if(arr[pmt1].adress == arr[pmt2].adress)break ;resetinit(pmt1);resetinit(pmt2);if(arr[pmt1].pre != -1){p = arr[pmt1].pre ;arr[p].next = -1 ;}else{p = arr[pmt1].adress ;arr[p].start =  -1;}arr[pmt1].adress = arr[pmt2].adress ;arr[pmt1].pre = pmt2 ;arr[pmt2].next = pmt1 ;break ;case 2 :if (arr[pmt1].adress == arr[pmt2].adress)break ;resetinit(pmt1);if(arr[pmt1].pre != -1){p = arr[pmt1].pre ;arr[p].next = -1 ;}else{p = arr[pmt1].adress ;arr[p].start =  -1;}arr[pmt1].adress = arr[pmt2].adress ;p = pmt2 ;while(p != -1){q = p ;p = arr[p].next ;}arr[q].next = pmt1 ;arr[pmt1].pre = q ;break;case 3 : if (arr[pmt1].adress == arr[pmt2].adress)break ;resetinit(pmt2);p = arr[pmt1].pre ;if(p != -1){arr[p].next = -1 ;}else{q = arr[pmt1].adress ;arr[q].start = -1 ;}p = pmt1 ;while(p != -1){arr[p].adress = arr[pmt2].adress ;p = arr[p].next ;}arr[pmt2].next = pmt1 ;arr[pmt1].pre = pmt2 ;break;case 4 : if (arr[pmt1].adress == arr[pmt2].adress)break ;p = arr[pmt1].pre ;if(p != -1){arr[p].next = -1 ;}else{q = arr[pmt1].adress ;arr[q].start = -1 ;}p = pmt1 ;while(p != -1){arr[p].adress = arr[pmt2].adress ;p = arr[p].next ;}p = pmt1 ;q = pmt2 ;while(arr[q].next != -1){q = arr[q].next ;}arr[q].next = p ;arr[p].pre = q ;break;}}for(int i = 0 ; i < T ; i ++){printf("%d:" , i) ;int tmp = arr[i].start ;while(tmp != -1){printf(" %d" , tmp) ;tmp = arr[tmp].next ;}printf("\n") ;}return 0 ;}

CT说用了cin，cout，又是c语言风格，乱七八糟的……苦口婆心劝说下，老老实实改成了scanf和printf，结果AC的时间减少了0.004~