UVa 10017 - The Never Ending Towers of Hanoi
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題目:輸入n塊盤子的漢諾塔,以及移動的步數m,模擬前m步的操作(輸出各柱子狀態)。
分析:模擬,搜索。直接利用遞歸求解漢諾塔的前m步,然後模擬輸出。
漢諾塔求解:定義遞歸函數 f(當前柱子X,借住柱子Y,目標柱子Z,移動X上的前n塊);
則可以生成遞歸關係:
1,首先借住Z把X上的前n-1塊盤子移動到Y上;
2,然後把地n塊盤子放在地Z柱子上;
3,最後借住X把Y上的n-1塊盤子移動到Z上;
利用set和get兩個函數不斷的取某個柱子的最上面的盤子,然後移動到另一個模擬輸出;
說明:注意輸出時problem和數據間有空行(包括自己的,即下面的)。
#include <stdio.h>#include <stdlib.h>#define PEG_ID_A 0#define PEG_ID_B 1#define PEG_ID_C 2// tower_of_hanoiint _hanoi_move_steps = 0; int _hanoi_max_steps = 0;int _hanoi_move_steps_list[100001];int _hanoi_peg[3][256];int _hanoi_peg_size[3];// set disk(id is disk_id) to peg(id is peg_id)void hanoi_disk_set(int peg_id, int disk_id){ _hanoi_peg[peg_id][_hanoi_peg_size[peg_id] ++] = disk_id;}// get disk(top disk) from peg(id is peg_id)int hanoi_disk_get(int peg_id){ if (_hanoi_peg_size[peg_id] > 0) { return _hanoi_peg[peg_id][-- _hanoi_peg_size[peg_id]]; }else { return -1; }}void hanoi_initital(int hanoi_size, int max_steps){ _hanoi_move_steps = 0; _hanoi_max_steps = max_steps; _hanoi_peg_size[PEG_ID_A] = 0; _hanoi_peg_size[PEG_ID_B] = 0; _hanoi_peg_size[PEG_ID_C] = 0; // put all disks on the peg A for (int i = hanoi_size; i >= 1; -- i) { hanoi_disk_set(PEG_ID_A, i); }}void hanoi_calculate(int peg_from, int peg_use, int peg_to, int n){ if (n >= 1 && _hanoi_move_steps < _hanoi_max_steps) { hanoi_calculate(peg_from, peg_to, peg_use, n-1); if (_hanoi_move_steps >= _hanoi_max_steps) { return; } // save in one variable, dist id is not necessary _hanoi_move_steps_list[_hanoi_move_steps ++] = peg_from*10 + peg_to; //printf("move %d from %c to %c\n",n,peg_from+'A',peg_to+'A'); // test hanoi_calculate(peg_use, peg_from, peg_to, n-1); }}void peg_output(int peg_id){ printf("%c=>",peg_id + 'A'); int peg_size = _hanoi_peg_size[peg_id]; if (peg_size > 0) { printf(" "); for (int i = 0; i < peg_size; ++ i) { printf(" %d",_hanoi_peg[peg_id][i]); } } puts("");}void hanoi_output(){ peg_output(PEG_ID_A); peg_output(PEG_ID_B); peg_output(PEG_ID_C); puts("");}// tower_of_hanoi endint main(){ int n, m, problem_id = 0; while (~scanf("%d%d",&n,&m) && n+m) { hanoi_initital(n, m); hanoi_calculate(PEG_ID_A, PEG_ID_B, PEG_ID_C, n); printf("Problem #%d\n\n", ++ problem_id); int peg_from, peg_to; hanoi_output(); for (int i = 0; i < _hanoi_move_steps; ++ i) { peg_from = _hanoi_move_steps_list[i]%100/10; peg_to = _hanoi_move_steps_list[i]%10; hanoi_disk_set(peg_to, hanoi_disk_get(peg_from)); hanoi_output(); } } return 0; }
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