UVA 10651 Pebble Solitaire(bfs + 哈希判重(记忆化搜索?))
Problem A
Pebble Solitaire
Input: standard input
Output: standard output
Time Limit: 1 second
Pebble solitaire is an interesting game. This is a game where you are given a board with an arrangement of small cavities, initially all but one occupied by a pebble each. The aim of the game is to remove as many pebbles as possible from the board. Pebbles disappear from the board as a result of a move. A move is possible if there is a straight line of three adjacent cavities, let us call them A, B, and C, with B in the middle, where A is vacant, but B and C each contain a pebble. The move constitutes of moving the pebble from C to A, and removing the pebble in Bfrom the board. You may continue to make moves until no more moves are possible.
In this problem, we look at a simple variant of this game, namely a board with twelve cavities located along a line. In the beginning of each game, some of the cavities are occupied by pebbles. Your mission is to find a sequence of moves such that as few pebbles as possible are left on the board.
Input
The input begins with a positive integer n on a line of its own. Thereafter n different games follow. Each game consists of one line of input with exactly twelve characters, describing the twelve cavities of the board in order. Each character is either '-' or 'o' (The fifteenth character of English alphabet in lowercase). A '-' (minus) character denotes an empty cavity, whereas a 'o' character denotes a cavity with a pebble in it. As you will find in the sample that there may be inputs where no moves is possible.
Output
For each of the n games in the input, output the minimum number of pebbles left on the board possible to obtain as a result of moves, on a row of its own.
Sample Input Output for Sample Input
5
---oo-------
-o--o-oo----
-o----ooo---
oooooooooooo
oooooooooo-o
1
2
3
12
1
题意：就是跳棋。比如-oo 可以第三个棋子跳到-上 消除掉中间那个o。
思路：bfs + 哈希判重。记录下每次状态。之后不考虑重复状态。
代码：
#include <stdio.h>#include <string.h>#include <limits.h>int t, vis[5555], ans;char str[15];struct Q {char str[15];int num;} q[5555];int hash(char *str) {int num = 0, i;for (i = 0; i < 12; i ++) {if (str[i] == 'o') {num += 1 << i;}}return num;}void bfs() {int i, head = 0, rear = 1;ans = INT_MAX;memset(q, 0, sizeof(q));memset(vis, 0, sizeof(vis));strcpy(q[0].str, str);vis[hash(q[0].str)] = 1;for (i = 0; i < 12; i ++)if (q[0].str[i] == 'o')q[0].num ++;while (head < rear) {if (q[head].num < ans)ans = q[head].num;for (i = 0; i < 10; i ++) {if (q[head].str[i] == '-' && q[head].str[i + 1] == 'o' && q[head].str[i + 2] == 'o') {char sb[15];strcpy(sb, q[head].str);sb[i] = 'o';sb[i + 1] = '-';sb[i + 2] = '-';if (!vis[hash(sb)]) {vis[hash(sb)] = 1;strcpy(q[rear].str, sb);q[rear].num = q[head].num - 1;rear ++;}}}for (i = 0; i < 10; i ++) {if (q[head].str[i] == 'o' && q[head].str[i + 1] == 'o' && q[head].str[i + 2] == '-') {char sb[15];strcpy(sb, q[head].str);sb[i] = '-';sb[i + 1] = '-';sb[i + 2] = 'o';if (!vis[hash(sb)]) {vis[hash(sb)] = 1;strcpy(q[rear].str, sb);q[rear].num = q[head].num - 1;rear ++;}}}head ++;}}int main () {scanf("%d%*c", &t);while (t --) {gets(str);bfs();printf("%d\n", ans);}return 0;}