uva 11210 Chinese Mahjong(暴力搜索）

Chinese Mahjong

Mahjong () is a game of Chinese origin usually played by four persons with tiles resembling dominoes and bearing various designs, which are drawn and discarded until one player wins with a hand of four combinations of three tiles each and a pair of matching tiles.

A set of Mahjong tiles will usually differ from place to place. It usually has at least 136 tiles, most commonly 144, although sets originating from America or Japan will have more. The 136-tile Mahjong includes:

Dots: named as each tile consists of a number of circles. Each circle is said to represent copper (tong) coins with a square hole in the middle. In this problem, they're represented by 1T, 2T, 3T, 4T, 5T, 6T, 7T, 8T and 9T.

Bams: named as each tile (except the 1 Bamboo) consists of a number of bamboo sticks. Each stick is said to represent a string (suo) that holds a hundred coins. In this problem, they're represented by 1S, 2S, 3S, 4S, 5S, 6S, 7S, 8S and 9S.

Craks: named as each tile represents ten thousand (wan) coins, or one hundred strings of one hundred coins. In this problem, they're represented by 1W, 2W, 3W, 4W, 5W, 6W, 7W, 8W and 9W.

Wind tiles: East, South, West, and North. In this problem, they're represented by DONG, NAN, XI, BEI.

Dragon tiles: red, green, and white. The term dragon tile is a western convention introduced by Joseph Park Babcock in his 1920 book introducing Mahjong to America. Originally, these tiles are said to have something to do with the Chinese Imperial Examination. The red tile means you pass the examination and thus will be appointed a government official. The green tile means, consequently you will become financially well off. The white tile (a clean board) means since you are now doing well you should act like a good, incorrupt official. In this problem, they're represented by ZHONG, FA, BAI.

There are 9*3+4+3=34 kinds, with exactly 4 tiles of each kind, so there are 136 tiles in total.

To who may be interested, the 144-tile Mahjong also includes:

Flower tiles:

typically optional components to a set of mahjong tiles, often contain artwork on their tiles. There are exactly one tile of each kind, so 136+8=144 tiles in total. In this problem, we don��t consider these tiles.

Chinese Mahjong is very complicated. However, we only need to know very few of the rules in order to solve this problem. A meldis a certain set of tiles in one's hand. There are three kinds of melds you need to know (to who knows Mahjong already, kong is not considered):

Pong: A set of three identical titles. Example: ;.

Chow: A set of three suited tiles in sequence. All three tiles must be of the same suites. Sequences of higher length are not permissible (unless it forms more than one meld). Obviously, wind tiles and dragon tiles can never be involved in chows. Example:;.

Eye: The pair, while not a meld, is the final component to the standard hand. It consists of any two identical tiles.

A player wins the round by creating a standard mahjong hand. That means, the hand consists of an eye and several (possible zero) pongs and chows. Note that each title can be involved in exactly one eye/pong/chow.

When a hand is one tile short of wining, the hand is said to be a ready hand, or more figuratively, 'on the pot'. The player holding a ready hand is said to be waiting for certain tiles. For example

is waiting for , and .

To who knows more about Mahjong: don't consider special winning hands such as ''.

Input

The input consists of at most 50 test cases. Each case consists of 13 tiles in a single line. The hand is legal (e.g. no invalid tiles, exactly 13 tiles). The last case is followed by a single zero, which should not be processed.

Output

For each test case, print the case number and a list of waiting tiles sorted in the order appeared in the problem description (1T~9T, 1S~9S, 1W~9W, DONG, NAN, XI, BEI, ZHONG, FA, BAI). Each waiting tile should be appeared exactly once. If the hand is not ready, print a message 'Not ready' without quotes.

Sample Input

1S 1S 2S 2S 2S 3S 3S 3S 7S 8S 9S FA FA
1S 2S 3S 4S 5S 6S 7S 8S 9S 1T 3T 5T 7T
0

Output for the Sample Input

Case 1: 1S 4S FA
Case 2: Not ready

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题目大意：给出13张麻将牌，问在取一张牌就可以胡牌的牌，所处所有满足的情况。这里的胡牌不需要考虑太多，只需要满足存在一个对子， 而其他的全是3个顺或者3个相同的就可以了，一些特殊的胡牌不需要考虑。

解题思路：将所有给出的麻将牌转化成数字进行处理，对应的用一个数组统计牌的个数cnt[i]表示标号为i的麻将牌有cnt[i]张。因为存在特殊的牌面，所以可以将特殊牌面的标号分开，这样在dfs的过程中就无需考虑连续的标号是否是顺子的问题。接下来就是枚举所有可能拿到的牌（出现过4次的不能再取），加入后用dfs判断是否可以胡牌，因为只有三种情况，dfs的时候直接写出来就可以了，注意对子只出现一次。

#include <stdio.h>
#include <string.h>
const int N = 105;

int cnt[N], answer[N];
const char sign[] = "TSW";
const char sen[][10] = {"BAI", "DONG", "NAN", " ",  "XI", "BEI", " ",  "ZHONG", "FA"};

int changeOne(char c) {
    if (c == 'T')   return 0;
    else if (c == 'S')	return 1;
    else if (c == 'W')	return 2;
}

int changeTwo(char a, char b) {
    if (a == 'D' && b == 'O')		return 31;
    else if (a == 'N' && b == 'A')	return 32;
    else if (a == 'X' && b == 'I')	return 34;
    else if (a == 'B' && b == 'E')	return 35;
    else if (a == 'Z' && b == 'H')	return 37;
    else if (a == 'F' && b == 'A')	return 38;
    else if (a == 'B' && b == 'A')	return 40;
    else    return -1;
}

void handle(char str[]) {
    memset(cnt, 0, sizeof(cnt));
    cnt[10] = cnt[20] = cnt[30] = cnt[33] = cnt[36] = cnt[39] = 5;
    int n = 0, len = strlen(str), cur;
    for (int i = 0; i < len - 1; i++) {
	if (str[i] > '0' && str[i] <= '9') {
	    cur = 0;
	    cur += str[i++] - '0';
	    cur += changeOne(str[i++]) * 10;
	    cnt[cur]++;
	}
	else {
	    cur = changeTwo(str[i], str[i + 1]);
	    if (cur > 0)
		cnt[cur]++;
	}
    }
}

bool dfs(int sum, int flag) {
    if (sum >= 14)  return true;

    for (int i = 1; i <= 40; i++) {
	if (cnt[i] >= 3 && cnt[i] < 5) {
	    cnt[i] -= 3;
	    if (dfs(sum + 3, flag)) {
		cnt[i] += 3;
		return true;
	    }
	    cnt[i] += 3;
	}
	if (!flag && cnt[i] >= 2 && cnt[i] < 5) {
	    cnt[i] -= 2;
	    if (dfs(sum + 2, 1)) {
		cnt[i] += 2;
		return true;
	    }
	    cnt[i] += 2;
	}

	if (cnt[i] && cnt[i + 1] && cnt[i + 2] && cnt[i] < 5 && cnt[i + 1] < 5 && cnt[i + 2] < 5) {
	    cnt[i]--, cnt[i + 1]--, cnt[i + 2]--;
	    if (dfs(sum + 3, flag)) {
		cnt[i]++, cnt[i + 1]++, cnt[i + 2]++;
		return true;
	    }
	    cnt[i]++, cnt[i + 1]++, cnt[i + 2]++;
	}
    }
    return false;
}

void put(int cur) {
    int a = cur / 10, b = cur % 10;
    if (a < 3)
	printf(" %d%c", b, sign[a]);
    else
	printf(" %s", sen[b]);
}

int main() {
    char str[N];
    int cas = 1;
    while (gets(str)) {
	if (strcmp(str, "0") == 0)
	    break;
	memset(answer, 0, sizeof(answer));
	int t = 0;

	handle(str);

	for (int i = 1; i <= 40; i++) {
	    if (cnt[i] >= 4)	continue;
	    cnt[i]++;
	    if (dfs(0, 0))
		answer[t++] = i;
	    cnt[i]--;
	}
	printf("Case %d:", cas++);
	if (t)
	    for (int i = 0; i < t; i++)
		put(answer[i]);
	else
	    printf(" Not ready");
	printf("\n");
    }
    return 0;
}