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?ˉt 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 meld is 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 ‘
‘.
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
Case 1: 1S 4S FA Case 2: Not ready
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），加入后用回溯法判断是否可以胡牌。
#include<stdio.h>
#include<string.h>
int mj[20], cnt[35];
const char* mahjong[]={"1T","2T","3T","4T","5T","6T","7T","8T","9T",
"1S","2S","3S","4S","5S","6S","7S","8S","9S",
"1W","2W","3W","4W","5W","6W","7W","8W","9W",
"DONG","NAN","XI","BEI",
"ZHONG","FA","BAI"};
int getNum(char *str) { //将牌面转换为编号
for (int i = 0; i < 34; i++) {
if (!strcmp(mahjong[i], str)) return i;
}
}
int check2(int n) {
for (int i = 0; i < 34; i++) { //刻子
if (cnt[i] >= 3) {
if (n == 3) return 1;
cnt[i] -= 3;
if (check2(n + 1)) return 1;
cnt[i] += 3;
}
}
for (int i = 0; i <= 24; i++) { //顺子
if (i % 9 <= 6 && cnt[i] && cnt[i + 1] && cnt[i + 2]) {
if (n == 3) return 1;
cnt[i]--;
cnt[i + 1]--;
cnt[i + 2]--;
if (check2(n + 1)) return 1;
cnt[i]++;
cnt[i + 1]++;
cnt[i + 2]++;
}
}
return 0;
}
int check() {
for (int i = 0; i < 34; i++) {
if (cnt[i] >= 2) {//将
cnt[i] -= 2;
if (check2(0)) return 1;
cnt[i] += 2;
}
}
return 0;
}
int main() {
char str[3];
int Case = 1;
while (scanf("%s", str) == 1) {
if (strcmp(str, "0") == 0) break;
printf("Case %d:", Case++);
mj[0] = getNum(str);
for (int i = 1; i < 13; i++) {
scanf("%s", str);
mj[i] = getNum(str);//将牌面转化为编号
}
int flag = 0;
for (int i = 0; i < 34; i++) {//枚举34种可能和的情况
memset(cnt, 0, sizeof(cnt));
for (int j = 0; j < 13; j++) {
cnt[mj[j]]++;//统计每张牌出现次数
}
if (cnt[i] >= 4) continue;//已出现四次则不再考虑这张牌
cnt[i]++;
if (check()) {
flag = 1;
printf(" %s", mahjong[i]);
}
}
if (!flag) printf(" Not ready\n");
else printf("\n");
}
return 0;
}
uva 11210 Chinese Mahjong(暴力枚举）
原文地址：http://blog.csdn.net/llx523113241/article/details/43163425
