hdu4431 枚举+模拟

Mahjong

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3889 Accepted Submission(s): 778

Problem Description

Japanese Mahjong is a four-player game. The game needs four people to sit around a desk and play with a set of Mahjong tiles. A set of Mahjong tiles contains four copies of the tiles described next:

One to nine Man, which we use 1m to 9m to represent;

One to nine Sou, which we use 1s to 9s to represent;

One to nine Pin, which we use 1p to 9p to represent;

Character tiles, which are:Ton, Nan, Sei, Pei, Haku, Hatsu, Chun, which we use 1c to 7c to represent.

A winning state means a set of 14 tiles that normally contains a pair of same tiles (which we call "eyes") and four melds. A meld is formed by either three same tiles(1m, 1m, 1m or 2c, 2c, 2c for example) or three continuous non-character tiles(1m, 2m, 3m or 5s, 6s, 7s for example).

However, there are two special winning states that are different with the description above, which are:
"Chii Toitsu", which means 7 different pairs of tiles;
"Kokushi Muso", which means a set of tiles that contains all these tiles: 1m, 9m, 1p, 9p, 1s, 9s and all 7 character tiles. And the rest tile should also be one of the 13 tiles above.

And the game starts with four players receiving 13 tiles. In each round every player must draw one tile from the deck one by one. If he reaches a winning state with these 14 tiles, he can say "Tsu Mo" and win the game. Otherwise he should discard one of his 14 tiles. And if the tile he throws out can form a winning state with the 13 tiles of any other player, the player can say "Ron" and win the game.

Now the question is, given the 13 tiles you have, does there exist any tiles that can form a winning state with your tiles?

(Notes: Some of the pictures and descriptions above come from Wikipedia.)

Input

The input data begins with a integer T(1≤T≤20000). Next are T cases, each of which contains 13 tiles. The description of every tile is as above.

Output

For each cases, if there actually exists some tiles that can form a winning state with the 13 tiles given, print the number first and then print all those tiles in order as the description order of tiles above. Otherwise print a line "Nooten"(without quotation marks).

Sample Input

21s 2s 3s 2c 2c 2c 2p 3p 5m 6m 7m 1p 1p1p 1p 2p 3p 4s 5s 6s 7c 7c 3s 3s 2m 2m

Sample Output

2 1p 4pNooten

Source

2012 Asia Tianjin Regional Contest 

#include<stdio.h>#include<string.h>#define N 40int mark[N],b[N],vis[N];int judge1(int x){    int num=0,i;    for(i=1;i<=34;i++)    {//必须是7对不相同的牌        if(mark[i]==2)            num++;    }    if(num==7)      return 1;    else       return 0;}int judge2(int x){//处理13妖的胡法    int j,i;    if(mark[1]==0||mark[9]==0||mark[10]==0||mark[18]==0||mark[19]==0||mark[27]==0)        return 0;//必须出现的数字    for(j=28;j<=34;j++)    {//必须出现的文字        if(mark[j]==0)            return 0;    }    for(i=1;i<=27;i++)    {//不能出现的数字        if(i!=1&&i!=9&&i!=10&&i!=18&&i!=19&&i!=27)        {            if(mark[i]!=0)                return 0;        }    }    return 1;}int f(int k){//处理是否越出当前牌的范围    if(k==9||k==18||k==27)        return 0;    if(k+1==9||k+1==18||k+1==27)        return 0;    return 1;}int judge3(int x){    int i,j,k,num,flag=0;    for(i=1;i<=34;i++)    {        if(mark[i]>=2)        {//取任意两张相同的牌，当对子。            num=0;            for(j=1;j<=34;j++)                vis[j]=mark[j];          vis[i]-=2;          for(j=1;j<=27;j++)          {//判断是数字的有没有三张相同的和顺子的              if(vis[j]>=3)              {//大于等于三张的先取出3张                  num++;                  vis[j]-=3;              }              while(f(j)&&vis[j]&&vis[j+1]&&vis[j+2])              {                  num++;                  vis[j]--; vis[j+1]--; vis[j+2]--;              }          }//判断文字的          for(j=28;j<=34;j++)          {              if(vis[j]>=3)                 num++;          }          if(num==4)          {              flag=1;              break;          }        }    }    if(flag==1)        return 1;    else        return 0;}int main(){    int t,i,a,k;    char str[10];    scanf("%d",&t);    while(t--)    {        memset(mark,0,sizeof(mark));        for(i=1;i<=13;i++)        {//把牌转换成数字的形式            scanf("%s",str);            a=str[0]-'0';            if(str[1]=='s')               a+=9;            else if(str[1]=='p')                a+=18;            else  if(str[1]=='c')                a+=27;            mark[a]++;        }       k=0;       for(i=1;i<=34;i++)       {           mark[i]++;//注意判断mark[i]<=4          if(mark[i]<=4&&(judge1(i)||judge2(i)||judge3(i)))            b[k++]=i;          mark[i]--;       }       if(k==0)       {           printf("Nooten\n");           continue;       }       printf("%d",k);       for(i=0;i<k;i++)       {           //printf("%d\n",b[i]);           if(b[i]%9==0)             printf(" 9");           else            printf(" %d",b[i]%9);           if(b[i]<=9)            printf("m");           else if(b[i]<=18)            printf("s");           else if(b[i]<=27)             printf("p");           else            printf("c");       }       printf("\n");    }    return 0;}/*1111s 1s 2s 2s 4s 4s 5s 5s 6s 6s 7s 7s 8s1s 9s 1m 9m 1p 9p 1c 2c 3c 4c 5c 6c 7c1s 9s 1m 9m 1p 9p 1c 2c 3c 4c 5c 6c 1m2s 3s 4s 5s 6s 7s 7s 1c 1c 1c 2c 2c 2c2s 3s 4s 5s 6s 7s 7s 7s 1c 1c 2c 2c 2c1s 1s 2s 2s 3s 4s 4s 5s 5s 6s 6s 1c 1c1s 1s 1s 2s 3s 4s 5s 6s 7s 8s 9s 9s 9s1s 1s 1s 1s 2s 3s 9s 9s 9s 2s 3s 1c 1c1s 1s 1s 1s 2s 2s 2s 3s 3s 3s 4s 4s 4s2s 2s 2s 3s 4s 4s 5s 5s 5s 6s 6s 7s 9s2s 2s 2s 3s 4s 4s 5s 5s 5s 6s 6s 7s 8s1p 1p 1p 1p 2p 2p 3p 3p 4p 4p 5p 5p 8s2p 2p 2p 3p 3p 4p 4p 4p 5p 5p 5p 6p 6p1c 2c 3c 4c 5c 6c 7c 1s 9s 1m 9m 1p 9p*/