HDU 4431 Mahjong 第37届ACM/ICPC 天津赛区现场赛A题 (枚举,判断麻将胡牌)

Mahjong

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 648    Accepted Submission(s): 111


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

 

Sample Output
2 1p 4p Nooten
 

 

Source
 

 

Recommend
zhoujiaqi2010
 
 
很有意思的题目。
就是给了13张牌。问增加哪些牌可以胡牌。
 
胡牌有以下几种情况:
1、一个对子 +  4组 3个相同的牌或者顺子。  只有m、s、p是可以构成顺子的。东西南北这样的牌没有顺子。
2、7个不同的对子。
3、1m,9m,1p,9p,1s,9s,1c,2c,3c,4c,5c,6c,7c.  这13种牌每种都有,而且仅有这13种牌。肯定是有一种2张。其他的1张。
 
 
首先是枚举18+7=34张牌,加进去构成14张牌,判断胡牌。
胡牌判断如下。
对于第一种情况:枚举每一个对子。然后按照顺序找3张相同或者顺子。如果有三种相同的,构成3张相同的。没有就看能不能和后面的构成顺子。一定要按照顺序从小到大找过去。 1c```7c只能构成3张一样的。然后判断是不是刚好找到4组。
对于第二种情况:就是要每一种牌的数量要么是0,要么是2,这样一定是7个不同的对子了。
对于第三种情况:就是要让这13种牌的数量不等于0,而且其他牌的数量为0;
 
 
这种做法是Fox 提示我之后才想到的,果然不错。
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;

int cnt[35];

bool judge4X3()
{
    int ret=0;
    int tmp[35];
    for(int i=0;i<34;i++)tmp[i]=cnt[i];

    for(int i=0;i<=18;i+=9)
      for(int j=0;j<9;j++)
      {
          if(tmp[i+j]>=3)
          {
              tmp[i+j]-=3;
              ret++;
          }
          while(j+2<9 && tmp[i+j] && tmp[i+j+1] &&tmp[i+j+2])
          {
              tmp[i+j]--;
              tmp[i+j+1]--;
              tmp[i+j+2]--;
              ret++;
          }
      }
    for(int j=0;j<7;j++)
    {
        if(tmp[27+j]>=3)
        {
            tmp[27+j]-=3;
            ret++;
        }
    }
    if(ret==4)return true;
    return false;
}

bool judge1()
{
    for(int i=0;i<34;i++)
    {
        if(cnt[i]>=2)
        {
            cnt[i]-=2;//枚举对子
            if(judge4X3())
            {
                cnt[i]+=2;
                return true;
            }
            cnt[i]+=2;
        }
    }
    return false;
}


bool judge2()
{
    for(int i=0;i<34;i++)
    {
        if(cnt[i]!=2 && cnt[i]!=0)
          return false;
    }
    return true;
}

bool judge3()
{
    for(int j=0;j<7;j++)
      if(cnt[j+27]==0)
        return false;
    for(int i=0;i<=18;i+=9)
    {
        if(cnt[i]==0 || cnt[i+8]==0)return false;
        for(int j=1;j<8;j++)
          if(cnt[i+j]!=0)
            return false;
    }
    return true;
}

bool judge()
{
    if(judge1() || judge2() || judge3())return true;
    return false;
}


int main()
{
    int T;
    char str[10];
    scanf("%d",&T);
    int ans[35],tol;
    while(T--)
    {
        memset(cnt,0,sizeof(cnt));
        for(int i=0;i<13;i++)
        {
            scanf("%s",&str);
            int t=str[0]-'1';
            if(str[1]=='m')t+=0;
            else if(str[1]=='s')t+=9;
            else if(str[1]=='p')t+=18;
            else t+=27;
            cnt[t]++;
        }
        tol=0;
        for(int i=0;i<34;i++)
        {
            cnt[i]++;
            if(cnt[i]<=4 && judge())
               ans[tol++]=i;
            cnt[i]--;
        }
        if(tol==0)printf("Nooten\n");
        else
        {
            printf("%d",tol);
            for(int i=0;i<tol;i++)
            {
                printf(" %d",(ans[i]%9)+1);
                if(ans[i]/9==0)printf("m");
                else if(ans[i]/9==1)printf("s");
                else if(ans[i]/9==2)printf("p");
                else printf("c");
            }
            printf("\n");
        }
    }
    return 0;
}

 

posted on 2012-10-27 22:36 kuangbin 阅读(...) 评论(...) 编辑 收藏

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