ZOJ 3671 Japanese Mahjong III

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3671

Japanese Mahjong III

Japanese mahjong is usually played with 136 tiles, which can be organized into several categories:

• Suited tiles. All suited tiles are of a rank and a suit.There are three suits of tiles, with ranks ranging from one to nine. There are four tiles of each rank and suit combination, thus there are 36 tiles in a suit, and 108 suited tiles in total.
  • The circle suit

    

  • The bamboo suit

    

  • The character suit

    

• Honor tiles. Honor Tiles are tiles that do not have a rank or suit. They are divided into two categories. There are four types of Wind tiles and three types of Dragon tiles, with four of each type of honor tile. Thus, there are 16 wind tiles and
  12 Dragon tiles for 28 honor tiles.
  • Wind tiles. The Wind tiles consist of four kinds of tile: East, South, West, and North.

    

  • Dragon tiles. The Dragon titles consist of three types of tile: Red, Green, White.

    

In this problem, we introduce two abnormal kinds of winning hand different from the normal winning hand, which consist of four melds and one eyes.

1. Seven Pairs: the hand consists of seven different pairs. No two pairs are the same. For example:
2. Thirteen Orphans: the hand consists of thirteen kinds of tiles: Circle One, Circle Nine, Bamboo One, Bamboo Nine, Character One, Character Nine, East, South, West, North, Red, Green and White. Only one kind of them are two tiles,
   while the others are all only one tile. For example:

Now, given a hand with 14 tiles, can you answer this hand is which kind of abnormal kind of winning hand?

Input

There are multiple cases. Each case consists of 28 characters in one line, describing 14 tiles. The manner for describing each type of tile is:

• Two characters stand for one tile.
• For Suited tiles, the first is a integer ranged from one to nine, the tile‘s rank, and the second is a character: ‘p‘ for the circle suit, ‘s‘ for the bamboo suit, and ‘m‘ for the charactersuit.
• For Honor tiles, the first is a integer: from one to seven stands for East, South, West, North, White, Green, and Red respectively. The second one is always ‘z‘.

We promise that the input is a legal hand, which means there isn‘t another type of tiles described above or there are more than four same tiles.

Output

For each case, if the hand is Seven Pairs, output "Seven Pairs". If the hand is Thirteen Orphans, output "Thirteen Orphans". If the hand is neither Seven Pairs nor Thirteen
Orphans, output "Neither!".

Sample Input

1z1z2z2z3z3z4z4z5z5z6z6z7z7z
1s9s1m9m1p9p1z2z3z4z5z6z7z7z
1s9s1m9m1p9p1z2z3z4z5z6z7z7p

Sample Output

Seven Pairs
Thirteen Orphans
Neither!

Contest: ZOJ Monthly, November 2012

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一道简单的模拟题。我们可以用一个二维数组 a[i][j] 来记录状态。a[0][j] 表示 jp  a[1][j]表示 js, a[2][j] 表示 jm ，a[3][j] 表示 jz 。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<string>
#include<sstream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<set>
#include<cstdlib>
#include<map>
using namespace std;
#define CLR(A) memset(A,0,sizeof(A))
const int MAX=50;
int a[10][10];
char str[MAX];
int main(){
    while(~scanf("%s",str)){
        CLR(a);int len=strlen(str);
        for(int i=0;i<len-1;i++){
            if(str[i+1]=='p') a[0][str[i]-'0']++;
            if(str[i+1]=='s') a[1][str[i]-'0']++;
            if(str[i+1]=='m') a[2][str[i]-'0']++;
            if(str[i+1]=='z') a[3][str[i]-'0']++;
        }
        bool flag=0;
        for(int i=0;i<4;i++){
            for(int j=1;j<10;j++){
                if(a[i][j]&&a[i][j]!=2){
                    flag=1;break;
                }
            }
            if(flag) break;
        }
        if(!flag) cout<<"Seven Pairs"<<endl;
        else{
            if(a[0][1]&&a[0][9]&&a[1][1]&&a[1][9]&&a[2][1]&&a[2][9]&&a[3][1]&&a[3][2]&&a[3][3]&&a[3][4]&&a[3][5]&&a[3][6]&&a[3][7]){
                if(a[0][1]==2||a[0][9]==2||a[1][1]==2||a[1][9]==2||a[2][1]==2||a[2][9]==2||a[3][1]==2||a[3][2]==2||a[3][3]==2||a[3][4]==2||a[3][5]==2||a[3][6]==2||a[3][7]==2){
                    cout<<"Thirteen Orphans"<<endl;
                }
                else cout<<"Neither!"<<endl;
            }
            else cout<<"Neither!"<<endl;
        }
    }
    return 0;
}