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Published on 2017-09-02 11:31 in 暂未分类 with qscqesze

Codeforces Round #573 (Div. 2) Tokitsukaze and Mahjong 水题

B. Tokitsukaze and Mahjong

time limit per test1 second
memory limit per test256 megabytes

Tokitsukaze is playing a game derivated from Japanese mahjong. In this game, she has three tiles in her hand. Each tile she owns is a suited tile, which means it has a suit (manzu, pinzu or souzu) and a number (a digit ranged from 1 to 9). In this problem, we use one digit and one lowercase letter, which is the first character of the suit, to represent a suited tile. All possible suited tiles are represented as 1m, 2m, …, 9m, 1p, 2p, …, 9p, 1s, 2s, …, 9s.

In order to win the game, she must have at least one mentsu (described below) in her hand, so sometimes she should draw extra suited tiles. After drawing a tile, the number of her tiles increases by one. She can draw any tiles she wants, including those already in her hand.

Do you know the minimum number of extra suited tiles she needs to draw so that she can win?

Here are some useful definitions in this game:

A mentsu, also known as meld, is formed by a koutsu or a shuntsu;
A koutsu, also known as triplet, is made of three identical tiles, such as [1m, 1m, 1m], however, [1m, 1p, 1s] or [1m, 4m, 7m] is NOT a koutsu;
A shuntsu, also known as sequence, is made of three sequential numbered tiles in the same suit, such as [1m, 2m, 3m] and [5s, 7s, 6s], however, [9m, 1m, 2m] or [1m, 2p, 3s] is NOT a shuntsu.
Some examples:

[2m, 3p, 2s, 4m, 1s, 2s, 4s] — it contains no koutsu or shuntsu, so it includes no mentsu;
[4s, 3m, 3p, 4s, 5p, 4s, 5p] — it contains a koutsu, [4s, 4s, 4s], but no shuntsu, so it includes a mentsu;
[5p, 5s, 9m, 4p, 1s, 7p, 7m, 6p] — it contains no koutsu but a shuntsu, [5p, 4p, 6p] or [5p, 7p, 6p], so it includes a mentsu.
Note that the order of tiles is unnecessary and you can assume the number of each type of suited tiles she can draw is infinite.

Input

The only line contains three strings — the tiles in Tokitsukaze's hand. For each string, the first character is a digit ranged from 1 to 9 and the second character is m, p or s.

Output

Print a single integer — the minimum number of extra suited tiles she needs to draw.

Examples

input
1s 2s 3s
output
0
input
9m 9m 9m
outputCopy
0
inputCopy
3p 9m 2p
outputCopy
1

Note

In the first example, Tokitsukaze already has a shuntsu.

In the second example, Tokitsukaze already has a koutsu.

In the third example, Tokitsukaze can get a shuntsu by drawing one suited tile — 1p or 4p. The resulting tiles will be [3p, 9m, 2p, 1p] or [3p, 9m, 2p, 4p].

题意

日麻,给你三张麻将牌,问你怎么最少增加多少张牌可以凑成出一碰,或者一shuntsu

题解

答案就3种,0,1,2;0张直接check,1就加一张check,2就剩下的。

代码

#include<bits/stdc++.h>
using namespace std;

pair<int,int> form(string s){
    pair<int,int> p;
    p.first=int(s[0]-'0');
    if(s[1]=='m'){
        p.second=0;
    }else if(s[1]=='p'){
        p.second=1;
    }else{
        p.second=2;
    }
    return p;
}
bool check(vector<pair<int,int> >v){
    sort(v.begin(),v.end());
    for(int i=0;i<v.size();i++){
        for(int j=i+1;j<v.size();j++){
            for(int k=j+1;k<v.size();k++){
                if(v[i].second==v[j].second&&v[j].second==v[k].second){
                    if(v[i].first==v[j].first&&v[j].first==v[k].first){
                        return true;
                    }
                    if(v[i].first+1==v[j].first&&v[j].first+1==v[k].first){
                        return true;
                    }
                }
            }
        }
    }
    return false;
}
int main(){
    vector<pair<int,int> >v;
    for(int i=0;i<3;i++){
        string s;
        cin>>s;
        v.push_back(form(s));
    }
    if(check(v)){
        cout<<"0"<<endl;
        return 0;
    }
    for(int i=1;i<10;i++){
        for(int j=0;j<3;j++){
            vector<pair<int,int> >d = v;
            d.push_back(make_pair(i,j));
            if(check(d)){
                cout<<"1"<<endl;
                return 0;
            }
        }
    }
    cout<<"2"<<endl;
    return 0;
}
posted @ 2019-07-29 01:29  qscqesze  阅读(207)  评论(0编辑  收藏