Week9 作业B - 东东学打牌 [计蒜客

原题链接

https://nanti.jisuanke.com/t/41408

题目描述

基本思路

本题的关键问题在于玩家的排序，首先根据手牌情况进行排序，若手牌相同，则根据name进行排序。为了更加方便的进行排序，可以将手牌情况映射到一个整数score，score大的排在前面。

求score的方法：可以使用THREE(1000000)、TWO(10000)、ONE(100)、1表示不同的优先级。score等于不同优先级的指标与对应优先级相乘再相加得到。

例如：
"两对"手牌的score=3(类型)×THREE+较大的对子的值×TWO+较小的对子的值×ONE+剩下的一张牌的值×1"两对"手牌的score = 3(类型) \times THREE + 较大的对子的值 \times TWO + 较小的对子的值 \times ONE + 剩下的一张牌的值\times 1"两对"手牌的score=3(类型)×THREE+较大的对子的值×TWO+较小的对子的值×ONE+剩下的一张牌的值×1
"龙顺"手牌的score=8(类型)×THREE"龙顺"手牌的score = 8(类型) \times THREE"龙顺"手牌的score=8(类型)×THREE

小技巧

对五张手牌进行排序后有利于区分其类型。

完整代码

#include<iostream>
#include<algorithm>
#define THREE 1000000
#define TWO  10000
#define ONE  100
using namespace std;
int n;
int v[200];
struct Player{string name;int score;int ps[5];bool operator < (const Player &p) const{if(score != p.score)  return score > p.score;return name < p.name;}
}pyr[100005];
int c;
void setType(Player &p)
{if(p.ps[0]==1 && p.ps[1] == 10 && p.ps[2] == 11 && p.ps[3] == 12 && p.ps[4]==13){p.score = 8 * THREE;}else if(p.ps[0]+1==p.ps[1] && p.ps[1]+1==p.ps[2] && p.ps[2]+1==p.ps[3] &&p.ps[3]+1==p.ps[4]){p.score = 7 *THREE + p.ps[4] * TWO; }else if(p.ps[0] == p.ps[3]){p.score = 6 * THREE + p.ps[0] * TWO + p.ps[4] * ONE;}else if(p.ps[1] == p.ps[4]){p.score = 6 * THREE + p.ps[1] * TWO + p.ps[0] * ONE;}else if(p.ps[0] == p.ps[2] && p.ps[3] == p.ps[4]){p.score = 5 * THREE + p.ps[0] * TWO + p.ps[3] * ONE;}else if(p.ps[2] == p.ps[4] && p.ps[0] == p.ps[1]){p.score = 5 * THREE + p.ps[2] * TWO + p.ps[0] * ONE;}else if(p.ps[0] == p.ps[2]){p.score = 4 * THREE + p.ps[0] * TWO + (p.ps[3]+p.ps[4])*ONE;}else if(p.ps[1] == p.ps[3]){p.score = 4 * THREE + p.ps[1] * TWO +  (p.ps[0]+p.ps[4])*ONE;} else if(p.ps[2] == p.ps[4]){p.score = 4 * THREE + p.ps[2] * TWO + (p.ps[0]+p.ps[1])*ONE;}else if(p.ps[0] == p.ps[1] && p.ps[2] == p.ps[3]){p.score = 3 * THREE + p.ps[2] * TWO + p.ps[1] * ONE + p.ps[4];}else if(p.ps[0] == p.ps[1] && p.ps[3] == p.ps[4]){p.score = 3 * THREE + p.ps[3] * TWO + p.ps[1] * ONE + p.ps[2];}else if(p.ps[1] == p.ps[2] && p.ps[3] == p.ps[4]){p.score = 3 * THREE + p.ps[3] * TWO + p.ps[2] * ONE + p.ps[0];}else if(p.ps[0] == p.ps[1]){p.score = 2 * THREE + p.ps[0] * TWO + (p.ps[2]+p.ps[3]+p.ps[4]) * ONE;}else if(p.ps[1] == p.ps[2]){p.score = 2 * THREE + p.ps[1] * TWO + (p.ps[0]+p.ps[3]+p.ps[4]) * ONE;}else if(p.ps[2] == p.ps[3]){p.score = 2 * THREE + p.ps[2] * TWO + (p.ps[0]+p.ps[1]+p.ps[4]) * ONE;}else if(p.ps[3] == p.ps[4]){p.score = 2 * THREE + p.ps[3] * TWO + (p.ps[0]+p.ps[1]+p.ps[2]) * ONE;}else p.score = THREE + (p.ps[0]+p.ps[1]+p.ps[2]+p.ps[3]+p.ps[4])*TWO;
}void readpyr()
{cin>>pyr[c].name;string poks;cin>>poks;int m=0;int len = poks.length();for(int i=0;i<len;i++){if(poks[i] == '1') i++, pyr[c].ps[m++]=10;else pyr[c].ps[m++]=v[poks[i]];}sort(pyr[c].ps, pyr[c].ps+m);setType(pyr[c]);c++;
}
int main ()
{ios::sync_with_stdio(false);for(int i='2';i<='9';i++)   v[i]=i-'0';v['A']=1;v['J']=11;v['Q']=12;v['K']=13;while(cin>>n){c=0;for(int i=0;i<n;i++)    readpyr();sort(pyr, pyr+c);for(int i=0;i<c;i++)  cout<<pyr[i].name<<endl; }return 0;
}