| Time Limit: 3000MS | 64bit IO Format: %lld & %llu |
Description
In 5-card draw poker, a player is dealt a hand of five cards (which may be looked at). The player may then discard between zero and five of his or her cards and have them replaced by the same number of cards from the top of the deck (which is face down). The object is to maximize the value of the final hand. The different values of hands in poker are given at the end of this problem.
Normally the player cannot see the cards in the deck and so must use probability to decide which cards to discard. In this problem, we imagine that the poker player is psychic and knows which cards are on top of the deck. Write a program which advises the player which cards to discard so as to maximize the value of the resulting hand.
Input and Output
Input will consist of a series of lines, each containing the initial five cards in the hand then the first five cards on top of the deck. Each card is represented as a two-character code. The first character is the face-value (A=Ace, 2-9, T=10, J=Jack, Q=Queen, K=King) and the second character is the suit (C=Clubs, D=Diamonds, H=Hearts, S=Spades). Cards will be separated by single spaces. Each input line will be from a single valid deck, that is there will be no duplicate cards in each hand and deck.
Each line of input should produce one line of output, consisting of the initial hand, the top five cards on the deck, and the best value of hand that is possible. Input is terminated by end of file.
Use the sample input and output as a guide. Note that the order of the cards in the player's hand is irrelevant, but the order of the cards in the deck is important because the discarded cards must be replaced from the top of the deck. Also note that examples of all types of hands appear in the sample output, with the hands shown in decreasing order of value.
Sample Input
TH JH QC QD QS QH KH AH 2S 6S 2H 2S 3H 3S 3C 2D 3D 6C 9C TH 2H 2S 3H 3S 3C 2D 9C 3D 6C TH 2H AD 5H AC 7H AH 6H 9H 4H 3C AC 2D 9C 3S KD 5S 4D KS AS 4C KS AH 2H 3C 4H KC 2C TC 2D AS AH 2C 9S AD 3C QH KS JS JD KD 6C 9C 8C 2D 7C 2H TC 4C 9S AH 3D 5S 2H QD TD 6S KH 9H AD QH
Sample Output
Hand: TH JH QC QD QS Deck: QH KH AH 2S 6S Best hand: straight-flush Hand: 2H 2S 3H 3S 3C Deck: 2D 3D 6C 9C TH Best hand: four-of-a-kind Hand: 2H 2S 3H 3S 3C Deck: 2D 9C 3D 6C TH Best hand: full-house Hand: 2H AD 5H AC 7H Deck: AH 6H 9H 4H 3C Best hand: flush Hand: AC 2D 9C 3S KD Deck: 5S 4D KS AS 4C Best hand: straight Hand: KS AH 2H 3C 4H Deck: KC 2C TC 2D AS Best hand: three-of-a-kind Hand: AH 2C 9S AD 3C Deck: QH KS JS JD KD Best hand: two-pairs Hand: 6C 9C 8C 2D 7C Deck: 2H TC 4C 9S AH Best hand: one-pair Hand: 3D 5S 2H QD TD Deck: 6S KH 9H AD QH Best hand: highest-card
| 6736778 | Accepted | 0 | 0 | 2923 |
题目大意如下：这是一几局德州扑克游戏。给定5张初始手牌和5张牌的牌堆（可预知牌堆中的牌类、数字、花色以及顺序，即玩家具有超能力），可以拿手牌换牌堆最上方的牌，之后手牌丢弃，换到的牌不允许再换。
思路如下：先确定枚举和搜索的对象，本题的对象应该是要弃的手牌，由于考虑到换牌的数量和制定牌不定，我们要用到子集枚举的算法。而这里我们用增量法。
用string类配合cin（string是类，不能用scanf或printf，所以果断用cin和cout）之后，预处理牌的点数，将2~9和A、T、J、Q、K统一转化为数字。之后，用增量法枚举要换的牌，之后使用judge函数对它的花色和点数进行判断，这里要外加一个sort优化方便直接用if语句判断，记录最高分手牌则用一个常数数组记录同花顺~大牌的降序字符串数组，用数组的下标值来比较，记录最小值ans，最后按题目要求输出。
代码如下：（2KB）
#include<string>
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
string cd[11];
string bthd[]={"straight-flush","four-of-a-kind","full-house","flush","straight","three-of-a-kind","two-pairs","one-pair","highest-card"};
int val[11],a[11],hd[6],ans;
int jud(int cur)
{
bool samecol=1;
for(int i=0;i<cur;i++){
hd[i]=val[a[i]];
if(cd[a[0]][1]!=cd[a[i]][1])samecol=0;
}
for(int i=cur;i<5;i++){
hd[i]=val[5+i-cur];
if(cd[a[0]][1]!=cd[5+i-cur][1])samecol=0;
}
sort(hd,hd+5);
bool flush=1;
for(int i=0;i<4;i++)
if((i||hd[0]!=1||hd[4]!=13)&&hd[i]+1!=hd[i+1])
{flush=0;break;}
if(samecol&&flush)return 0;
if((hd[0]==hd[1]||hd[3]==hd[4])&&hd[1]==hd[2]&&hd[2]==hd[3])return 1;
if((hd[0]==hd[1]&&hd[1]==hd[2]&&hd[3]==hd[4])||(hd[0]==hd[1]&&hd[2]==hd[3]&&hd[3]==hd[4]))return 2;
if(samecol)return 3;
if(flush)return 4;
if((hd[0]==hd[1]&&hd[1]==hd[2])||(hd[1]==hd[2]&&hd[2]==hd[3])||(hd[2]==hd[3]&&hd[3]==hd[4]))return 5;
if(((hd[0]==hd[1])&&(hd[2]==hd[3]||hd[3]==hd[4]))||(hd[1]==hd[2]&&hd[3]==hd[4]))return 6;
if(hd[0]==hd[1]||hd[1]==hd[2]||hd[2]==hd[3]||hd[3]==hd[4])return 7;
return 8;
}
void sub(int cur)
{
ans=min(ans,jud(cur));
int s=cur?a[cur-1]+1:0;
for(int i=s;i<5;i++){
a[cur]=i;
sub(cur+1);
}
}
void solve()
{
for(int i=0;i<10;i++){
if(cd[i][0]=='T')val[i]=10;
else if(cd[i][0]=='J')val[i]=11;
else if(cd[i][0]=='Q')val[i]=12;
else if(cd[i][0]=='K')val[i]=13;
else if(cd[i][0]=='A')val[i]=1;
else val[i]=cd[i][0]-'0';
}
ans=8;
sub(0);
}
void out()
{
cout<<"Hand: ";
for(int i=0;i<5;i++)
cout<<cd[i]<<" ";
cout<<"Deck: ";
for(int i=5;i<10;i++)
cout<<cd[i]<<" ";
cout<<"Best hand: "<<bthd[ans]<<endl;
}
int main()
{
while(cin>>cd[0]){
for(int i=1;i<10;i++)
cin>>cd[i];
solve();
out();
}
return 0;
}