UVA 131 - The Psychic Poker Player

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最新推荐文章于 2026-04-16 10:06:26 发布 292 阅读 AI 写同款· GEO 优化 ›

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

下面列出从大到小的扑克牌面. 这是所有的扑克通用规则。

1	Royal Flush 同花大顺又称皇家同花顺它是所有德州扑克中的王牌，即使您经常玩扑克，也很少见到这样的牌。好比打高尔夫球一杆进洞一样。它是由T（10）到Ace的清一色同花组成。

2	Straight Flush 同花顺除了由最大同花所组成的同花大顺以外的同花组成的顺子。

3	Four-of-a-Kind 四条四张同样的牌+任意一张牌。

4	Full House 俘虏或船牌或葫芦三条带一对，即三张同样的牌带两张同样的牌。如都是Full House,则先比较谁的三条大，如三条一样大，则比谁的两对大。如：

5	Flush 五张同花用五张同一花色但不相连的牌型组成，如都是五张同花，则谁的同花牌大谁赢。

6	Straight 五张顺子由五张相连但不同花色的牌组成，在连牌中，Ace是既可作最大也可以作最小的牌。

7	Three-of-a-Kind 三条即三张同样的牌。它有两种叫法，取决于一对牌是在您手中还是在桌上。一对在手中，桌上有一张，称之为“set”;v如手中有一张，桌上有一对，则称之为“Three of A Kind”。

8	Two Pair 两对由五张牌中的两对牌组成。如果都有两对，则先比大对，再比小对。

9	One Pair 一对当不止一人有同样的一对牌时，则要比一对后面的牌，称之为“Kickers”。记住，德州扑克是挑选最好的五张牌去比。

10	High Card 大牌无以上任何牌型时，决定牌的大小。

可以用0-2^5 32个二进制数生成所有弃牌情况用0，1分别代表第x张牌弃还是留，然后从牌堆补到5张；

至于比大小，其实只有三大类，顺子（10 J Q K A特殊），同花，对子（包含多张相同）；

统计对子可以将每次产生的五张牌分别统计与每一张相同的张数（包含自己）然后排序

#define RUN
#ifdef RUN

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <vector>
#include <list>
#include <cctype> 
#include <algorithm>
#include <utility>
#include <math.h>
#include <ctime>

using namespace std;

#define MAXN 110

//输入序列
char numIn[11];	//牌的大小，标示值，如T，J，Q，K，A
int num[11];	//牌的大小，真实值
char suit[11];	//花色

int make[6];	//存放产生的二进制数

char name[10][MAXN] = {"highest-card","one-pair","two-pairs","three-of-a-kind","straight",
	"flush","full-house","four-of-a-kind","straight-flush"};

//手中牌的大小和花色
int Num[6];
int Suit[6];

int Win;

void printout(){
	for(int i=1; i<=10; i++){
		printf("%c%c ", numIn[i], suit[i]);
	}
	printf("\n");
	
	for(int i=1; i<=10; i++){
		printf("%d%c ", num[i], suit[i]);
	}
	printf("\n");
	printf("\n");
}

void printout2(){
	for(int i=1; i<=5; i++){
		printf("%d", make[i]);
	}
	printf("\n");
}

void printout3(){
	for(int i=1; i<=5; i++){
		printf("%d%c ", Num[i], Suit[i]);
	}
	printf("\n");
}

//判断牌大小和花色来确定什么牌
void play(){
	int win=0, shunzi=1, tonghua=1;
	int sum[6];	//存放重复的个数

	memset(sum, 0, sizeof(sum));

	//判断是否顺子
	for(int i=1; i<5; i++){
		if(Num[i]+1 != Num[i+1]){
			shunzi = 0;
			break;
		}
	}

	//顺子的特殊情况:(A,10,J,Q,K)
	if(Num[1]==1 && Num[2]==10 && Num[3]==11 && Num[4]==12 && Num[5]==13){
		shunzi = 1;
	}

	//判断是否同花
	for(int i=1; i<5; i++){
		if(Suit[i] != Suit[5]){
			tonghua = 0;
		}
	}

	//准备计算对子的数目
	//牌面大小为i共有sum[i]张
	for(int i=1; i<=5; i++){
		for(int j=1; j<=5; j++){
			if(Num[i] == Num[j]){
				sum[i]++;
			}
		}
	}

	//升序冒泡排序sum数组
	for(int i=1; i<5; i++){
		for(int j=i+1; j<=5; j++){
			if(sum[i] > sum[j]){
				int t = sum[i];
				sum[i] = sum[j];
				sum[j] = t;
			}
		}
	}

	//根据定义来确定最佳的牌面
	//对子 one-pair
	if(sum[5] == 2){
		win = 1;
	}

	//两对 two-pairs
	if(sum[2]+sum[3] == 4){
		win = 2;
	}

	//三条 three-of-a-kind
	if(sum[5] == 3){
		win = 3;
	}

	//顺子 straight
	if(shunzi){
		win = 4;
	}

	//五张同花 flush
	if(tonghua){
		win = 5;
	}

	//三条且一对 full-house
	if(sum[2]+sum[3] == 5){
		win = 6;
	}

	//4张同牌 four-of-a-kind
	if(sum[2] == 4){
		win = 7;
	}

	//同花顺 straight-flush
	if(tonghua+shunzi == 2){
		win = 8;
	}

	if(win > Win){
		Win = win;
	}
}

int main(){

#ifndef ONLINE_JUDGE
	freopen("131.in", "r", stdin);
	freopen("out.out", "w", stdout);
#endif
	
	while(scanf("%c%c", &numIn[1],&suit[1]) != EOF){
		for(int i=2; i<=10; i++){
			scanf(" %c%c", &numIn[i], &suit[i]);
		}
		getchar();
		
		Win = 0;

		//把牌值的T,J,Q,K,A转为数字，存入num
		for(int i=1; i<=10; i++){
			if(numIn[i] == 'T'){
				num[i] = 10;
			}
			else if(numIn[i] == 'J'){
				num[i] = 11;
			}
			else if(numIn[i] == 'Q'){
				num[i] = 12;
			}
			else if(numIn[i] == 'K'){
				num[i] = 13;
			}
			else if(numIn[i] == 'A'){
				num[i] = 1;
			}
			else if(isdigit(numIn[i])){
				num[i] = numIn[i] - '0';
			}
		}

		//printout();

		//产生5位二进制数的所有可能，从0到2^5-1
		for(int i=0; i<(1<<5); i++){
			int binDigit = 5;
			int tmp = i;
			memset(make, 0, sizeof(make));
			while(tmp != 0){
				make[binDigit] = tmp % 2;
				tmp /= 2;
				binDigit--;
			}

			//printout2();

			//根据产生的二进制序列来选择该丢弃哪一张牌，序列中0表示抛弃，1表示保留
			int dumpCnt = 0;
			int dump[6];
			memset(dump, 0, sizeof(dump));
			for(int i=1; i<=5; i++){
				if(make[i] == 0){
					dumpCnt++;
					//摸最顶端的牌来填补到前面被抛弃牌的位置
					Num[i] = num[5+dumpCnt];
					Suit[i] = suit[5+dumpCnt];
				}
				else{
					//没被抛弃的牌保持不变
					Num[i] = num[i];
					Suit[i] = suit[i];
				}
			}
			
			//printout3();

			//从小到大冒泡排序
			for(int i=1; i<5; i++){
				for(int j=i+1; j<=5; j++){
					if(Num[i] > Num[j]){
						int t1 = Num[i];
						Num[i] = Num[j];
						Num[j] = t1;
						char t2 = Suit[i];
						Suit[i] = Suit[j];
						Suit[j] = t2;
					}
				}
			}

			//printout3();

			play();

			


		}//End of generating binary

		

		printf("Hand: ");
		for(int i=1; i<=5; i++){
			printf("%c%c ", numIn[i], suit[i]);
		}

		printf("Deck: ");
		for(int i=6; i<=10; i++){
			printf("%c%c ", numIn[i], suit[i]);
		}

		printf("Best hand: %s\n", name[Win]);

	}

}


#endif