ZOJ1111：Poker Hands(模拟题)

A poker deck contains 52 cards - each card has a suit which is one of clubs, diamonds, hearts, or spades (denoted C, D, H, S in the input data). Each card also has a value which is one of 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, king, ace (denoted 2, 3, 4, 5, 6, 7, 8, 9, T, J, Q, K, A). For scoring purposes, the suits are unordered while the values are ordered as given above, with 2 being the lowest and ace the highest value.

A poker hand consists of 5 cards dealt from the deck. Poker hands are ranked by the following partial order from lowest to highest

• High Card. Hands which do not fit any higher category are ranked by the value of their highest card. If the highest cards have the same value, the hands are ranked by the next highest, and so on.
• Pair. 2 of the 5 cards in the hand have the same value. Hands which both contain a pair are ranked by the value of the cards forming the pair. If these values are the same, the hands are ranked by the values of the cards not forming the pair, in decreasing order.
• Two Pairs. The hand contains 2 different pairs. Hands which both contain 2 pairs are ranked by the value of their highest pair. Hands with the same highest pair are ranked by the value of their other pair. If these values are the same the hands are ranked by the value of the remaining card.
• Three of a Kind. Three of the cards in the hand have the same value. Hands which both contain three of a kind are ranked by the value of the 3 cards.
• Straight. Hand contains 5 cards with consecutive values. Hands which both contain a straight are ranked by their highest card.
• Flush. Hand contains 5 cards of the same suit. Hands which are both flushes are ranked using the rules for High Card.
• Full House. 3 cards of the same value, with the remaining 2 cards forming a pair. Ranked by the value of the 3 cards.
• Four of a kind. 4 cards with the same value. Ranked by the value of the 4 cards.
• Straight flush. 5 cards of the same suit with consecutive values. Ranked by the highest card in the hand.

Your job is to compare several pairs of poker hands and to indicate which, if either, has a higher rank.

Input Specification

Several lines, each containing the designation of 10 cards: the first 5 cards are the hand for the player named "Black" and the next 5 cards are the hand for the player named "White."

Output Specification

For each line of input, print a line containing one of:

   Black wins.
   White wins.
   Tie.

Sample Input

2H 3D 5S 9C KD 2C 3H 4S 8C AH
2H 4S 4C 2D 4H 2S 8S AS QS 3S
2H 3D 5S 9C KD 2C 3H 4S 8C KH
2H 3D 5S 9C KD 2D 3H 5C 9S KH

Sample Output

White wins.
Black wins.
Black wins.
Tie.

题意：又是一道模拟题，好久没敲过模拟题了，这次敲了蛮久，还记得暑假由此做某个地方的省赛，也是一道模拟题，那道题是打麻将，这次变成玩扑克了。

题意很简单，看过赌神的人都知道，每人手中5张排，比牌面大小，牌面由大到小分别是（这里花色无大小）

同花顺：牌面一样，只比较最大的看谁大，一样大则平手

四条：四个一样的，看这四个一样的中谁大谁赢

葫芦：三条+一对，看三条中谁的牌面大

同花：都是同花就按从大到小比较，看谁的更大

顺子：都是顺子看最大的谁大，一样则平手

三条：三条看三条中谁的牌面大

两对:两对就看谁的对子大，都一样大比单牌

一对:比对子，一样则比单牌

单排：按顺序比较大小，大者胜

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

struct node
{
    int num,mark;
} white[10],black[10];

int cmp(node a,node b)
{
    return a.num<b.num;
}

int  set(char c)
{
    if(c<='9' && c>='2')
        return c-'0';
    if(c == 'T')
        return 10;
    if(c == 'J')
        return 11;
    if(c == 'Q')
        return 12;
    if(c == 'K')
        return 13;
    if(c == 'A')
        return 14;
    if(c == 'C')
        return 1;
    if(c == 'D')
        return 2;
    if(c == 'H')
        return 3;
    if(c == 'S')
        return 4;
}

int up[10];

int same(node *a)
{
    int i,l = 0;
    for(i = 2; i<=5; i++)
        if(a[i].num!=a[i-1].num)
            up[l++] = i;
    return l;
}
/*
我们用数字表示牌面
1-同花顺
2-四条
3-葫芦
4-同花
5-顺子
6-三条
7-两对
8-一对
9-单牌
*/
int judge(node *a)
{
    int flag1 = 0;
    int flag2 = 0;
    if(a[1].mark == a[2].mark && a[2].mark == a[3].mark &&a[3].mark == a[4].mark &&a[4].mark == a[5].mark)
        flag1 = 1;
    if(a[1].num+1 == a[2].num && a[2].num+1 == a[3].num &&a[3].num+1 == a[4].num &&a[4].num+1 == a[5].num)
        flag2 = 1;
    if(flag1 && flag2)
        return 1;
   else  if(flag1 && !flag2)
        return 4;
    else if(flag2 && !flag1)
        return 5;
    int k = same(a);
    if(k == 0)
        return 2;
    else if(k == 1)
    {
        if(up[0] == 2 || up[0] == 5)
            return 2;
        if(up[0] == 3 || up[0] == 4)
            return 3;
    }
    else if(k == 2)
    {
        if(up[0] == 2 && up[1] == 3 || up[0] == 4 && up[1] == 5 || up[0] == 2 && up[1] == 5)
            return 6;
        else
            return 7;
    }
    else if(k == 3)
        return 8;
    else if(k == 4)
        return 9;
    return 0;
}

int find_one(node *a)
{
    int i;
    for(i = 1; i<=5; i++)
    {
        if(i == 1 && a[i].num!=a[i+1].num)
            return 1;
        if(i == 5 && a[i].num!=a[i-1].num)
            return 5;
        if(a[i].num!=a[i-1].num && a[i].num!=a[i+1].num)
            return i;
    }
    return 0;
}

int find_pair(node *a)
{
    int i;
    for(i = 1; i<=5; i++)
    {
        if(i == 1 && a[i].num == a[i+1].num)
            return 2;
        if(a[i].num == a[i-1].num)
            return i;
    }
    return 0;
}

int compare(int x)
{
    int i,j,a,b;
    if(x == 1 || x == 5)
    {
        if(white[5].num>black[5].num)
            return -1;
        else if(white[5].num<black[5].num)
            return 1;
        return 0;
    }
    else if(x == 2 || x == 3 || x == 6)
    {
        if(white[3].num>black[3].num)
            return -1;
        else if(white[3].num<black[3].num)
            return 1;
        return 0;
    }
    else if(x == 4 || x == 9)
    {
        for(i = 5; i>=1; i--)
        {
            if(black[i].num<white[i].num)return -1;
            else if(black[i].num>white[i].num)
                return 1;
        }
        return 0;
    }
    else if(x == 7)
    {
        if(white[4].num>black[4].num)
            return -1;
        else if(white[4].num<black[4].num)
            return 1;
        else
        {
            if(white[2].num>black[2].num)
                return -1;
            else if(white[2].num<black[2].num)
                return 1;
            int a = find_one(white);
            int b = find_one(black);
            if(black[b].num<white[a].num)
                return -1;
            else if(black[b].num>white[a].num)
                return 1;
            return 0;
        }
    }
    else if(x == 8)
    {
        int a = find_pair(white);
        int b = find_pair(black);
        int tem_w[10],tem_b[10],i,lb = 1,lw = 1;
        if(white[a].num>black[b].num)
            return -1;
        else if(white[a].num<black[b].num)
            return 1;
        for(i = 1; i<=5; i++)
        {
            if(i!=a && i!=a-1)
                tem_w[lw++] = white[i].num;
            if(i!=b && i!=b-1)
                tem_b[lb++] = black[i].num;
        }
        for(i = 3; i>=1; i--)
        {
            if(tem_w[i]>tem_b[i])
                return -1;
            else if(tem_b[i]>tem_w[i])
                return 1;
        }
        return 0;
    }
    return 0;
}

int main()
{
    char str[5];
    int sumb,sumw,j,k,i;
    while(~scanf("%2s",str))
    {
        j = 1;
        k = 1;
        black[j].num = set(str[0]);
        black[j++].mark = set(str[1]);
        for(i = 2; i<=10; i++)
        {
            scanf("%2s",str);
            if(i<6)
            {
                black[j].num = set(str[0]);
                black[j++].mark = set(str[1]);
            }
            else
            {
                white[k].num = set(str[0]);
                white[k++].mark = set(str[1]);
            }
        }
        sort(black+1,black+6,cmp);
        sort(white+1,white+6,cmp);
        sumb = judge(black);
        sumw = judge(white);
        if(sumb>sumw)
            printf("White wins.\n");
        else if(sumb<sumw)
            printf("Black wins.\n");
        else
        {
            int flag = compare(sumb);
            if(flag == -1)
                printf("White wins.\n");
            else if(flag == 1)
                printf("Black wins.\n");
            else
                printf("Tie.\n");
        }
    }

    return 0;
}