题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1111
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.
题意：两手扑克牌，一手各5张牌，按照接下来的规则比大小。下面的规则是按照一手牌从大到小介绍的。
规则一：5张牌花色一样，大小连着的，俗称“同花顺”。同花顺的大小根据最大值的判断。
规则二：有4张牌一样大的。就是说5张牌中能组成炸弹。这样的牌通过那4张牌的大小再来进行比较大小。比如（A，3，3，3，3）和（9，4，4，4，4）就是后者大。
规则三：3张牌一样大，另外两张牌组成一对。满足规则三的两手牌根据那3张牌再进行大小比较。
规则四：5张牌同色，大小比较就看牌的大小了。
规则五：5张牌大小连着组成了顺子，大小看最大值。
规则六：3张牌相同，大小看3张牌的值的大小。
规则七：组成两对和一张单牌，大小看那两对的大小，如果还相同的就看那单牌。
规则八：组成一队，大小先看这一对，然后再比较剩下的三张散牌。
规则九：5张牌没什么组合的，散牌，就单纯的比较大小。
解法：按照这几个规则模拟即可。
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<cstdlib>
5 #include<cmath>
6 #include<algorithm>
7 using namespace std;
8 struct node
9 {
10 int value;
11 char color;
12 friend bool operator <(node a,node b)
13 {
14 return a.value > b.value;
15 }
16 }an[6],bn[6];
17 int cmp(int i,int j) {return i<j; }
18 int getvalue(char *str)
19 {
20 if (str[0]>=‘2‘&&str[0]<=‘9‘) return str[0]-‘0‘;
21 if (str[0]==‘T‘) return 10;
22 if (str[0]==‘J‘) return 11;
23 if (str[0]==‘Q‘) return 12;
24 if (str[0]==‘K‘) return 13;
25 if (str[0]==‘A‘) return 14;
26 }
27 int color_ok(node *cn,int k)
28 {
29 int flag=1;
30 for (int i=2 ;i<=5 ;i++) if (cn[i].color!=cn[1].color) flag=0;
31 for (int i=1 ;i<=4 ;i++) if (cn[i].value!=cn[i+1].value+1) flag=0;
32 if (cn[1].value!=k) flag=0;
33 if (flag) return 1;
34 return 0;
35 }
36 int four_ok(node *cn,int k)
37 {
38 int cnt=0;
39 for (int i=1 ;i<=5 ;i++) if (cn[i].value==k) cnt++;
40 if (cnt==4) return 1;
41 return 0;
42 }
43 int full_ok(node *cn,int k)
44 {
45 int cnt=0;
46 int u=0,v=0;
47 for (int i=1 ;i<=5 ;i++)
48 {
49 if (cn[i].value==k) cnt++;
50 else if (!u) u=i;
51 else v=i;
52 }
53 if (cnt==3)
54 {
55 if (cn[u].value==cn[v].value) return 1;
56 else return 0;
57 }
58 else return 0;
59 }
60 int same_color(node *cn)
61 {
62 for (int i=2 ;i<=5 ;i++) if (cn[i].color!=cn[1].color) return 0;
63 return 1;
64 }
65 int straight_ok(node *cn,int k)
66 {
67 int flag=0;
68 for (int i=1 ;i<=4 ;i++) if (cn[i].value!=cn[i+1].value+1) flag=1;
69 if (cn[1].value!=k) flag=1;
70 if (flag) return 0;
71 return 1;
72 }
73 int three_ok(node *cn,int k)
74 {
75 int cnt=0;
76 for (int i=1 ;i<=5 ;i++) if (cn[i].value==k) cnt++;
77 if (cnt==3) return 1;
78 return 0;
79 }
80 int first_ok(node *cn,int k)
81 {
82 int cnt=0;
83 for (int i=1 ;i<=5 ;i++) if (cn[i].value==k) cnt++;
84 if (cnt==2) return 1;
85 return 0;
86 }
87 int main()
88 {
89 char str[4];
90 while (scanf("%s",str)!=EOF)
91 {
92 an[1].value=getvalue(str);an[1].color=str[1];
93 for (int i=2 ;i<=5 ;i++)
94 {
95 scanf("%s",str);
96 an[i].value=getvalue(str);
97 an[i].color=str[1];
98 }
99 for (int i=1 ;i<=5 ;i++)
100 {
101 scanf("%s",str);
102 bn[i].value=getvalue(str);
103 bn[i].color=str[1];
104 }
105 sort(an+1,an+5+1);sort(bn+1,bn+5+1);
106 int OK=0;
107 int flag1=0,flag2=0;
108 ///Straight flush
109 for (int i=14 ;i>=6 ;i--)
110 {
111 flag1=flag2=0;
112 if (color_ok(an,i)) flag1=1;
113 if (color_ok(bn,i)) flag2=1;
114 if (flag1 && flag2) {printf("Tie.\n");OK=1;break; }
115 if (flag1) {printf("Black wins.\n");OK=1;break; }
116 if (flag2) {printf("White wins.\n");OK=1;break; }
117 }
118 if (OK) continue;
119 ///Four of a kind
120 for (int i=14 ;i>=2 ;i--)
121 {
122 flag1=flag2=0;
123 if (four_ok(an,i)) flag1=1;
124 if (four_ok(bn,i)) flag2=1;
125 if (flag1 && flag2) {printf("Tie.\n");OK=1;break; }
126 if (flag1) {printf("Black wins.\n");OK=1;break; }
127 if (flag2) {printf("White wins.\n");OK=1;break; }
128 }
129 if (OK) continue;
130 ///Full House
131 for (int i=14 ;i>=2 ;i--)
132 {
133 flag1=flag2=0;
134 if (full_ok(an,i)) flag1=1;
135 if (full_ok(bn,i)) flag2=1;
136 if (flag1 && flag2) {printf("Tie.\n");OK=1;break; }
137 if (flag1) {printf("Black wins.\n");OK=1;break; }
138 if (flag2) {printf("White wins.\n");OK=1;break; }
139 }
140 if (OK) continue;
141 ///Flush
142 flag1=flag2=0;
143 if (same_color(an)) flag1=1;
144 if (same_color(bn)) flag2=1;
145 if (flag1 && flag2)
146 {
147 int ok=0;
148 for (int i=1 ;i<=5 ;i++)
149 {
150 if (an[i].value>bn[i].value) {printf("Black wins.\n");ok=1;OK=1;break; }
151 else if (an[i].value<bn[i].value) {printf("White wins.\n");ok=1;OK=1;break; }
152 }
153 if (!ok) {printf("Tie.\n");OK=1;continue;}
154 }
155 if (flag1&&!OK) {printf("Black wins.\n");OK=1;continue; }
156 if (flag2&&!OK) {printf("White wins.\n");OK=1;continue; }
157 if (OK) continue;
158 ///Straight
159 for (int i=14 ;i>=6 ;i--)
160 {
161 flag1=flag2=0;
162 if (straight_ok(an,i)) flag1=1;
163 if (straight_ok(bn,i)) flag2=1;
164 if (flag1 && flag2) {printf("Tie.\n");OK=1;break; }
165 if (flag1) {printf("Black wins.\n");OK=1;break; }
166 if (flag2) {printf("White wins.\n");OK=1;break; }
167 }
168 if (OK) continue;
169 ///Three of a Kind
170 for (int i=14 ;i>=2 ;i--)
171 {
172 flag1=flag2=0;
173 if (three_ok(an,i)) flag1=1;
174 if (three_ok(bn,i)) flag2=1;
175 if (flag1 && flag2) {printf("Tie.\n");OK=1;break; }
176 if (flag1) {printf("Black wins.\n");OK=1;break; }
177 if (flag2) {printf("White wins.\n");OK=1;break; }
178 }
179 if (OK) continue;
180 ///Two Pairs
181 for (int i=14 ;i>=2 ;i--)
182 {
183 flag1=flag2=0;
184 int k1=0,k2=0;
185 if (first_ok(an,i))
186 {
187 for (int j=i-1 ;j>=2 ;j--)
188 {
189 if (first_ok(an,j)) {flag1=1;k1=j;break;}
190 }
191 }
192 if (first_ok(bn,i))
193 {
194 for (int j=i-1 ;j>=2 ;j--)
195 {
196 if (first_ok(bn,j)) {flag2=1;k2=j;break;}
197 }
198 }
199 if (flag1 && flag2 && k1==k2)
200 {
201 int u=0,v=0;
202 for (int j=1 ;j<=5 ;j++) if (an[j].value!=i && an[j].value!=k1) {u=j;break; }
203 for (int j=1 ;j<=5 ;j++) if (bn[j].value!=i && bn[j].value!=k2) {v=j;break; }
204 if (an[u].value==bn[v].value) {printf("Tie.\n");OK=1;break; }
205 else if (an[u].value>bn[v].value) {printf("Black wins.\n");OK=1;break; }
206 else {printf("White wins.\n");OK=1;break; }
207 }
208 if (flag1 && flag2 && !OK && k1!=k2)
209 {
210 if (k1>k2) {printf("Black wins.\n");OK=1;break; }
211 else {printf("White wins.\n");OK=1;break; }
212 }
213 if (flag1) {printf("Black wins.\n");OK=1;break; }
214 if (flag2) {printf("White wins.\n");OK=1;break; }
215 }
216 if (OK) continue;
217 ///Pair
218 for (int i=14 ;i>=2 ;i--)
219 {
220 flag1=flag2=0;
221 if (first_ok(an,i)) flag1=1;
222 if (first_ok(bn,i)) flag2=1;
223 if (flag1 && flag2)
224 {
225 int a[4],b[4];
226 int c=0,d=0;
227 for (int j=1 ;j<=5 ;j++) if (an[j].value!=i) a[c++]=an[j].value;
228 for (int j=1 ;j<=5 ;j++) if (bn[j].value!=i) b[d++]=bn[j].value;
229 sort(a,a+c,cmp);sort(b,b+d,cmp);
230 int flag=0;
231 for (int j=0 ;j<3 ;j++)
232 {
233 if (a[j]>b[j]) {printf("Black wins.\n");flag=1;OK=1;break; }
234 else if (a[j]<b[j]) {printf("White wins.\n");flag=1;OK=1;break; }
235 }
236 if (flag==0) {printf("Tie.\n");OK=1;break; }
237 }
238 if (flag1&&!OK) {printf("Black wins.\n");OK=1;break; }
239 if (flag2&&!OK) {printf("White wins.\n");OK=1;break; }
240 }
241 if (OK) continue;
242 ///High Card
243 int flag=0;
244 for (int i=1 ;i<=5 ;i++)
245 {
246 if (an[i].value>bn[i].value) {printf("Black wins.\n");flag=1;OK=1;break; }
247 else if (an[i].value<bn[i].value) {printf("White wins.\n");flag=1;OK=1;break; }
248 }
249 if (!flag) {printf("Tie.\n");continue; }
250 }
251 return 0;
252 }
时间： 2024-12-10 01:39:20