Card Game Cheater
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1822 Accepted Submission(s): 998
Problem Description
Adam
and Eve play a card game using a regular deck of 52 cards. The rules
are simple. The players sit on opposite sides of a table, facing each
other. Each player gets k cards from the deck and, after looking at
them, places the cards face down in a row on the table. Adam’s cards are
numbered from 1 to k from his left, and Eve’s cards are numbered 1 to k
from her right (so Eve’s i:th card is opposite Adam’s i:th card). The
cards are turned face up, and points are awarded as follows (for each i ∈
{1, . . . , k}):
If Adam’s i:th card beats Eve’s i:th card, then Adam gets one point.
If Eve’s i:th card beats Adam’s i:th card, then Eve gets one point.
A
card with higher value always beats a card with a lower value: a three
beats a two, a four beats a three and a two, etc. An ace beats every
card except (possibly) another ace.
If the two i:th cards
have the same value, then the suit determines who wins: hearts beats all
other suits, spades beats all suits except hearts, diamond beats only
clubs, and clubs does not beat any suit.
For example, the ten of spades beats the ten of diamonds but not the Jack of clubs.
This
ought to be a game of chance, but lately Eve is winning most of the
time, and the reason is that she has started to use marked cards. In
other words, she knows which cards Adam has on the table before he turns
them face up. Using this information she orders her own cards so that
she gets as many points as possible.
Your task is to, given Adam’s and Eve’s cards, determine how many points Eve will get if she plays optimally.
Input
There
will be several test cases. The first line of input will contain a
single positive integer N giving the number of test cases. After that
line follow the test cases.
Each test case starts with a line
with a single positive integer k <= 26 which is the number of cards
each player gets. The next line describes the k cards Adam has placed on
the table, left to right. The next line describes the k cards Eve has
(but she has not yet placed them on the table). A card is described by
two characters, the first one being its value (2, 3, 4, 5, 6, 7, 8 ,9,
T, J, Q, K, or A), and the second one being its suit (C, D, S, or H).
Cards are separated by white spaces. So if Adam’s cards are the ten of
clubs, the two of hearts, and the Jack of diamonds, that could be
described by the line
TC 2H JD
Output
For
each test case output a single line with the number of points Eve gets
if she picks the optimal way to arrange her cards on the table.
Sample Input
3
1
JD
JH
2
5D TC
4C 5H
3
2H 3H 4H
2D 3D 4D
Sample Output
1
1
2
Source
题意：
甲乙两个人进行扑克比赛，乙已经知道甲每次要出什么牌，问乙最多能赢几局。每次只出一张牌，大小顺序是2,3,4,5,6,7,8,9,10，J,Q,K,A,大小相同的牌红桃>黑桃>方块>梅花。
代码：
1 //跟以前做过的田忌赛马相似，从最小的开始比较，乙能赢就赢掉；如果一样大就比较最大的最大的能赢就赢掉，不能赢就用最小的输掉对方最大的；如果小就用它输掉对方最大的。
2 //二分图最大匹配数，分为甲乙两个点集，将乙能赢甲的之间连条线，求最大匹配数。
3 #include<iostream>
4 #include<cstdio>
5 #include<cstring>
6 #include<algorithm>
7 using namespace std;
8 int num[120];
9 struct desk
10 {
11 int x,y;
12 };
13 bool cmp(desk a,desk b)
14 {
15 if(a.x==b.x) return a.y<b.y;
16 return a.x<b.x;
17 }
18 int main()
19 {
20 int t,n;
21 char ch[3];
22 scanf("%d",&t);
23 while(t--)
24 {
25 desk p1[30],p2[30];
26 scanf("%d",&n);
27 for(int i=0;i<n;i++)
28 {
29 scanf("%s",ch);
30 if(ch[0]>=‘1‘&&ch[0]<=‘9‘)
31 p1[i].x=ch[0]-‘0‘+;
32 else switch(ch[0])
33 {
34 case ‘T‘:p1[i].x=10;break;
35 case ‘J‘:p1[i].x=11;break;
36 case ‘Q‘:p1[i].x=12;break;
37 case ‘K‘:p1[i].x=13;break;
38 case ‘A‘:p1[i].x=14;break;
39 }
40 switch(ch[1])
41 {
42 case ‘C‘:p1[i].y=1;break;
43 case ‘D‘:p1[i].y=2;break;
44 case ‘S‘:p1[i].y=3;break;
45 case ‘H‘:p1[i].y=4;break;
46 }
47 }
48 for(int i=0;i<n;i++)
49 {
50 scanf("%s",ch);
51 if(ch[0]>=‘1‘&&ch[0]<=‘9‘)
52 p2[i].x=ch[0]-‘0‘;
53 else switch(ch[0])
54 {
55 case ‘T‘:p2[i].x=10;break;
56 case ‘J‘:p2[i].x=11;break;
57 case ‘Q‘:p2[i].x=12;break;
58 case ‘K‘:p2[i].x=13;break;
59 case ‘A‘:p2[i].x=14;break;
60 }
61 switch(ch[1])
62 {
63 case ‘C‘:p2[i].y=1;break;
64 case ‘D‘:p2[i].y=2;break;
65 case ‘S‘:p2[i].y=3;break;
66 case ‘H‘:p2[i].y=4;break;
67 }
68 }
69 sort(p1,p1+n,cmp);
70 sort(p2,p2+n,cmp);
71 int s1=0,s2=0,e1=n-1,e2=n-1,ans=0;
72 while(s1<=e1)
73 {
74 if((p2[s2].x>p1[s1].x)||((p2[s2].x==p1[s1].x)&&(p2[s2].y>p1[s1].y)))
75 {
76 s2++;s1++;
77 ans++;
78 }
79 else if((p2[s2].x==p1[s1].x)&&(p2[s2].y==p1[s1].y))
80 {
81 if((p2[e2].x>p1[e1].x)||((p2[e2].x==p1[e1].x)&&(p2[e2].y>p1[e1].y)))
82 {
83 e2--;e1--;
84 ans++;
85 }
86 else
87 {
88 s2++;e1--;
89 }
90 }
91 else
92 {
93 s2++;e1--;
94 }
95 }
96 printf("%d\n",ans);
97 }
98 return 0;
99 }