HDU 1528&&ZOJ 2223 Card Game Cheater（二分图匹配）

Card Game Cheater
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Time Limit: 10 Seconds Memory Limit: 32768 KB
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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
题目意思其实比较简单：两个人在一起打牌，每张牌由两个字符组成。第一个字符有
(2, 3, 4, 5, 6, 7, 8 ,9, T, J, Q, K, or A)这些，第二个字符有(C, D, S, or H)这些，
都是越到后面越大。两张牌比大小，第一个字符大的牌就大，如果相同，看第二个字符的大小。
现在两个手上都是N张牌，开始Adam把牌依次放在座子上，现在Eve可以根据Adam牌的顺序调整自己牌的顺序，使得自己尽可能多赢分（赢一张牌的一分）

那么我们可以构建一个二分图，把Eve可能赢的边都连接出来，在求一遍最大二分匹配

这个就是答案了

#include<iostream>
#include<cstdio>
#include<functional>
#include<algorithm>
#include<cstring>
using namespace std;
string s1[30],s2[30];
int match[30],book[30];
int e[30][30];
int t,num;
int getnum(char s)//把字符转换为数字大小
{
	//第一个字符
	if('0'<=s&&s<='9') return s-'0';
	if(s=='T') return 10;
	else if(s=='J') return 11;
	else if(s=='Q') return 12;
	else if(s=='K') return 13;
	else if(s=='A') return 14;
	//第二个字符
	else if(s=='C') return 20;
	else if(s=='D') return 21;
	else if(s=='S') return 22;
	else if(s=='H') return 23;


}
int dfs(int u)
{
	int i;
	for(i=1;i<=num;i++)
	{
		if(book[i]==0&&e[u][i]==1)
		{
			book[i]=1;
			if(match[i]==0||dfs(match[i]))
			{
				match[i]=u;
				return 1;
			}
		}
	}
	return 0;
}
int main()
{
	int i,j,sum;
	int a,b;
	scanf("%d",&t);
	while(t--)
	{
		sum=0;
		memset(e,0,sizeof(e));
		scanf("%d",&num);
		for(i=1;i<30;i++)
		{
			s1[i].clear();
			s2[i].clear();
		}
		for(i=1;i<=num;i++)
			cin>>s1[i];
		for(i=1;i<=num;i++)
			cin>>s2[i];

		for(i=1;i<=num;i++)//建图过程
			for(j=1;j<=num;j++)
		{
			a=getnum(s2[i][0]);//Eve牌面大小
			b=getnum(s1[j][0]);//Adam牌面大小
			if(a>b)//第一个字符就大
				e[i][j]=1;
			else if(a==b)//第一个字符相同
			{
				a=getnum(s2[i][1]);
				b=getnum(s1[j][1]);
				if(a>b)
					e[i][j]=1;
			}
		}
		for(i=1;i<=num;i++)//二分图最大匹配过程
			match[i]=0;
		for(i=1;i<=num;i++)
		{
			for(j=1;j<=num;j++)
				book[j]=0;
			if(dfs(i))
				sum++;
		}
		printf("%d\n",sum);
	}
	return 0;
}