题意：Adam和Eve打牌，Eve可以看到Adam的牌.现在Adam已经把牌按顺序摆好了,Eve知道Adam所有的牌,所以Eve可以最优化自己牌的摆放顺序,使得自己每个位置上的牌赢Adam对应位置上的牌的个数最多.
其中每张牌由值+类型构成,值是2, 3, 4, 5, 6, 7, 8 ,9, T, J, Q, K, A,这些值依次增加其中2最小,A最大.类型为: C, D, S, H. 类型同样依次增大,H类最大.
问题Eve最多能拿几分?(一张牌赢拿1分,输了或平局不拿分)
思路：能看懂题目就很好做了
构建二分图:左点集是Eve对应的n张牌,右点集是Adam对应的n张牌. 如果左i牌>右j牌,那么就连一条左i右j的边.
本题要获得最多的分,就是要找到一个最大匹配.每条匹配边就是1分.所以本题最终解== 二分图最大匹配边数.
本题我把每张牌的直接转化为分数,然后比较分数即可.
2 -A 代表20分,30分,…140分. 而 CDSH分别代表1,2,3,4分.
#include<cstdio>
#include<cstring>
#include<vector>
#include<string>
#include<iostream>
using namespace std;
const int maxn = 26+5;
struct Max_Match
{
int n;
vector<int> g[maxn];
bool vis[maxn];
int left[maxn];
void init(int n)
{
this->n=n;
for(int i=1;i<=n;i++) g[i].clear();
memset(left,-1,sizeof(left));
}
bool match(int u)
{
for(int i=0;i<g[u].size();i++)
{
int v=g[u][i];
if(!vis[v])
{
vis[v]=true;
if(left[v]==-1 || match(left[v]))
{
left[v]=u;
return true;
}
}
}
return false;
}
int solve()
{
int ans=0;
for(int i=1;i<=n;i++)
{
memset(vis,0,sizeof(vis));
if(match(i)) ++ans;
}
return ans;
}
}MM;
int get_score(string& s)
{
int ans=0,i=0;
if(s[i]>='2'&&s[i]<='9') ans += (s[i]-'0')*10;
else if(s[i]=='T') ans += 100;
else if(s[i]=='J') ans += 110;
else if(s[i]=='Q') ans +=120;
else if(s[i]=='K') ans +=130;
else if(s[i]=='A') ans +=140;
i=1;//C, D, S, or H
if(s[i]=='C') ans+=1;
else if(s[i]=='D') ans+=2;
else if(s[i]=='S') ans+=3;
else if(s[i]=='H') ans+=4;
return ans;
}
int main()
{
int T; scanf("%d",&T);
while(T--)
{
int n;
scanf("%d",&n);
MM.init(n);
int s1[maxn],s2[maxn];
string s;
for(int i=1;i<=n;i++)
{
cin>>s;
s2[i]=get_score(s);
}
for(int i=1;i<=n;i++)
{
cin>>s;
s1[i]=get_score(s);
}
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
if(s1[i]>s2[j])
MM.g[i].push_back(j);
printf("%d\n",MM.solve());
}
return 0;
}
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
