Card Game Cheater(贪心+二分匹配)
Card Game Cheater
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1566 Accepted Submission(s): 822
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.
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
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.
1
JD
JH
2
5D TC
4C 5H
3
2H 3H 4H
2D 3D 4D
1
2
题目大意:
a和b手上都有n张牌,b的一张牌赢了a的一张牌,b就得一分,问b能得多少分。
二分最大匹配代码:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define mem(x,y) memset(x,y,sizeof(x))
using namespace std;
const int INF=0x3f3f3f3f;
const int MAXN=;
struct Node{
int v;
}a[MAXN],b[MAXN];
int mp[MAXN][MAXN];
int vis[MAXN],usd[MAXN];
int N;
int getnum(char *s,Node &temp){
int x=;
if(s[]=='A')x=;
else if(s[]=='T')x=;
else if(s[]=='J')x=;
else if(s[]=='Q')x=;
else if(s[]=='K')x=;
else if(s[]>=''&&s[]<='')x=s[]-'';
x=x*;
if(s[]=='C')x+=;
else if(s[]=='D')x+=;
else if(s[]=='S')x+=;
else if(s[]=='H')x+=;
temp.v=x;
}
bool dfs(int u){
for(int i=;i<=N;i++){
if(!vis[i]&&mp[u][i]){
vis[i]=;
if(!usd[i]||dfs(usd[i])){
usd[i]=u;
return true;
}
}
}
return false;
}
int main(){
int T;
char s[];
scanf("%d",&T);
while(T--){
mem(mp,);
scanf("%d",&N);
for(int i=;i<=N;i++){
scanf("%s",s);
getnum(s,a[i]);
}
for(int i=;i<=N;i++){
scanf("%s",s);getnum(s,b[i]);
for(int j=;j<=N;j++)
if(b[i].v>a[j].v)mp[i][j]=;
}
mem(usd,);
int ans=;
for(int i=;i<=N;i++){
mem(vis,);
if(dfs(i))ans++;
}
printf("%d\n",ans);
}
return ;
}
贪心:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define mem(x,y) memset(x,y,sizeof(x))
using namespace std;
typedef long long LL;
const int INF=0x3f3f3f3f;
int a[],b[];
char s[];
int getnum(){
int x=;
if(s[]=='A')x=;
else if(s[]=='T')x=;
else if(s[]=='J')x=;
else if(s[]=='Q')x=;
else if(s[]=='K')x=;
else if(s[]>=''&&s[]<='')x=s[]-'';
x=x*;
if(s[]=='C')x+=;
else if(s[]=='D')x+=;
else if(s[]=='S')x+=;
else if(s[]=='H')x+=;
return x;
}
int main(){
int T,N;
scanf("%d",&T);
while(T--){
scanf("%d",&N);
for(int i=;i<N;i++){
scanf("%s",s);
a[i]=getnum();
}
for(int i=;i<N;i++){
scanf("%s",s);
b[i]=getnum();
}
// for(int i=0;i<N;i++)printf("%d ",a[i]);puts("");
// for(int i=0;i<N;i++)printf("%d ",b[i]);puts("");
sort(a,a+N);sort(b,b+N);
int ans=;
for(int i=,j=;i<N&&j<N;){
if(b[i]>=a[j]){
i++;j++;ans++;
}
else i++;
}
printf("%d\n",ans);
}
return ;
}
Card Game Cheater(贪心+二分匹配)的更多相关文章
- hdu 1528 Card Game Cheater (二分匹配)
Card Game Cheater Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others ...
- (hdu step 6.3.5)Card Game Cheater(匹配的最大数:a与b打牌,问b赢a多少次)
称号: Card Game Cheater Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Other ...
- (简单匹配)Card Game Cheater -- hdu --1528
http://acm.hdu.edu.cn/showproblem.php?pid=1528 Card Game Cheater Time Limit: 2000/1000 MS (Java/Othe ...
- hdu----(1528)Card Game Cheater(最大匹配/贪心)
Card Game Cheater Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others ...
- 【HDU 2255】奔小康赚大钱 (最佳二分匹配KM算法)
奔小康赚大钱 Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Subm ...
- Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) Problem D (Codeforces 831D) - 贪心 - 二分答案 - 动态规划
There are n people and k keys on a straight line. Every person wants to get to the office which is l ...
- HDU 6178 Monkeys(树上的二分匹配)
http://acm.hdu.edu.cn/showproblem.php?pid=6178 题意:现在有一n个顶点的树形图,还有k只猴子,每个顶点只能容纳一只猴子,而且每只猴子至少和另外一只猴子通过 ...
- HDOJ 1528 Card Game Cheater
版权声明:来自: 码代码的猿猿的AC之路 http://blog.csdn.net/ck_boss https://blog.csdn.net/u012797220/article/details/3 ...
- 【DFS求树的最大二分匹配+输入外挂】HDU 6178 Monkeys
http://acm.hdu.edu.cn/showproblem.php?pid=6178 [题意] 给定一棵有n个结点的树,现在有k个猴子分布在k个结点上,我们可以删去树上的一些边,使得k个猴子每 ...
随机推荐
- ios学习Day3
bool 数据类型 #define TRUE 1// #define FALAE 0 #define BOOL int Bool flag=1; bool型 实质上是 int型 c89没有提供 c99 ...
- JavaMail收发邮件的一般流程与主要方法
1.Properties属性类 Properties p = new Properties(); p.put(key, value); key -| mail.smtp.host -| mail.sm ...
- iOS 中多线程的简单使用
iOS中常用的多线程操作有( NSThread, NSOperation GCD ) 为了能更直观的展现多线程操作在SB中做如下的界面布局: 当点击下载的时候从网络上下载图片: - (void)loa ...
- If the server requires more time, try increasing the timeout in the server editor
双击服务器,在overview下的Timeouts中的Start选项,改成10000或者较大就可以了.防止服务器自启动频繁.
- html 浮动元素
在CSS布局中分为内联元素(display:inline)和块状元素(display:block),块状元素默认会占据一行,可设置高度宽度以及边距,而内联元素不会也不能设置.常见的内联元素有:a.sp ...
- RAC ORA-12170 ora-12535/tns-12535
现象:开发人员抱怨RAC数据库出现了时连得上时连不上的情况,用SQLPLUS一试,果然有这样的情况: SQL> conn system/*******@bjyd 已连接. SQL> con ...
- 深究带PLL的错误复位设计
PLL复位通常犯的错误 或者是像上一篇文章 FPGA知识大梳理(四)FPGA中的复位系统大汇总 中的图一样,也是错误设计.为何呢?看ALTPLL (Phase-Locked Loop) IP Cor ...
- javascript closure 闭包 事件绑定
先来一个基本的例子 <!-- 实现一段脚本,使得点击对应链接alert出相应的编号 --> <meta http-equiv="Content-Type" con ...
- 64位系统未注册"MSDAORA.1"提供程序
原因:如错误,64位系统未注册"MSDAORA.1"提供程序 解决:在IIS应用程序池中找到自己的网站,打开高级设置,设置“启用32位应用程序”为“True”即可. 另外还有其他解 ...
- QT小技巧—更好管理项目(增加预编译头文件,并且指定moc文件的生成位置)good
预编译加速编译 QT也可以像VS那样使用预编译头文件来加速编译器的编译速度.首先在.pro文件中加入: CONFIG += precompiled_header 然后定义需要预编译的头文件: PREC ...