二分图最大匹配问题 扑克题还是用map比较方便 #include<bits/stdc++.h> using namespace std; #define MAXI 52 ]; ]; int used[MAXI]; int vis[MAXI]; int n,m; map<char,int>mm; bool judge(int a,int b) { ]]>mm[ yd[b][] ] ) return true; ]]==mm[ yd[b][] ]) ]]>mm[ yd[b]…

Card Game Cheater Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1272    Accepted Submission(s): 675 Problem Description Adam and Eve play a card game using a regular deck of 52 cards. The rule…

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 Problem Description Adam and Eve play a card game using a regular deck of 52 cards. The rule…

称号: Card Game Cheater Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 103 Accepted Submission(s): 74   Problem Description Adam and Eve play a card game using a regular deck of 52 cards. The rules…

http://acm.hdu.edu.cn/showproblem.php?pid=1528 Card Game Cheater Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1559    Accepted Submission(s): 820 Problem Description Adam and Eve play a card…

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/35236457 简单二分图匹配.... Card Game Cheater Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1073   …

Card Game Cheater Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1072    Accepted Submission(s): 564 Problem Description Adam and Eve play a card game using a regular deck of 52 cards. The rule…

Card Game Cheater Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1576    Accepted Submission(s): 830 Problem Description Adam and Eve play a card game using a regular deck of 52 cards. The rule…

题意与分析 题意是这样的:有\(n\)张牌,然后第一行是Adam的牌,第二行是Eve的牌:每两个字符代表一张牌,第一个字符表示牌的点数,第二个表示牌的花色.Adam和Eve每次从自己的牌中选出一张牌进行比较,先看大小再看花色,花色顺序是C,D,S,H(依次增大),谁的牌大谁就加一分,问Eve最多能得到多少分. 思路:最大二分匹配,Eve的牌为集合1,Adam的牌为集合2,集合1中的牌与集合2中比它小的牌建立联系,找最大匹配.这种两个集合的要想到试试二分图. 代码 /* ACM Code writ…

两组牌中两张牌相比能赢的就连,后求最大匹配. #include <cmath> #include <cstdlib> #include <cstdio> #include <cstring> #include <algorithm> #include <fstream> #include <iostream> #define rep(i, l, r) for(int i=l; i<=r; i++) #define c…