题目连接：
https://www.hackerrank.com/contests/ieeextreme-challenges/challenges/kabloom
Description
The card game Kabloom is played with multiple decks of playing cards. Players are dealt 2 n cards, face up and arranged in two rows of n cards. The players must discard some of the cards, so that the cards that remain in the first row match the rank of the cards that remain in the second row. The cards match only in rank (e.g. an Ace of Hearts matches any other Ace regardless of suit), but they must appear in the same order in each row. The players are not able to rearrange the order in which the cards appear. Note also that a Joker can match any card including another Joker .
The goal is to maximize the sum of the point value of the cards that remain. Aces are worth 20 points, face cards are worth 15 points, and the numbered cards are worth the number on the card (e.g. the Seven of Clubs is worth 7 points).The value of a Joker is equal to the card with which it is matched, e.g. a Joker matched with an Ace is worth 20 points, a Joker matched with a face card is worth 15 points, etc. If two Jokers are matched with each other, they are worth 50 points each.
Input
The input is made up of multiple test cases (#test cases<=30, if 1<=n<=10 or #test cases<=10 if 10<=n<=1000). Each test case contains three lines of input.
The first line in each test case is an integer n , 1 <= n <= 1,000, indicating how many cards are in each row.
The second line of the test case will contain n symbols representing the ranks of the cards in the first row. Each symbol will be chosen from the list {A, 2, 3, 4, 5, 6, 7, 8, 9, T, J, Q, K, R}. The symbols in the list represent the following ranks, respectively, {Ace, Two, Three, Four, Five, Six, Seven, Eight, Nine, Ten, Jack, Queen, King, Joker}. Similarly, the third line of the test case will contain the n symbols of the cards in the second row.
The input will end with a 0 on a line by itself.
Output
For each test case, output the value of the best Kabloom hand on a line by itself. Note that the cards that comprise the best Kabloom hand may not be unique for a test case.
Note: Every line of output should end in a newline character .
Sample Input
9
6 3 7 4 2 A K R T
3 5 4 7 R A Q K T
0
Sample Output
140
Hint
题意
给你2n个扑克牌，每行n张牌
你需要扔掉一些牌，使得上下两层牌一一对应。
如果两个A对应，那么可以得20分，如果是脸牌的话，那么就可以得15分，其他就是牌的分值。
王可以替代任意牌，如果是两张王牌对应的话，那么可以得50分。
问你最多可以得多少分，答案需要乘以2
题解
比较裸的dp，带权的最长公共子序列
代码
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e3+6;
int add(char a,char b){
if(a=='R'&&b=='R')return 50;
if(a=='R'){
if(b=='A')return 20;
if(b=='Q')return 15;
if(b=='K')return 15;
if(b=='J')return 15;
if(b=='T')return 10;
return b-'0';
}
if(b=='R'){
if(a=='A')return 20;
if(a=='Q')return 15;
if(a=='K')return 15;
if(a=='J')return 15;
if(a=='T')return 10;
return a-'0';
}
if(a=='A')return 20;
if(a=='Q')return 15;
if(a=='K')return 15;
if(a=='J')return 15;
if(a=='T')return 10;
return a-'0';
}
char a[maxn][5],b[maxn][5];
int n,dp[maxn][maxn];
int main()
{
while(scanf("%d",&n)!=EOF)
{
if(n==0)break;
memset(dp,0,sizeof(dp));
for(int i=1;i<=n;i++)
scanf("%s",a[i]);
for(int i=1;i<=n;i++)
scanf("%s",b[i]);
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
if(a[i][0]==b[j][0]||a[i][0]=='R'||b[j][0]=='R'){
dp[i][j]=max(dp[i][j],dp[i-1][j-1]+add(a[i][0],b[j][0]));
}
}
}
cout<<dp[n][n]*2<<endl;
}
}
