Fighting the Landlords

Problem Description

Fighting the Landlords is a card game which has been a heat for years in China. The game goes with the 54 poker cards for 3 players, where the “Landlord” has 20 cards and the other two (the “Farmers”) have 17. The Landlord wins if he/she has no cards left, and the farmer team wins if either of the Farmer have no cards left. The game uses the concept of hands, and some fundamental rules are used to compare the cards. For convenience, here we only consider the following categories of cards:

1.Solo: a single card. The priority is: Y (i.e. colored Joker) > X (i.e. Black & White Joker) > 2 > A (Ace) > K (King) > Q (Queen) > J (Jack) > T (10) > 9 > 8 > 7 > 6 > 5 > 4 > 3. It’s the basic rank of cards.

2.Pair : two matching cards of equal rank (e.g. 3-3, 4-4, 2-2 etc.). Note that the two Jokers cannot form a Pair (it’s another category of cards). The comparison is based on the rank of Solo, where 2-2 is the highest, A-A comes second, and 3-3 is the lowest.

3.Trio: three cards of the same rank (e.g. 3-3-3, J-J-J etc.). The priority is similar to the two categories above: 2-2-2 > A-A-A > K-K-K > . . . > 3-3-3.

4.Trio-Solo: three cards of the same rank with a Solo as the kicker. Note that the Solo and the Trio should be different rank of cards (e.g. 3-3-3-A, 4-4-4-X etc.). Here, the Kicker’s rank is irrelevant to the comparison, and the Trio’s rank determines the priority. For example, 4-4-4-3 > 3-3-3-2.

5.Trio-Pair : three cards of the same rank with a Pair as the kicker (e.g. 3-3- 3-2-2, J-J-J-Q-Q etc.). The comparison is as the same as Trio-Solo, where the Trio is the only factor to be considered. For example,4-4-4-5-5 > 3-3-3-2-2. Note again, that two jokers cannot form a Pair.

6.Four-Dual: four cards of the same rank with two cards as the kicker. Here, it’s allowed for the two kickers to share the same rank. The four same cards dominates the comparison: 5-5-5-5-3-4 > 4-4-4-4-2-2.

In the categories above, a player can only beat the prior hand using of the same category but not the others. For example, only a prior Solo can beat a Solo while a Pair cannot. But there’re exceptions:

7.Nuke: X-Y (JOKER-joker). It can beat everything in the game.

8.Bomb: 4 cards of the same rank. It can beat any other category except Nuke or another Bomb with a higher rank. The rank of Bombs follows the rank of individual cards: 2-2-2-2 is the highest and 3-3-3-3 is the lowest.

Given the cards of both yours and the next player’s, please judge whether you have a way to play a hand of cards that the next player cannot beat you in this round.If you no longer have cards after playing, we consider that he cannot beat you either. You may see the sample for more details.

Input

The input contains several test cases. The number of test cases T (T<=20) occurs in the first line of input.

Each test case consists of two lines. Both of them contain a string indicating your cards and the next player’s, respectively. The length of each string doesn’t exceed 17, and each single card will occur at most 4 times totally on two players’ hands except that the two Jokers each occurs only once.

Output

For each test case, output Yes if you can reach your goal, otherwise output No.

Sample Input

4 33A 2 33A 22 33 22 5559T 9993

Sample Output

Yes No Yes Yes

题意：斗地主游戏规则，但是只有两个人玩这个游戏，当第一个人出一次牌第二个人无法出就算赢。

sL :很水的模拟了。直接存下单张的。对的。3个的。4个的。 开4个数组就可以了。

1 // by caonima
  2 // hehe
  3 #include <cstdio>
  4 #include <cstring>
  5 #include <algorithm>
  6 #include <vector>
  7 #include <queue>
  8 #include <set>
  9 using namespace std;
 10 const int MAX = 100;
 11 vector<int> tot1[5],tot2[5];
 12 char str1[MAX];
 13 char str2[MAX];
 14 int a[MAX],b[MAX];
 15 int cnt1[MAX],cnt2[MAX];
 16 int cmp(int a,int b) {
 17     return a>b;
 18 }
 19 
 20 int check(int n1,int n2) {
 21     if(n1==1) return true;
 22     if(n1==2) {
 23         if(tot1[2].size()==1) return true;
 24     }
 25    // printf("%d %d\n",cnt1[56],cnt1[55]);
 26     if(cnt1[56]==1&&cnt1[55]==1) return true;
 27     if(n1==3) {
 28         if(tot1[3].size()==1) return true;
 29     }
 30     if(n1==4) {
 31         if(tot1[3].size()==1&&tot1[1].size()==2) return true;
 32     }
 33     if(n1==5) {
 34         if((tot1[3].size()==1&&tot1[2].size()==2)) return true;
 35     }
 36     if(n1==6) {
 37         if((tot1[4].size()==1&&tot1[2].size()==2)||(tot1[4].size()==1&&tot1[1].size()==3))
 38         return true;
 39     }
 40     if(cnt2[55]==1&&cnt2[56]==1) return false;
 41 
 42     if(tot1[1].size()!=0) {
 43         int t=tot1[1].size();
 44         int x=tot1[1][t-1];
 45         if(tot2[4].size()==0) {
 46             if(tot2[1].size()==0) return true;
 47             else if(tot2[1].size()!=0){
 48                 int m=tot2[1].size();
 49                 int y=tot2[1][m-1];
 50                 if(x>y) return true;
 51             }
 52         }
 53     }
 54     if(tot1[2].size()!=0) {
 55         int t=tot1[2].size();
 56         int x=tot1[2][t-1];
 57         if(tot2[4].size()==0) {
 58             if(tot2[2].size()==0) return true;
 59             else if(tot2[2].size()!=0){
 60                 int m=tot2[2].size();
 61                 int y=tot2[2][m-1];
 62                 if(x>y) return true;
 63             }
 64         }
 65     }
 66     if(tot1[3].size()!=0) {
 67         int t=tot1[3].size();
 68         int x=tot1[3][t-1];
 69         if(tot2[4].size()==0) {
 70             if(tot2[3].size()==0) return true;
 71             else if (tot2[3].size()!=0){
 72                 int m=tot2[3].size();
 73                 int y=tot2[3][m-1];
 74                 if(x>y) return true;
 75             }
 76         }
 77     }
 78     if(tot1[4].size()!=0) {
 79         int t=tot1[4].size();
 80         int x=tot1[4][t-1];
 81         if(tot2[4].size()==0) return true;
 82         else {
 83              int m=tot2[4].size();
 84              int y=tot2[4][m-1];
 85              if(x>y) return true;
 86         }
 87     }
 88 
 89     return false;
 90 }
 91 void init() {
 92     for(int i=0;i<=4;i++) {
 93         tot1[i].clear();
 94         tot2[i].clear();
 95     }
 96 }
 97 int main() {
 98     int cas;
 99     scanf("%d",&cas);
100     while(cas--) {
101         init();
102         scanf("%s %s",str1,str2);
103         memset(cnt1,0,sizeof(cnt1));
104         memset(cnt2,0,sizeof(cnt2));
105         int n1=strlen(str1);
106         int n2=strlen(str2);
107         for(int i=0;i<n1;i++) {
108             if(str1[i]==‘Y‘) a[i]=56;
109             else if(str1[i]==‘X‘) a[i]=55;
110             else if(str1[i]==‘2‘) a[i]=54;
111             else if(str1[i]==‘A‘) a[i]=53;
112             else if(str1[i]==‘K‘) a[i]=52;
113             else if(str1[i]==‘Q‘) a[i]=51;
114             else if(str1[i]==‘J‘) a[i]=50;
115             else if(str1[i]==‘T‘) a[i]=49;
116             else a[i]=str1[i]-‘0‘;
117         }
118         for(int i=0;i<n2;i++) {
119             if(str2[i]==‘Y‘) b[i]=56;
120             else if(str2[i]==‘X‘) b[i]=55;
121             else if(str2[i]==‘2‘) b[i]=54;
122             else if(str2[i]==‘A‘) b[i]=53;
123             else if(str2[i]==‘K‘) b[i]=52;
124             else if(str2[i]==‘Q‘) b[i]=51;
125             else if(str2[i]==‘J‘) b[i]=50;
126             else if(str2[i]==‘T‘) b[i]=49;
127             else b[i]=str2[i]-‘0‘;
128         }
129         sort(a,a+n1,cmp);
130         sort(b,b+n2,cmp);
131         for(int i=0;i<n1;i++) {
132             cnt1[a[i]]++;
133         }
134         for(int i=0;i<n2;i++) {
135             cnt2[b[i]]++;
136         }
137         for(int i=0;i<=57;i++) {
138             if(cnt1[i]) {
139                 int num=cnt1[i];
140                 while(num) {
141                     tot1[num].push_back(i);
142                     num--;
143                 }
144             }
145         }
146         for(int i=0;i<=57;i++) {
147             if(cnt2[i]) {
148                 int num=cnt2[i];
149                 while(num) {
150                     tot2[num].push_back(i);
151                     num--;
152                 }
153             }
154         }
155 
156         for(int i=0;i<=4;i++) {
157             sort(tot1[i].begin(),tot1[i].end());
158             sort(tot2[i].begin(),tot2[i].end());
159         }
160         int ans=check(n1,n2);
161         if(ans) printf("Yes\n");
162         else printf("No\n");
163 
164     }
165     return 0;

166 }

HDU 4903 （模拟+贪心）