标签：des style blog http color os io strong
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
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 }
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HDU 4903 （模拟+贪心）
标签：des style blog http color os io strong
原文地址：http://www.cnblogs.com/acvc/p/3897918.html
