hdu 4930 Fighting the Landlords--2014 Multi-University Training Contest 6

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4930

Fighting the Landlords

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 312    Accepted Submission(s): 100

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

 

Source

2014 Multi-University Training Contest 6

 

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Statistic | Submit | Discuss | Note

这题事实上就是斗地主，一道比較麻烦的模拟题，但假设分析好了事实上也不难。

我分情况进行讨论。

首先推断牌能否出完，然后推断炸弹，接着是1根，2根，3根，3带1,3带对，4带2,。。。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<string>
#include<vector>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<map>
using namespace std;
#define CLR(A) memset(A,0,sizeof(A))
const int MAX=20;
char stra[MAX],strb[MAX];
int va[MAX],vb[MAX];
int na,nb;
void change(char s[],int v[]){
    for(int i=0;i<strlen(s);i++){
        if(s[i]>='3'&&s[i]<='9') v[i]=s[i]-'0';
        if(s[i]=='T') v[i]=10;
        if(s[i]=='J') v[i]=11;
        if(s[i]=='Q') v[i]=12;
        if(s[i]=='K') v[i]=13;
        if(s[i]=='A') v[i]=14;
        if(s[i]=='2') v[i]=15;
        if(s[i]=='X') v[i]=16;
        if(s[i]=='Y') v[i]=17;
    }
}
int Bomb(int v[],int len){
    int cnt=1,i;
    for(i=len-1;i>=0;i--){
        if(v[i]<16) break;
    }
    if(i<3) return 0;
    i--;
    for(;i>=0;i--){
        if(v[i]==v[i+1])
            cnt++;
        else
            cnt=1;
        if(cnt==4) return v[i];
    }
    return 0;
}

int Nuke(int v[],int len){
    if(len<2) return 0;
    if(v[len-1]==17&&v[len-2]==16) return 1;
    return 0;
}

int Pair(int v[],int len){
    if(len<2) return 0;
    for(int i=len-2;i>=0;i--){
        if(v[i]==v[i+1]) return v[i];
    }
    return 0;
}

int Trio(int v[],int len){
    if(len<3) return 0;
    int cnt=1;
    for(int i=len-2;i>=0;i--){
        if(v[i]==v[i+1]) cnt++;
        else cnt=1;
        if(cnt==3) return v[i];
    }
    return 0;
}

int Trio_Solo(int v[],int len){
    if(len<4) return 0;
    int x=Trio(v,len);
    if(x && x!=v[0] && x!=v[len-1])
    return Trio(v,len);
}

int Trio_Pair(int v[],int len){
    if(len<5) return 0;
    int x=Trio(v,len);
    int pos;
    if(x!=0){
        pos=lower_bound(v,v+len,x)-v;
        int t1=Pair(v,pos);
        if(t1!=0) return x;
        pos++;//防止溢出
        for(pos;pos<len;pos++){
            if(v[pos]!=v[pos-1]) break;
        }
        int t2=Pair(v+pos,len-pos);
        if(t2!=0) return x;
    }
    return 0;
}

int Four_Dual(int v[],int len){
    if(len<6) return 0;
    return Bomb(v,len);
}

bool Drop_All(int v[],int len){
    if(len>7) return false;
    if(len==6&&Four_Dual(v,len)!=0) return 1;
    if(len==5&&Trio_Pair(v,len)!=0) return 1;
    if(len==4&&(Trio_Solo(v,len)!=0||Bomb(v,len)!=0)) return 1;
    if(len==3&&Trio(v,len)!=0) return 1;
    if(len==2&&Pair(v,len)!=0&&Nuke(v,len)) return 1;
    if(len==1) return 1;
    return 0;
}

bool check(int v1[],int len1,int v2[],int len2){
    int x,y;
    if(Drop_All(v1,len1)) return true;
    if(Nuke(v1,len1)!=0) return true;
    if(Nuke(v2,len2)!=0) return false;
    x=Bomb(v1,len1);y=Bomb(v2,len2);
    if(x && y==0) return true;
    if(x==0 && y) return false;
    if(x && x>=y) return true;
    if(y>x)    return false;
    if(v1[len1-1]>=v2[len2-1]) return true;
    x=Pair(v1,len1);y=Pair(v2,len2);
    if(x && x>=y) return true;
    x=Trio(v1,len1);y=Trio(v2,len2);
    if(x && x>=y) return true;
    x=Trio_Solo(v1,len1);y=Trio_Solo(v2,len2);
    if(x && x>=y) return true;
    x=Trio_Pair(v1,len1);y=Trio_Pair(v2,len2);
    if(x && x>=y) return true;
    x=Four_Dual(v1,len1);y=Four_Dual(v2,len2);
    if(x && x>=y) return true;
    return false;
}

int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%s%s",stra,strb);
        na=strlen(stra);nb=strlen(strb);
        change(stra,va);change(strb,vb);
        sort(va,va+na);sort(vb,vb+nb);
        if(check(va,na,vb,nb)) cout<<"Yes"<<endl;
        else cout<<"No"<<endl;
    }
    return 0;
}