hdu4930 Fighting the Landlords（模拟 多校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): 546 Accepted Submission(s): 188

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, theKicker’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 youin 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, andeach 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

433A233A2233225559T9993

Sample Output

YesNoYesYes

Author

BUPT

Source

2014 Multi-University Training Contest 6 

题目意思：
两个人设为A和B，A和B在打斗地主，上面一行是A手里的牌，下面一行是B手里的牌，若A第一次出牌B压不住或者A一次就把牌出完了，那么输出Yes，否则若A牌没出完而且被B压住了那么输出No。

代码如下：

#include <cstdio>#include <iostream>#include <algorithm>#include <cstring>using namespace std;#define M 26char s1[M], s2[M];int c1[M], c2[M];int a1[M], a2[M];int b1[M], b2[M];int len1, len2;int max(int a, int b){    return a > b ? a:b;}void init(){    memset(a1,0,sizeof(a1));    memset(b1,0,sizeof(b1));    memset(a2,0,sizeof(a1));    memset(b2,0,sizeof(b2));    memset(c1,0,sizeof(c1));    memset(c2,0,sizeof(c2));}int f(char c){    if(c >= '3' && c <= '9')        return c -'2';    if(c == 'T')        return 8;    if(c == 'J')        return 9;    if(c == 'Q')        return 10;    if(c == 'K')        return 11;    if(c == 'A')        return 12;    if(c == '2')        return 13;    if(c == 'X')        return 14;    if(c == 'Y')        return 15;}void Find(int *c, int flag){    int i;    int a[6], b[6];    memset(a,0,sizeof(a));    memset(b,0,sizeof(b));    for(i = 1; i <= 13; i++)    {        if(c[i])        {            b[1] = max(b[1],i);            if(c[i] == 1)//单张            {                a[1]++;            }            else if(c[i] == 2)//对子            {                a[2]++;                b[2] = max(b[2],i);            }            else if(c[i] == 3)//三个的            {                a[3]++;                b[3] = max(b[3],i);            }            else if(c[i] == 4)//炸弹            {                a[4]++;                b[4] = max(b[4],i);            }        }    }    if(c[14] && c[15])//双王    {        b[1] = 15;        a[5]++;    }    else if(c[15])//大王    {        b[1] = 15;    }    else if(c[14])//小王    {        b[1] = 14;    }        for(i = 1; i <= 5; i++)    {        if(flag == 1)        {            a1[i] = a[i];            b1[i] = b[i];        }        else if(flag == 2)        {            a2[i] = a[i];            b2[i] = b[i];        }    }}void solve(){    int i, j;    int flag1, flag2;    int n1=strlen(s1);    int n2=strlen(s2);    if((n1==1)||(n1==2&&a1[2])||(n1==3&&a1[3])||(n1==4&&(a1[4]||a1[3]))||(n1==5&&a1[2]&&a1[3])||(n1==6&&a1[2]&&a1[4]))    {        printf("Yes\n");//一次出完     }    else if(a1[5])    {        printf("Yes\n"); //A手里有王炸     }    else if(a2[5])    {        printf("No\n");   //B手里有王炸     }    else if(b1[4]>b2[4])    {        printf("Yes\n");   //A手里的炸弹比B手里的炸弹大     }    else if(b1[4]<b2[4])    //反之     {          printf("No\n");    }    else if(b1[1]>b2[1])    //出单，且单比B的大    {           printf("Yes\n");    }    else if(b1[2]>b2[2])    //出对，且对比B的大    {           printf("Yes\n");    }    else if(b1[3]&&(b1[3]>b2[3]||(b1[1]&&!b2[1])||(b1[2]&&!b2[2])))    {  //出3张时，可以带1张也可以带2张也可以不带，依次判断         printf("Yes\n");    }    else     {                 printf("No\n");    }}int main(){    int t;    int i, j;    scanf("%d",&t);    while(t--)    {        init();        scanf("%s",s1);        scanf("%s",s2);        len1 = strlen(s1);        len2 = strlen(s2);        for(i = 0; i < len1; i++)        {            c1[f(s1[i])]++;        }        for(i = 0; i < len2; i++)        {            c2[f(s2[i])]++;        }        Find(c1, 1);        Find(c2, 2);        solve();    }    return 0;}