HDU 4930 Fighting the Landlords --多Trick，较复杂模拟

题意：两个人A和B在打牌，只有题目给出的几种牌能出若A第一次出牌B压不住或者A一次就把牌出完了，那么A赢，输出Yes，否则若A牌没出完而且被B压住了，那么A输，输出No。

解法：知道规则，看清题目，搞清有哪些Trick，就可以直接模拟搞了。详见代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
#define N 100102
#define M 22

char s1[24],s2[24];
int A[24],B[24];
int cnt1[20],cnt2[20];

int main()
{
    int n,i,j;
    int n1,n2;
    scanf("%d",&n);
    while(n--)
    {
        priority_queue<int> Single,Pair,Tri,Four,Nuke;
        priority_queue<int> Single2,Pair2,Tri2,Four2,Nuke2;
        scanf("%s",s1);
        scanf("%s",s2);
        n1 = strlen(s1);
        n2 = strlen(s2);
        for(i=0;i<n1;i++)
        {
            if(s1[i] >= '3' && s1[i] <= '9')
                A[i] = s1[i]-'0';
            else if(s1[i] == 'T')
                A[i] = 10;
            else if(s1[i] == 'J')
                A[i] = 11;
            else if(s1[i] == 'Q')
                A[i] = 12;
            else if(s1[i] == 'K')
                A[i] = 13;
            else if(s1[i] == 'A')
                A[i] = 14;
            else if(s1[i] == '2')
                A[i] = 15;
            else if(s1[i] == 'X')
                A[i] = 16;
            else if(s1[i] == 'Y')
                A[i] = 17;
        }
        for(i=0;i<n2;i++)
        {
            if(s2[i] >= '3' && s2[i] <= '9')
                B[i] = s2[i]-'0';
            else if(s2[i] == 'T')
                B[i] = 10;
            else if(s2[i] == 'J')
                B[i] = 11;
            else if(s2[i] == 'Q')
                B[i] = 12;
            else if(s2[i] == 'K')
                B[i] = 13;
            else if(s2[i] == 'A')
                B[i] = 14;
            else if(s2[i] == '2')
                B[i] = 15;
            else if(s2[i] == 'X')
                B[i] = 16;
            else if(s2[i] == 'Y')
                B[i] = 17;
        }
        sort(A,A+n1);
        sort(B,B+n2);
        memset(cnt1,0,sizeof(cnt1));
        memset(cnt2,0,sizeof(cnt2));
        for(i=0;i<n1;i++)   //计算A各种牌的个数
            cnt1[A[i]]++;
        for(i=0;i<n2;i++)   //计算B各种牌的个数
            cnt2[B[i]]++;
        for(i=3;i<=15;i++)
        {
            if(cnt1[i] == 4)         //出现四个，可以做四个出，可以做三个出，也可以做两个或一个出
                Four.push(i),Tri.push(i),Pair.push(i),Single.push(i);
            else if(cnt1[i] == 3)    
                Tri.push(i),Pair.push(i),Single.push(i);
            else if(cnt1[i] == 2)
                Pair.push(i),Single.push(i);
            else if(cnt1[i] == 1)
                Single.push(i);
        }
        for(i=3;i<=15;i++)
        {
            if(cnt2[i] == 4)
                Four2.push(i),Tri2.push(i),Pair2.push(i),Single2.push(i);
            else if(cnt2[i] == 3)
                Tri2.push(i),Pair2.push(i),Single2.push(i);
            else if(cnt2[i] == 2)
                Pair2.push(i),Single2.push(i);
            else if(cnt2[i] == 1)
                Single2.push(i);
        }
        if(cnt1[16])    //有王，可以做单个出
            Nuke.push(16),Single.push(16);
        if(cnt1[17])
            Nuke.push(17),Single.push(17);
        if(cnt2[16])
            Nuke2.push(16),Single2.push(16);
        if(cnt2[17])
            Nuke2.push(17),Single2.push(17);
        if(Nuke.size() >= 2)    //双王，直接赢
        {
            puts("Yes");
            continue;
        }
        //-------------------------------下面判断能否一次出完
        if(n1 == 1)
        {
            puts("Yes");
            continue;
        }
        if(n1 == 2)
        {
            if(A[0] == A[1])
            {
                puts("Yes");
                continue;
            }
        }
        if(n1 == 3)
        {
            if(A[0] == A[1] && A[1] == A[2])
            {
                puts("Yes");
                continue;
            }
        }
        if(n1 == 4)
        {
            if(A[0] == A[1] && A[1] == A[2] && A[2] == A[3])
            {
                puts("Yes");
                continue;
            }
            if(A[0] != A[1] && A[1] == A[2] && A[2] == A[3])
            {
                puts("Yes");
                continue;
            }
            if(A[2] != A[3] && A[0] == A[1] && A[1] == A[2])
            {
                puts("Yes");
                continue;
            }
        }
        if(n1 == 5)
        {
            if(A[0] == A[1] && A[1] != A[2] && A[2] == A[3] && A[3] == A[4])
            {
                puts("Yes");
                continue;
            }
            if(A[3] == A[4] && A[2] != A[3] && A[0] == A[1] && A[1] == A[2])
            {
                puts("Yes");
                continue;
            }
        }
        if(n1 == 6)
        {
            int tag = 0;
            for(i=0;i<=2;i++)
            {
                if(A[i] == A[i+1] && A[i+1] == A[i+2] && A[i+2] == A[i+3])
                {
                    tag = 1;
                    break;
                }
            }
            if(tag)
            {
                puts("Yes");
                continue;
            }
        }
        //-----------------------------如果不能一次出完
        if(Nuke2.size() >= 2)                //对方有双王，必输
        {
            puts("No");
            continue;
        }
        if(!Nuke.empty() && Nuke2.empty())   //A有王，B没王
        {
            puts("Yes");
            continue;
        }
        if(!Nuke.empty() && !Nuke2.empty())  //都有王，看谁的大，如果A小，则不选择出王，继续
        {
            if(Nuke.top() > Nuke2.top())
            {
                puts("Yes");
                continue;
            }
        }
        if(Four.empty() && !Four2.empty())  //炸弹，如果不能一次出完又没炸弹，那么必会被炸，输
        {
            puts("No");
            continue;
        }
        if(!Four.empty() && Four2.empty())  //有炸弹出炸弹
        {
            puts("Yes");
            continue;
        }
        if(!Four.empty() && !Four2.empty())  //都有炸弹，A的如果小，因为不能一次出完，必输
        {
            if(Four.top() >= Four2.top())
            {
                puts("Yes");
                continue;
            }
            else
            {
                puts("No");
                continue;
            }
        }
        if(!Tri.empty() && Tri2.empty())    //三个的情况
        {
            puts("Yes");
            continue;
        }
        if(!Tri.empty() && !Tri2.empty())
        {
            if(Tri.top() >= Tri2.top())
            {
                puts("Yes");
                continue;
            }
            else if(n1 >= 4 && n2 <= 3)   //A有的带，B没得带
            {
                puts("Yes");
                continue;
            }
        }
        if(!Pair.empty() && Pair2.empty())     //对子
        {
            puts("Yes");
            continue;
        }
        if(!Pair.empty() && !Pair2.empty())
        {
            if(Pair.top() >= Pair2.top())
            {
                puts("Yes");
                continue;
            }
        }
        if(Single.empty() && !Single2.empty())   //单个牌
        {
            puts("No");
            continue;
        }
        if(!Single.empty() && Single2.empty())
        {
            puts("Yes");
            continue;
        }
        if(!Single.empty() && !Single2.empty())
        {
            if(Single.top() >= Single2.top())
            {
                puts("Yes");
                continue;
            }
        }
        puts("No");                                //如果以上都不满足，那么A输了，gg。
    }
    return 0;
}