hdu 4930 Fighting the Landlords(模拟)

题目连接：hdu 4930 Fighting the Landlords

题目大意：就是两个人玩斗地主，有8种牌型，单张，一对，三张，三带一，三带对，四带二，四炸，王炸。要求上家这一轮出牌下家管不上或者上家将牌走完则输出yes。

解题思路：总共就20张牌，枚举220种出牌方法，然后保留每种牌型的最大值，判断一下就可以了，注意细节。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxl = 20;

int N, A[10], B[10];

inline int bitcount (int x) {
    return x == 0 ? 0 : bitcount(x>>1) + (x&1);
}

inline int change (char ch) {
    if (ch >= '3' && ch <= '9')
        return ch - '0';
    switch (ch) {
        case 'T':
            return 10;
        case 'J':
            return 11;
        case 'Q':
            return 12;
        case 'K':
            return 13;
        case 'A':
            return 14;
        case '2':
            return 15;
        case 'X':
            return 16;
        case 'Y':
            return 17;
    }
    return -1;
}

int get (char* str, int n, int s, int& v) {
    int bit = bitcount(s);
    if (bit > 6)
        return 0;

    int c[maxl], type = 0;
    memset(c, 0, sizeof(c));

    for (int i = 0; i < n; i++) {
        if (s&(1<<i)) {
            int tmp = change(str[i]);
            ++c[tmp];

            if (c[tmp] > type || (c[tmp] == type && tmp > v)) {
                type = c[tmp];
                v = tmp;
            }
        }
    }

    // 王炸;
    if (c[17] && c[16] && bit == 2) {
        v = 17;
        return 7;
    }

    // ;
    if (type == 4 && bit == 4)
        return 7;

    // 4带2;
    if (type == 4 && bit == 6)
        return 6;

    // 3带1;
    if (type == 3 && bit == 4)
        return 4;

    // 3;
    if (type == 3 && bit == 3)
        return 3;

    // 对;
    if (type == 2 && bit == 2)
        return 2;

    // 单;
    if (type == 1 && bit == 1)
        return 1;

    // 3带2;
    if (type == 3 && bit == 5) {
        for (int i = 3; i <= 17; i++)
            if (c[i] == 2)
                return 5;
    }
    return 0;
}

void solve (char* s, int* t) {
    int n = strlen(s), v;

    for (int i = 1; i < (1<<n); i++) {
        v = 0;
        int k = get(s, n, i, v);
        t[k] = max(t[k], v);
    }
}

bool judge () {

    if (N <= 6 && A[N])
        return true;

    if (N == 4 && A[7])
        return true;

    if (A[7] && B[7])
        return A[7] > B[7];

    if (B[7])
        return false;

    for (int i = 1; i <= 7; i++) {
        if (A[i] && A[i] >= B[i])
            return true;
    }
    return false;
}

void init () {
    memset(A, 0, sizeof(A));
    memset(B, 0, sizeof(B));

    char a[maxl], b[maxl];
    scanf("%s%s", a, b);
    solve(a, A);
    solve(b, B);
    N = strlen(a);
}

int main () {
    int cas;
    scanf("%d", &cas);

    while (cas--) {
        init();
        printf("%s\n", judge() ? "Yes" : "No");
    }
    return 0;
}