hdu 4930 Fighting the Landlords(類比)

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題目串連：hdu 4930 Fighting the Landlords

題目大意：就是兩個人玩鬥地主，有8種牌型，單張，一對，三張，三帶一，三帶對，四帶二，四炸，王炸。要求上家這一輪出牌下家管不上或者上家將牌走完則輸出yes。

解題思路：總共就20張牌，枚舉220種出牌方法，然後保留每種牌型的最大值，判斷一下就可以了，注意細節。

#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;    }    // 炸彈;    if (type == 4 && bit == 4)        return 7;    // 4帶2;    if (type == 4 && bit == 6)        return 6;    // 3帶1;    if (type == 3 && bit == 4)        return 4;    // 3;    if (type == 3 && bit == 3)        return 3;    // 對;    if (type == 2 && bit == 2)        return 2;    // 單;    if (type == 1 && bit == 1)        return 1;    // 3帶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;}