HDU 4930 Fighting the Landlords

题目链接

题意：就是题中那几种牌型。假设先手能一步走完。或者一步让后手无法管上，就赢

思路：先枚举出两个人全部可能的牌型的最大值。然后再去推断就可以

代码：

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

struct Player {

    int rank[15];

} p1, p2;

int t, hash[205], cnt[(1<<20)], vis[20];

char a[25], b[25];

int bitcount(int x) {

    int ans = 0;

    while (x) {

	ans += (x&1);

	x >>= 1;

    }

    return ans;

}

void add(int num, Player &p) {

    if (num == 2) {

	if (vis[14] && vis[15]) {

	    p.rank[7] = 14;

	    return;

	}

    }

    if (num == 5) {

	int a = 0, b = 0;

	for (int i = 15; i >= 1; i--) {

	    if (vis[i] == 3) a = i;

	    if (vis[i] == 2) b = i;

	}

	if (a && b) {

	    p.rank[5] = max(p.rank[5], a);

	}

	return;

    }

    for (int i = 15; i >= 1; i--) {

	if (num == 4 && vis[i] == 4) {

	    p.rank[7] = max(p.rank[7], i);

	    return;

	}

	if (num == 4 && vis[i] == 3) {

	    p.rank[4] = max(p.rank[4], i);

	    return;

	}

	if (num == 6 && vis[i] == 4) {

	    p.rank[6] = max(p.rank[6], i);

	    return;

	}

	if (vis[i] == num) {

	    p.rank[num] = max(p.rank[num], i);

	    return;

	}

    }

}

void build(char *a, Player &p) {

    memset(p.rank, 0, sizeof(p.rank));

    int n = strlen(a);

    int maxs = (1<<n);

    memset(vis, 0, sizeof(vis));

    for (int i = 0; i < maxs; i++) {

	if (cnt[i] > 6) continue;

	memset(vis, 0, sizeof(vis));

	for (int j = 0; j < n; j++) {

	    if (i&(1<<j)) {

		vis[hash[a[j]]]++;

	    }

	}

	add(cnt[i], p);

    }

}

bool solve() {

    int n = strlen(a);

    if (n == 4) {

	if (p1.rank[7]) return true;

    }

    if (n <= 6) {

	if (p1.rank[n]) return true;

    }

    if (p1.rank[7] && p2.rank[7]) return p1.rank[7] > p2.rank[7];

    if (p1.rank[7] && !p2.rank[7]) return true;

    if (!p1.rank[7] && p2.rank[7]) return false;

    for (int i = 1; i < 7; i++) {

	if (p1.rank[i] > p2.rank[i]) return true;

    }

    return false;

}

int main() {

    for (int i = 0; i < (1<<20); i++)

	cnt[i] = bitcount(i);

    for (int i = 3; i <= 9; i++)

	hash[i + '0'] = i - 2;

    hash['T'] = 8; hash['J'] = 9; hash['Q'] = 10; hash['K'] = 11;

    hash['A'] = 12; hash['2'] = 13; hash['X'] = 14; hash['Y'] = 15;

    scanf("%d", &t);

    while (t--) {

	scanf("%s%s", a, b);

	build(a, p1);

	build(b, p2);

	if (solve()) printf("Yes\n");

	else printf("No\n");

    }

    return 0;

}