Codeforces 443A Borya and Hanabi(暴力)

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题目链接：Codeforces 443A Borya and Hanabi

题目大意：有若干个牌，每张牌有花色和数字两个值，现在问说至少询问多少次才能区分出所有的牌，每次询问可以确定一种花色牌的位置，或者是一种数字牌的位置。

解题思路：暴力枚举需要问的花色和数字，210,然后枚举两两判断是否可以被区分。

#include <cstdio>
#include <cstring>
#include <algorithm>using namespace std;
const int N = 105;int n, l[N], r[N];inline int cal (char ch) {if (ch == 'B')return 0;else if (ch == 'Y')return 1;else if (ch == 'W')return 2;else if (ch == 'G')return 3;else if (ch == 'R')return 4;return -1;
}int bit (int i) {return 1<<i;
}int bitCount (int x) {return x == 0 ? 0 : bitCount(x/2) + (x&1);
}bool judge (int s) {for (int i = 0; i < n; i++) {for (int j = i + 1; j < n; j++) {if (l[i] == l[j]) {if (r[i] != r[j] && (s&bit(r[i]+5)) == 0 && (s&bit(r[j]+5)) == 0)return false;} else {if ((s&bit(l[i])) || (s&bit(l[j])))continue;if (r[i] != r[j] && ( (s&bit(r[i]+5)) || (s&bit(r[j]+5)) ))continue;return false;}}}return true;
}int main () {char str[N];scanf("%d", &n);for (int i = 0; i < n; i++) {scanf("%s", str);l[i] = cal(str[0]);r[i] = str[1] - '1';}int ans = 10;for (int i = 0; i < (1<<10); i++) {int cnt = bitCount(i);if (cnt >= ans)continue;if (judge(i))ans = cnt;}printf("%d\n", ans);return 0;
}

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