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;
}

Codeforces 443A Borya and Hanabi(暴力),布布扣,bubuko.com

Codeforces 443A Borya and Hanabi(暴力)

标签：style   http   color   width   os   for   

原文地址：http://blog.csdn.net/keshuai19940722/article/details/35981919