Codeforces Round 927 (Div. 3) D. Card Game 题解 贪心

Card Game

题目描述

Two players are playing an online card game. The game is played using a 32-card deck. Each card has a suit and a rank. There are four suits: clubs, diamonds, hearts, and spades. We will encode them with characters ‘C’, ‘D’, ‘H’, and ‘S’, respectively. And there are 8 ranks, in increasing order: ‘2’, ‘3’, ‘4’, ‘5’, ‘6’, ‘7’, ‘8’, ‘9’.

Each card is denoted by two letters: its rank and its suit. For example, the 8 of Hearts is denoted as 8H.

At the beginning of the game, one suit is chosen as the trump suit.

In each round, players make moves like this: the first player places one of his cards on the table, and the second player must beat this card with one of their cards. After that, both cards are moved to the discard pile.

A card can beat another card if both cards have the same suit and the first card has a higher rank than the second. For example, 8S can beat 4S. Additionally, a trump card can beat any non-trump card, regardless of the rank of the cards, for example, if the trump suit is clubs (‘C’), then 3C can beat 9D. Note that trump cards can be beaten only by the trump cards of higher rank.

There were $n$ rounds played in the game, so the discard pile now contains $2n$ cards. You want to reconstruct the rounds played in the game, but the cards in the discard pile are shuffled. Find any possible sequence of $n$ rounds that might have been played in the game.

输入描述

The first line contains integer $t$ ($1\le t\le 100$) — the number of test cases. Then $t$ test cases follow.

The first line of a test case contains the integer number $n$ ($1\le n\le 16$).

The second line of a test case contains one character, the trump suit. It is one of “CDHS”.

The third line of a test case contains the description of $2n$ cards. Each card is described by a two-character string, the first character is the rank of the card, which is one of “23456789”, and the second one is the suit of the card, which is one of “CDHS”. All cards are different.

输出描述

For each test case print the answer to it:

• Print $n$ lines. In each line, print the description of two cards, in the same format as in the input: the first card that was played by the first player, and then the card that was used by the second player to beat it.
• If there is no solution, print a single line “IMPOSSIBLE”.

If there are multiple solutions, print any of them.

样例输入 #1

8
3
S
3C 9S 4C 6D 3S 7S
2
C
3S 5D 9S 6H
1
H
6C 5D
1
S
7S 3S
1
H
9S 9H
1
S
9S 9H
1
C
9D 8H
2
C
9C 9S 6H 8C

样例输出 #1

3C 4C
6D 9S
3S 7S
IMPOSSIBLE
IMPOSSIBLE
3S 7S
9S 9H
9H 9S
IMPOSSIBLE
6H 9C
9S 8C

原题

CF——传送门
洛谷——传送门

代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);

    int t;
    cin >> t;
    while (t--)
    {
        int n;
        char c;
        cin >> n >> c;
        vector<pair<string, string>> ans;
        vector<string> v(2 * n);
        for (int i = 0; i < 2 * n; i++)
            cin >> v[i];
        auto cmp = [&](string a, string b) -> bool
        {
            // 同为王牌花色时，点数小在前
            if (a[1] == c && b[1] == c) // 事实上该判断可与第四条判断合并，但这样更加清晰
                return a[0] < b[0];
            // 判断2和判断3是使王牌花色在排序后位于其他花色的后面
            if (a[1] == c)
                return 0;
            if (b[1] == c)
                return 1;
            // 相同花色，点数小在前
            if (a[1] == b[1])
                return a[0] < b[0];
            // 不同花色，按字典序(只要给定任意一种规则排序即可，这边为了方便采用字典序)
            if (a[1] != b[1])
                return a[1] < b[1];
            return 1;
        };
        sort(v.begin(), v.end(), cmp); // 按贪心顺序排序，即先消去王牌花色外的可消去牌对，再用王牌花色消去剩下的牌
        int small = 0;                 // 被击败的牌即小牌的索引
        int big = 1;                   // 击败小牌的大牌的索引
        vector<bool> del(2 * n, 0);
        // 操作一：先消去王牌花色外的可消去牌对
        while (big <= 2 * n - 1)
        {
            if (v[big][1] == c) // 遇到王牌花色则退出循环
                break;
            // 可消去
            if (v[small][1] == v[big][1])
            {
                del[small] = 1;
                del[big] = 1;
                ans.push_back({v[small], v[big]});
                small += 2;
                big += 2;
            }
            // 不可消去
            else
            {
                small++;
                big++;
            }
        }
        vector<string> v2; // 保存完成操作一后仍剩下的牌，供操作二消去
        for (int i = 0; i < 2 * n; i++)
        {
            if (del[i] == 0)
                v2.push_back(v[i]);
        }
        // 操作二：再用王牌花色消去剩下的牌
        for (int i = 0; i < v2.size() / 2; i++)
        {
            int j = v2.size() - 1 - i;
            if (v2[j][1] == c)
                ans.push_back({v2[i], v2[j]});
            else
                break;
        }
        if (ans.size() == n)
        {
            for (int i = 0; i < ans.size(); i++)
            {
                cout << ans[i].first << ' ' << ans[i].second << '\n';
            }
        }
        else
            cout << "IMPOSSIBLE\n";
    }

    return 0;
}