【ZOJ3921 2016年浙大2月月赛G】【爆搜+智商贪心剪枝 好题】Guan Dan 2副牌取27张最少出光牌次数
Guan Dan
Time Limit: 2 Seconds
Memory Limit: 65536 KB
Bob is playing a poker game called GuanDan. The rules are similar to PaoDeKuai and DouDiZhu. Basically, the game is played on two decks of poker with four jokers in total. And there are some specific patterns of hands that you can play. They will be listed below.
Notation
- A card contains two characters
- The first character is its rank, the second character is its suit
- For an ordinary card, rank can be A, 2, 3, 4, 5, 6, 7, 8, 9, 0, J, Q, K (A for Ace, 0 for 10)
- J, Q, K can be taken as 11, 12, 13 respectively. Ace can be taken as either 1 or 14.
- Suit can be D, S, H, C (Diamond, Spade, Heart, Club)
- So there are 13 * 4 = 52 different ordinary cards
- Special cards: BJ, RJ (BJ for Black Joker, RJ for Red Joker)
- So all cards are [A-K][DSHC]*2 + BJ*2 + RJ*2, in total 108 cards.
Hands you can play
| Hand | Description | Example(s) | Counter Example(s) | Difficulty |
|---|---|---|---|---|
| Solo | Any single card | AS (Spade A) | None | 1 |
| Pair | Two cards of the same rank | 0D 0C (Diamond 10 and Club 10) | BJ RJ (Jokers of different colors doesn't count as the same rank) | 1 |
| Three Pairs | Three pairs with continuous rank | AH AS 2S 2H 3D 3C, QH QH KH KH AH AH | KH KH AH AH 2H 2H | 1 |
| Trio | Three cards of the same rank | AH AH AS, 9H 9S 9C | BJ BJ RJ | 1 |
| Full House | A composite of a Trio and a Pair | AH AH AS 2S 2H | None | 1 |
| Airplane | Two Trios with continuous rank | AH AH AD 2H 2H 2D, KH KH KD AH AH AD | None | 1 |
| Straight | Five cards with continuous rank | AH 2S 3H 4C 5H, 0C JD QH KH AS | JD QH KH AS 2S | 1 |
| Straight Flush | Five cards with continuous rank and the same suit | AH 2H 3H 4H 5H, 0H JH QH KH AH | QH KH AH 2H 3H | 0 |
| Bomb | Four or more cards of the same rank | 5H 5H 5S 5S, 8H 8H 8S 8S 8C 8C 8D 8D | None | 0 |
Clarification
- Jokers' rank is not continuous with ordinary cards
- RJs have the same rank, BJs have the same rank, but an RJ and a BJ do not have the same rank.
- Ranks being continuous means they form a sequence of natural numbers such that each number is 1 plus the number before it
Special rules
- There is a main rank for each game. It can only be one of A, 2, 3, ..., K.
- The two cards of the main rank with a Heart suit are wildcards.
- A wildcard can be used as any card except the jokers.
Goal
Your goal is to play your cards as fast as possible and win the game by first playing all of your cards. Therefore, Bob would like to know the least sum of difficulty value if he combine his cards into valid hands optimally.
Input
The first line is an integer of the number of test cases.
For each test case, the first line contains a character denoting the main rank (rank is in {'A', '2', '3', '4', '5', '6', '7', '8', '9', '0', 'J', 'Q', 'K'}). The second line contains 54 characters, two characters is a card. Format is the same as descripted in Notation section. Test cases are randomly "drawn" from random shuffled "decks".
Output
For each test case, output the minimum difficulty if you arrange the cards optimally into hands.
Sample Input
2
3
AD3D8H6HKCKS4DJC3H3C9H2C5H9SJD3CKH7CKCAH5H2C4C3SBJAC5S
6
KC0D4HKD6DJCKC2S8HADJD4D9C7SJH3SBJRJ2DQH5H8S9D5S0S2D5CSample Output
6
9Hint
Here's one possible solution for sample 1 and sample 2.
Sample 1: 3H is a wildcard.
3H -> 3C
| AC AD AH 5H 5S || 3C 3D 3S JC JD || 9S || **KC KC KH KS** || BJ || 2C 2C 3C 3C 4C 4D || 5H 6H 7C 8H 9H |KKKK is a bomb, it has difficulty 0. Others each have difficulty 1. Sum is 6.
Sample 2:
| 2D 2S || 4H || 5H 5S || JD JH KC KC KD || BJ || RJ || 8S 9D 0S JC QH || 6D 7S 8H 9C 0D || AD 2D 3S 4D 5C |No bomb. Sum is 9.
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }
const int N = 0, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;
int casenum, casei;
char V; int O;
pair<int, int>w[30];
int ans;
int a[15][4];int cnt[17],cntt[17];
int K(char ch)
{
if (ch == 'A')return 1;
else if (ch == '0')return 10;
else if (ch == 'J')return 11;
else if (ch == 'Q')return 12;
else if (ch == 'K')return 13;
else return ch - 48;
}
void DataIsWeak()
{
if (a[V][2] > 1)while (1);//最多只有1个万能牌
}
void check(int val)
{
int d[4]; MS(d, 0);
for (int i = 1; i <= 15; ++i)if (cnt[i] < 4)++d[cnt[i]];
int FullHouse = min(d[3], d[2]);
val += FullHouse;
d[3] -= FullHouse;
d[2] -= FullHouse;
val += d[3] + d[2] + d[1];
gmin(ans, val);
}
//dfs只搜特殊搭配 顺子，2+2+2，3+3这些
void dfs(int p,int val)
{
if (val >= ans)return;
if (p > 27){check(val);return;}
int v = w[p].first;
if ( v > 13 || cnt[v] == 0) { dfs(p + 1, val); return; }
//尝试1，12345；有同花顺肯定考虑同花顺；没有同花顺可以普通的顺子
int nxt = v + 4; if (nxt == 14)nxt = 1;
if (v <= 10 && cnt[v] && cnt[v + 1] && cnt[v + 2] && cnt[v + 3] && cnt[nxt])
{
--cnt[v]; --cnt[v + 1]; --cnt[v + 2]; --cnt[v + 3]; --cnt[nxt];
int c = w[p].second;
if (a[v][c] && a[v + 1][c] && a[v + 2][c] && a[v + 3][c] && a[nxt][c])
{
--a[v][c]; --a[v + 1][c]; --a[v + 2][c]; --a[v + 3][c]; --a[nxt][c];
dfs(p + 1,val);
++a[v][c]; ++a[v + 1][c]; ++a[v + 2][c]; ++a[v + 3][c]; ++a[nxt][c];
}
else dfs(p + 1, val + 1);
++cnt[v]; ++cnt[v + 1]; ++cnt[v + 2]; ++cnt[v + 3]; ++cnt[nxt];
}
//尝试2，2+2+2
nxt = v + 2;
if (nxt == 14)nxt = 1;
if (v <= 12 && cnt[v] >= 2 && cnt[v + 1] >= 2 && cnt[nxt] >= 2)
{
cnt[v] -= 2; cnt[v + 1] -= 2; cnt[nxt] -= 2;
dfs(p + 1, val + 1);
cnt[v] += 2; cnt[v + 1] += 2; cnt[nxt] += 2;
}
//尝试3，3+3
nxt = v + 1;
if (nxt == 14)nxt = 1;
if (cnt[v] >= 3 && cnt[nxt] >= 3)
{
cnt[v] -= 3;cnt[nxt] -= 3;
dfs(p + 1, val + 1);
cnt[v] += 3;cnt[nxt] += 3;
}
dfs(p + 1, val);
}
void solve()
{
ans = 27;
if (a[V][2] == 1)//有特殊牌，数据中特殊牌最多只会有1张
{
a[V][2] = 0; --cnt[V];
for (int v = 1; v <= 13; ++v)
{
for (int c = 0; c < 4; ++c)
{
++a[v][c]; ++cnt[v];
w[O] = MP(v,c);
dfs(1, 0);
--a[v][c]; --cnt[v];
}
}
++cnt[14]; w[O] = MP(14, 0); dfs(1, 0); --cnt[14];
++cnt[15]; w[O] = MP(15, 0); dfs(1, 0); --cnt[15];
a[V][2] = 1; ++cnt[V];
}
else dfs(1, 0);
printf("%d\n", ans);
}
int main()
{
scanf("%d", &casenum);
for (int casei = 1; casei <= casenum; ++casei)
{
MS(a, 0); MS(cnt, 0); O = 0;
scanf(" %c", &V); V = K(V);
for (int i = 1; i <= 27; ++i)
{
char v, c; scanf(" %c %c", &v, &c);
if (v == 'B'&&c == 'J') { w[i] = MP(14, 0); ++cnt[14]; continue; }
if (v == 'R'&&c == 'J') { w[i] = MP(15, 0); ++cnt[15]; continue; }
v = K(v);
if (c == 'D')c = 0;else if (c == 'S')c = 1;else if (c == 'H')c = 2;else c = 3;
w[i] = MP(v, c);
++a[v][c]; ++cnt[v];
if (v == V&&c == 2)O = i;
}
DataIsWeak();
solve();
}
return 0;
}
/*
【trick&&吐槽】
1，最好不要因为题目说数据随机化就从一上来就考虑错误的程序。这可能将你带入万丈深渊！
2，做题要有智商啊！爆搜怎么剪枝？用智商剪枝！
3，如果想到用智商剪枝，比赛的时候就6题碾压全场了>_< 呜呜呜！
【题意】
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5672
自己看吧= =
【类型】
搜索+贪心 智商题 好题
【分析】
这道题需要智商。
不能一味爆搜。
我们枚举
1，12345+同花顺
2，3+3
3，2+2+2
剩下的牌可以走
1，炸弹
2，3+2
3，3
4，2
5，1
我们发现——
对于剩下的牌
1，能走炸弹肯定走炸弹。因为它拆了带别人的话，对答案也依然是没有贡献的。
2，有3张的话，先走3+2，走不了3+2的话走3
3，没有3张的话走2和走1
【时间复杂度&&优化】
搜索的时候只搜索作为开始的牌，最多能把我们带入3个分支，看似是27^k的复杂度，实际因为k平均很小，所以复杂度非常小，可以0msAC。
【数据】
input
100
0
KC KC KH KS
AC AD AH 5H 5S
3C 3D 3S JC JD
9S
BJ
2C 2C 3C 3C 4C 4D
5H 6H 7C 8H 9H
outout
6
input
100
7
4C5C6C7C8C
7C7D7S7D
BJBJ6C6D6D
AH2C3D4S5H
8C8D9D9D0D7H
ACAD
output
4
input
100
K
4C5C6C7C8C
7C7D7S7D
BJBJ6C6D6D
AH2C3D4S5H
8C8D9D9D0D0H
ACAD
output
4
*/