"Accordian" Patience
Time Limit: 3000 MS
You are to simulate the playing of games of ``Accordian‘‘ patience, the rules for which are as follows:
Deal cards one by one in a row from left to right, not overlapping. Whenever the card matches its immediate neighbour on the left, or matches the third card to the left, it may
be moved onto that card. Cards match if they are of the same suit or same rank. After making a move, look to see if it has made additional moves possible. Only the top card of each pile may be moved at any given time. Gaps between piles should be closed up
as soon as they appear by moving all piles on the right of the gap one position to the left. Deal out the whole pack, combining cards towards the left whenever possible. The game is won if the pack is reduced to a single pile.
Situations can arise where more than one play is possible. Where two cards may be moved, you should adopt the strategy of always moving the leftmost card possible. Where a card may be moved either one position
to the left or three positions to the left, move it three positions.
Input
Input data to the program specifies the order in which cards are dealt from the pack. The input contains pairs of lines, each line containing 26 cards separated by single space characters. The final line
of the input file contains a # as its first character. Cards are represented as a two character code. The first character is the face-value (A=Ace, 2-9, T=10, J=Jack, Q=Queen, K=King) and the second character is the suit (C=Clubs, D=Diamonds, H=Hearts,
S=Spades).
Output
One line of output must be produced for each pair of lines (that between them describe a pack of 52 cards) in the input. Each line of output shows the number of cards in each of the piles remaining after
playing ``Accordian patience‘‘ with the pack of cards as described by the corresponding pairs of input lines.
Sample Input
QD AD 8H 5S 3H 5H TC 4D JH KS 6H 8S JS AC AS 8D 2H QS TS 3S AH 4H TH TD 3C 6S 8C 7D 4C 4S 7S 9H 7C 5D 2S KD 2D QH JD 6D 9D JC 2C KH 3D QC 6C 9S KC 7H 9C 5C AC 2C 3C 4C 5C 6C 7C 8C 9C TC JC QC KC AD 2D 3D 4D 5D 6D 7D 8D TD 9D JD QD KD AH 2H 3H 4H 5H 6H 7H 8H 9H KH 6S QH TH AS 2S 3S 4S 5S JH 7S 8S 9S TS JS QS KS #
Sample Output
6 piles remaining: 40 8 1 1 1 1 1 pile remaining: 52
题目大意：
52张扑克，从左到右在平面上排列，按着如下规则处理：
1.按照从左到右的顺序，如果一张牌和左边的第一张或者第三张匹配，就把它放到对应的牌上面。
2.如果可以移动到多个位置，移动到最左端的牌上面。（匹配：花色或者数值相同）
注意：每次只移动每叠牌的最顶上的牌。
解题思路：vector / list、栈、模拟。对于每叠牌建立一个栈，进行模拟即可。
代码一：( list + stack )
#include<iostream>
#include<list>
#include<stack>
#include<cstdio>
using namespace std;
struct card{
char x,y;
card(char x0,char y0){x=x0,y=y0;}
};
list< stack<card> >cards;
list <stack<card> >::iterator it;
bool judge(card a,card b){//判断可以调换的情况
if(a.x==b.x||a.y==b.y) return true;
return false;
}
list< stack<card> >::iterator pre1(const list< stack<card> >::iterator i ){
list < stack<card> >::iterator it=i;//函数作用：当左边第一张等时返回左边第一个的地址
return --it;
}
list <stack<card> >::iterator pre3(const list< stack<card> >::iterator i){
list <stack<card> >::iterator it=i;//函数作用：当左边第三张等时
return ------it;//返回前三个的地址
}
bool input(){
char s[3];
while(~scanf("%s",s)){
if(s[0]=='#') return false;
card c(s[0],s[1]);
stack<card> temp;//栈是push
temp.push(c);//列表是push_back
cards.push_back(temp);
if(cards.size()==52) return true;
}
}
void solve(){
bool ismoved = true;
while(ismoved){
ismoved = false;
size_t cnt;//unsigned int;
for(it=cards.begin(),cnt=0;it!=cards.end();it++,cnt++){
if(cnt>2&&judge(it->top(),pre3(it)->top())){
pre3(it)->push(it->top());
it->pop();
ismoved=true;
if(it->empty()){
cards.erase(it);
}
break;
}
if(cnt>0&&judge(it->top(),pre1(it)->top())){
pre1(it)->push(it->top());
it->pop();
ismoved=true;
if(it->empty()){
cards.erase(it);
}
break;
}
}
}
}
void outResult(){
if(cards.size()==1){
printf("%d pile remaining:",cards.size());
}
else printf("%d piles remaining:",cards.size());
for(it=cards.begin();it!=cards.end();it++){
printf(" %d",it->size());//只能用迭代器访问，不能用下标
}
printf("\n");
cards.clear();
}
int main(){
while(input()){
solve();
outResult();
}
return 0;
}
代码二：( vector+stack )
#include<iostream>
#include<vector>
#include<stack>
#include<cstdio>
using namespace std;
struct card{
char x,y;
card(char x0,char y0){x=x0,y=y0;}
};
vector< stack<card> >cards;
bool judge(card a,card b){//判断可以调换的情况
if(a.x==b.x||a.y==b.y) return true;
return false;
}
int pre1(int i ){
int it=i;//函数作用：当左边第一张等时返回左边第一个的地址
return --it;
}
int pre3(int i ){
int it=i;//函数作用：当左边第三张等时
return ------it;//返回前三个的地址
}
bool input(){
char s[3];
while(~scanf("%s",s)){
if(s[0]=='#') return false;
card c(s[0],s[1]);
stack<card> temp;//栈是push
temp.push(c);//列表是push_back
cards.push_back(temp);
if(cards.size()==52) return true;
}
}
void solve(){
bool ismoved = true;
while(ismoved){
ismoved = false;
for(int it=0;it!=cards.size();it++){
if(it>2&&judge(cards[it].top(),cards[it-3].top())){
cards[it-3].push(cards[it].top());
cards[it].pop();
ismoved=true;
if(cards[it].empty()){
cards.erase(cards.begin()+it);
}
break;
}
if(it>0&&judge(cards[it].top(),cards[it-1].top())){
cards[it-1].push(cards[it].top());
cards[it].pop();
ismoved=true;
if(cards[it].empty()){
cards.erase(cards.begin()+it);
}
break;
}
}
}
}
void outResult(){
if(cards.size()==1){
printf("%d pile remaining:",cards.size());
}
else printf("%d piles remaining:",cards.size());
for(int it=0;it<cards.size();it++){
printf(" %d",cards[it].size());//只能用迭代器访问，不能用下标
}
printf("\n");
cards.clear();
}
int main(){
while(input()){
solve();
outResult();
}
return 0;
}