标签：des style blog class code tar
| ``Accordian‘‘ Patience |
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 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).
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.
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 #
6 piles remaining: 40 8 1 1 1 1 1 pile remaining: 52
解决方案：
首先题意要仔细理解，然后模拟即可，不过，由于处理大量的字符，必须用scanf来输入，而且scanf不能对string类型进行输入，所以不能用string，要用字符数组，如果用string，就必须把字符赋给string，这样很耗时，由于这个，这里tle了几次。然后是模拟，用vector，要2000多毫秒，用list可能比较少点。
代码：
#include<iostream>
#include<cstdio>
#include<string>
#include<vector>
#include<stack>
using namespace std;
struct node
{
char x,y;
};
bool cmp(node a,node b)
{
if(a.x==b.x) return true;
if(a.y==b.y) return true;
return false;
}
int main(){
while(1)
{
vector<stack<node > >card;//定义vector（容器），包含的元素类型为stack<node>;
char temp[3];
scanf("%c",&temp[0]);
if(temp[0]==‘#‘) break;
scanf("%c",&temp[1]);
stack<node > p;
node c;
c.x=temp[0];c.y=temp[1];
p.push(c);
card.push_back(p);//把栈存入card容器。
for(int i=1;i<52;i++)
{
getchar();
scanf("%c%c",&temp[0],&temp[1]);
node c;
c.x=temp[0];c.y=temp[1];
stack<node > t;
t.push(c);
card.push_back(t);
}
/* int len=card.size();
for(int i=0;i<len;i++)
{
cout<<card[i].top()<<endl;
}
getchar();*/
int len=card.size();//求容器的大小
for(int i=0;i<len;i++)
{
int j=i;
if(j-3>=0){
if(cmp(card[j-3].top(),card[j].top())){ //容器也可像数组一样通过下标来使用。
card[j-3].push(card[j].top());
card[j].pop();
if(card[j].empty())
{
card.erase(card.begin()+j);//删除某个位置的栈，参数必须为指向该位置的指针
len=card.size();
}
i=-1;
continue;
}
}
if(j-1>=0){
if(cmp(card[j-1].top(),card[j].top())){
card[j-1].push(card[j].top());
card[j].pop();
if(card[j].empty())
{
card.erase(card.begin()+j);
len=card.size();
}
i=-1;
continue;
}
}
}
int l=card.size();
if(l>1)
{
printf("%d piles remaining:",l);
for(int i=0;i<l;i++)
printf(" %d",card[i].size());
}
else {
printf("%d pile remaining:",l);
printf(" %d",card[0].size());
}
printf("\n");
getchar();
}
return 0;
}标签：des style blog class code tar
原文地址：http://blog.csdn.net/wujunokay/article/details/25086957
