UVa 127: "Accordian" patience data structure topics

Topic Link:

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=63

Type of topic: data structure, linked list

Meaning

Send a deck of cards (52 pieces, two lines), starting from left to right, for the current card, if the value (face-value) or shape (suit) of the first or the left is the same, move the card to the first or third row on the left (if the left 1 and the left 3 can be moved, Move to the left 3, and then continue moving until it is moved, if the move condition continues to be satisfied. When a stack of cards is empty, "empty" is eliminated.

Ideas for solving problems:

This problem is simple simulation, for each stack of cards, with a stack to save. The key to disintegration is the process of eliminating "empty stacks". Start with an array, but because the array to delete an element to move a lot, inefficient, resulting in tle. And then ready to use the linked list to solve. I was going to use the STL list at first, but I found it very uncomfortable. So I wrote the list of the implementation of the array to do, successful AC.

10269644

127

"Accordian" patience

Accepted

C++

0.412

2012-06-29 04:26:42

This paper url:http://www.bianceng.cn/programming/sjjg/201410/45628.htm

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
#
Output:

6 Piles remaining:40 8 1 1 1 1

1 Pile remaining:52

#include <iostream> #include <cstdio> #include <stack> using namespace std;  
    
int nsize;  
struct string{char face[3];  
    
};  
struct pile{stack<string>face;  
}ARR[100];  
    
int prev[100], next[100];  
    Eliminates "empty stack" void Delgaps () {int i,j;  
            for (i=0 i!=nsize; I=next[i]) {if (Arr[i].face.empty ()) {Next[prev[i]] = next[i];  
            Prev[next[i]] = prev[i];  
        return;  
    ()}///Move the token bool moves () {int cnt=1;  
        for (int i=next[0]; i!=nsize i=next[i],++cnt) {int last3, last1; if (cnt>2) {//If the current position is above 3, the first decision can be moved to the left 3 Last3 = Prev[prev[prev[i]]];//Get the coordinates of left 3 if (arr[i). Face.top (). Face[0]==arr[last3].face.top (). face[0] | |  
                Arr[i].face.top (). Face[1]==arr[last3].face.top (). Face[1]) {Arr[last3].face.push (Arr[i].face.top ());  
  Arr[i].face.pop ();              return true;  }} last1= Prev[i]; Gets the coordinates of the left 1 if (Arr[i].face.top (). Face[0]==arr[last1].face.top (). face[0] | | arr[i].face.top (). f  
            Ace[1]==arr[last1].face.top (). Face[1]) {Arr[last1].face.push (Arr[i].face.top ());  
            Arr[i].face.pop ();  
        return true;  
}//Do not continue to move, return false returned false;  } void Solve () {//As long as you can move, keep moving while (move ()) delgaps ();  
    After each move, to the process of eliminating the empty heap} int main () {freopen ("Input.txt", "R", stdin);  
    nsize=0;  
    String temp;  
        while (scanf ("%s", Temp.face)) {if (temp.face[0]== ' @ ') break;  
        while (!arr[nsize].face.empty ()) Arr[nsize].face.pop ();  
            
        
        Arr[nsize].face.push (temp);  
        Prev[nsize] = nSize-1;  
        Next[nsize] = nsize+1;  
            
        ++nsize; if (nsize==52) {soLve ();  
            int num=0;  
               
            for (int i=0; i!=nsize; i=next[i]) ++num;   if (num==1) printf ("1 pile remaining:");  
                
            POJ This problem, this pile also has plus s else printf ("%d piles remaining:", num);  
                for (int i=0; i!=nsize; I=next[i]) {if (i) printf ("");  
            printf ("%d", arr[i].face.size ());  
            printf ("\ n");  
        nsize=0;  
} return 0; }