UVa 127 - "Accordian" Patience POJ 1214 链表题解

UVa和POJ都有这道题。

不同的是UVa要求区分单复数，而POJ不要求。

使用STL做会比较简单，这里纯粹使用指针做了，非常麻烦的指针操作，一不小心就错。调试起来还是非常费力的

本题理解起来也是挺费力的，要搞清楚如何模拟也不容易啊，读题要很仔细。

纯指针的操作挺快的吧。不过POJ 0ms，而UVa就0.2左右了。

三相链表：

1 只要有叠起来的牌，那么就使用一个down指针指向下面的牌就可以了。

2 使用双向链表，可以方便前后遍历。

3 记得有了更新牌之后，又要重新开始检查是否需要更新牌，这是模拟的需要，不然或WA的。

#include <stdio.h>struct Node{int size;Node *pre, *post;Node *down;char a, b;Node () : pre(NULL), post(NULL), down(NULL), size(1) {}};//insert n to m positioninline void insertNode(Node *&m, Node *&n){n->post = m->post;n->pre = m->pre;if (m->post) m->post->pre = n;//小心断链if (m->pre) m->pre->post = n;//小心断链}inline void takeoutNode(Node *&n){if (n->down){Node *down = n->down;insertNode(n, down);return;}if (n->pre) n->pre->post = n->post;if (n->post) n->post->pre = n->pre;}inline void inStackNode(Node *&m, Node *&n){n->size = m->size+1;insertNode(m, n);n->down = m;}inline bool checkMovable(Node *n, Node *m){return n->a == m->a || n->b == m->b;}inline void pre3(Node *&n){if (n->pre) n = n->pre;if (n->pre) n = n->pre;if (n->pre) n = n->pre;}inline void pre1(Node *&n){if (n->pre) n = n->pre;}inline void deleteNodes(Node *&n){while (n){Node *p = n->post;while (n){Node *d = n->down;delete n; n = NULL;n = d;}n = p;}}int main(){Node *head = new Node; //Dummy headwhile (true){Node *it = new Node;it->a = getchar();if (it->a == '#') break;it->b = getchar();getchar();head->post = it;//initialize chainit->pre = head;for (int i = 1; i < 52; i++){Node *p = new Node;p->a = getchar();p->b = getchar();getchar();it->post = p;p->pre = it;it = p;}bool checkMove = true;while (checkMove){checkMove = false;it = head->post;while (it){Node *post = it->post;Node *p = it;pre3(p);if (p && p != head && checkMovable(p, it)){checkMove = true;takeoutNode(it);inStackNode(p, it);//调用参数不要乱了break;}p = it;pre1(p);if (p && p != head && checkMovable(p, it)){checkMove = true;takeoutNode(it);inStackNode(p, it);break;//题意，理解}it = post;}//while (it)}//while (checkMove && piles > 1)it = head->post;int piles = 0;while (it){piles++;it = it->post;}if (piles == 1) printf("%d pile remaining:", piles);else printf("%d piles remaining:", piles);it = head->post;while (it){printf(" %d", it->size);it = it->post;}putchar('\n');deleteNodes(head->post);}//while (true)delete head;return 0;}