习题6-9 纸牌游戏（“Accordian” Patience, UVa 127）

这道题最终还是用vector(水过)AC了。。耗时0.67s。。好慢。。
思路挺简单。。就是按着题意做，每移动一次后，迭代判断左边是否可以移动。。

#include <iostream>#include <string>#include <vector>#include <stack>#include <queue>#include <deque>#include <set>#include <map>#include <algorithm>#include <sstream>#include <utility>#include <cstring>#include <cstdio>#include <cstdlib>#include <ctime>#include <cmath>#include <cctype>#define CLEAR(a, b) memset(a, b, sizeof(a))#define IN() freopen("in.txt", "r", stdin)#define OUT() freopen("out.txt", "w", stdout)#define LL long long#define maxn 105#define maxm 10005#define mod 1000000007#define INF 1000000007#define eps 1e-5#define PI 3.1415926535898#define N 52using namespace std;//-------------------------CHC------------------------------//char card[N][3];stack<char *> pile[N];bool _match(int i, int j) {//match，名字竟然有问题。。char *a = pile[i].top(), *b = pile[j].top();return a[0] == b[0] || a[1] == b[1];}void put(int i, int j) {char *a = pile[i].top(); pile[i].pop();//printf("put %s to %s\n", a, pile[j].top());//pile[j].push(a);}void move(int i) {if (pile[i].empty()) return;int cnt = 0;for (int j = i - 1; j >= 0; --j) { //先考虑左三步if (pile[j].size()) cnt++;if (cnt == 3) {if (_match(i, j)) {put(i, j);for (int k = 0; k <= i; ++k)move(k);return;}break;}}for (int j = i - 1; j >= 0; --j) { //再考虑左一步if (pile[j].size()) {if (_match(i, j)) {put(i, j);for (int k = 0; k <= i; ++k)move(k);}return;}}}int main() {while (scanf("%s", &card[0]) && card[0][0] != '#') {for (int i = 0; i < 52; ++i)while (pile[i].size()) pile[i].pop();pile[0].push(card[0]);for (int i = 1; i < 52; ++i) {scanf("%s", card[i]);pile[i].push(card[i]);}for (int i = 0; i < 52; ++i) //从左开始找可以移动的。。move(i);int cnt = 0;for (int i = 0; i < 52; ++i)if (pile[i].size()) cnt++;printf("%d", cnt);printf(" pile");if (cnt > 1) putchar('s');putchar(' ');printf("remaining:");for (int i = 0; i < 52; ++i) {if(pile[i].size()) printf(" %d", pile[i].size());}puts("");}return 0;}