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

原题链接：https://vjudge.net/problem/UVA-127
分类：链表
备注：vector模拟链表
这题有点水，放到这章估计本该直接用链表的，但是我不太熟还是用vector代替了。

代码如下：

#include<string>
#include<iostream>
#include<vector>
using namespace std;
vector<string>pile[52];
bool match(int x, int y) {
	if (pile[x][pile[x].size() - 1][0] == pile[y][pile[y].size() - 1][0])return true;
	if (pile[x][pile[x].size() - 1][1] == pile[y][pile[y].size() - 1][1])return true;
	return false;
}
void push(int x, int y) {
	pile[x].push_back(pile[y][pile[y].size() - 1]);
	pile[y].pop_back();
}
int main(void) {
	string s;
	while (cin >> s && s != "#") {
		for (int i = 0; i < 52; i++)pile[i].clear();
		pile[0].push_back(s);//s[0]是点数,s[1]是花色
		for (int i = 1; i < 52; i++) {
			cin >> s; pile[i].push_back(s);
		}
		for (int i = 1; i < 52; i++) {
			if (pile[i].empty())continue;
			int flag = 0;
			int k = i;
			while (k > 0) {
				int pos1 = k, cnt = 0;
				while (cnt < 1) {
					pos1--;
					if (pos1 < 0)break;
					if (!pile[pos1].empty())cnt++;
				}
				int pos3 = pos1;
				while (cnt < 3) {
					pos3--;
					if (pos3 < 0)break;
					if (!pile[pos3].empty())cnt++;
				}
				if (cnt == 3) {
					if (match(pos3, k)) {
						push(pos3, k); k = pos3; flag = 1; 
					}
					else if (match(pos1, k)) {
						push(pos1, k); k = pos1; flag = 1;
					}
					else break;
				}
				else if (cnt > 0) {
					if (match(pos1, k)) {
						push(pos1, k); k = pos1; flag = 1; 
					}
					else break;
				}
				else break;
			}
			if (flag)i = 0;
		}
		vector<int>ans;
		for (int i = 0; i < 52; i++)
			if (!pile[i].empty())
				ans.push_back(pile[i].size());
		if (ans.size() == 1)printf("1 pile remaining: 52\n");
		else {
			printf("%d piles remaining: ", ans.size());
			for (int i = 0; i < ans.size(); i++)
				printf("%d%c", ans[i], i == ans.size() - 1 ? '\n' : ' ');
		}
	}
	return 0;
}