soj.1003 hit or miss

问题介绍

一个纸牌游戏，多名玩家数1~13，如果所数的数与最上面的牌的值相同，则传给下一个玩家，否则，将最上面的牌放至队尾。下一个玩家进行同样的操作，最后一名玩家遇到可以传递的牌，则丢弃而不是传给第一个玩家。胜利条件为最后一名玩家丢弃所有的牌。

输入：局数、玩家数、牌的顺序

输出：所有玩家最后丢弃（传递）的牌，或者unwinnable

问题剖析：

1、这道题用到的数据结构，queue（队列），即FIFO（先进先出），来模拟牌的状况。

2、这题的边界判断：什么时候为输----无线循环时，我采用的方法为，用一个steps数组记录下所有玩家，在丢弃下一张牌所需要花多少步，如果所花的步数，超过他自己手中牌乘以13的数值（即最差情况下，每一张手里牌都需要数13次才能数到这个数），这时，肯定是无法丢弃下一张牌，即输了。

下面是我的代码

#include<iostream>#include<queue>using namespace std;int main(void) {  int j, t;  // j for count, t for times.  cin >> t;  for (j = 0; j < t; ++j) {    // over is used to notice whether the loop is over.    // flag is used to notice whether the player is win.    int n, i, temp, over = 1, flag = 0, totalcard = 52;    // counter for each player's count number    // steps is to calculate how many step until we discard the next card    int counter[10], steps[10], answer[10];    queue<int> player[10];    // we create the queue array in this loop, so that after each loop it will automatically release the memory    // initialize    for (i = 0; i < 10; ++i) {      counter[i] = 1;      steps[i] = 0;    }    cin >> n;    // to store all 52 card number    for (i = 0; i < totalcard; ++i) {      cin >> temp;      player[0].push(temp);    }    while (over) {      for (i = 0; i < n; ++i) {if (player[i].empty())  // if the queue is empty, we need not to do this loop   continue;if (counter[i] == player[i].front()) {  // is time to deal with the card  if (player[i].size() == 1)  // if it is the last card of the queue    answer[i] = player[i].front();  // we need to store this card for answer  if (i != n - 1)  // if it is not the last player    player[i + 1].push(counter[i]);  // we pass the card to next one  else    --totalcard;  // to count whether we discard all the card  player[i].pop();  // discard  steps[i] = 0;  // count from begining} else {  // if our count not match the number of the front card  player[i].push(player[i].front());  player[i].pop();  // put the front card to the end}// add the counter and make sure it is not out of rangeif (counter[i] >= 13)  counter[i] = 1;else  ++counter[i];// if the steps that we need to discard one card is more than the steps that I can count all the card in the desk, it means we loseif (steps[i] > (int)player[i].size() * 13) {  over = 0;  break;}++steps[i];  // add steps      }      if (totalcard == 0) {  // we winflag = 1;over = 0;      }    }    cout << "Case " << j + 1 << ": ";    if (flag) {      for (i = 0; i < n - 1; ++i)cout << answer[i] << ' ';      cout << answer[i] << endl;    }    else      cout << "unwinnable" << endl;  }}