求比赛名次 【微软面试100题 第三十六题】 - tractorman - 博客园


求比赛名次 【微软面试100题 第三十六题】 - tractorman - 博客园
n支队伍比赛,分别编号为0,1,2,...,n-1,已知它们之间的实力对比关系存储在一个二维数组w[n][n]中,w[i][j]的值代表编号为i,j的队伍中更强的一支,所以w[i][j] = i或者j.
  现在给出它们的出场顺序,并存储在数组order[n]中,比如order[n] = {4,3,5,8,1......},那么第一轮比赛就是4对3,5对8.胜者晋级,败者淘汰,同一轮淘汰的所有队伍排名不再细分,即可以随便排,下一轮由上一轮的胜者按照顺序,再一次两两比,比如可能是4对5,直至出现第一名。
  编程实现,给出二维数组w,一维数组order和用于输出比赛名次的数组result[n],求result.
题目分析:
  假设出场次序为:1 3 4 2 0 5,实力图和过程讲解如下图:
常见的思路肯定是模拟。
用一个临时数组保存本轮winner胜利的人,loser存到result中,倒着存。

特别注意w比较的是w[tmp[i][[tmp[i+1]]而不是w[i][i+1]。
另外一定要先存loser,再改tmp,否则。。你懂的。

void match(int w[MAX][MAX], int* order, int* result, int n)
{
    int tb = n;
    int tmp[MAX];
    int lose = n-1;
    int win = 0;
    int i;
    memcpy(tmp, order, sizeof(int)*n);
    while(tb>1)
    {
        win = 0;
        for(i=0;i<tb && (i+1)<tb;i+=2)
        {
            if(w[tmp[i]][tmp[i+1]]==tmp[i])
            {
                result[lose--] = tmp[i+1];
                tmp[win++] = tmp[i];
            }else
            {
                result[lose--] = tmp[i];
                tmp[win++] = tmp[i+1];
            }
        }
        if(tb%2==1)
        {
            tmp[win++] = tmp[tb-1];
        }
        tb = win;
        for(i=0;i<tb;i++)
        {
            printf("%d ", tmp[i]);
        }
        printf("\n");
    }
    result[0] = tmp[0];
}

void Calc(int ppW[][N],int *pOrder,int *pResult,int n)
{
    if(ppW==NULL || *ppW==NULL || pOrder==NULL || pResult==NULL || n<=0)
        return ;

    int nCurPos = n-1;
    vector<int> vectOrder;

    for(int i = 0;i<n;i++)
        vectOrder.push_back(pOrder[i]);

    while(vectOrder.size()>1)
    {
        for(vector<int>::iterator j = vectOrder.begin();j!=vectOrder.end();j++)
        {
            if(j+1 != vectOrder.end())
            {
                if(ppW[*j][*(j+1)] == *j)
                {
                    pResult[nCurPos--] = *(j+1);
                    vectOrder.erase(j+1);
                }
                else
                {
                    pResult[nCurPos--] = *j;
                    //这句话的意思就是删除j,返回值是j的后一位,把该位重新赋值给j
                    j = vectOrder.erase(j);
                }
            }
        }
    }
    if(vectOrder.size()==1)
        pResult[nCurPos--] = *(vectOrder.begin());
}
xfinitytv.comcast.net
(1)        建立order的一头一尾两个指针,ij
(2)        头尾指针每次移动两位,移动后指向的队伍需要跟下一个队伍比赛。也就是order[i]order[i+1]比赛; order[j]order[j-1]比赛。然后将尾指针两队中的获胜者跟头指针中两队的失败者互换。即是:loser[order[i]]win[order[i+1]] win[order[j]]win[order[j+1]]
在这里根据参加比赛的队伍总数n的不同需要讨论几种特殊情况:
a)n4的倍数。停止条件为:i>j
i   i+1                         i   i+1  j-1  j                         j-1  j





















b). n是偶数但不是4的倍数。停止条件为i>j
                i  j=i+1
 










c). n是奇数字。那么最初的尾指针为j=n-1,也就是把最后一支队轮空。
最后要把最后一个元素再换到前面去(黑色),因为它轮空了。











(3)        i个元素再次进行相似的运算。直到n=1
(4)        注意当j=i+1的时候需要保证胜利的队在序列的前面。
(5)        注意当数组元素为3的时候,i指向的元素跟轮空的元素(第三个元素)是相等的。而我们期望的是失败的元素(第二个元素)跟第三个元素进行互换。这种情况比较特殊,当且仅当只剩下3个元素的时候才会发生,所以需要用一段程序进行一定的修改。

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