【编程题目】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，

/*
36.引用自网友：longzuo（运算）
谷歌笔试： 
19
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，再依次两两比，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]，
求出 result。
*/
#include <iostream>
#include <vector>
#include <string.h>
using namespace std;

//假设n一定是2的n次方，即每一轮不会出现轮空的情况
void getRank(int *w[], int * order, int * result, int n)
{
    memset(result, 0, n * sizeof(int));
    vector<int> stack;
    int round = 0;
    int n2 = n;
    while(n2 != 1)
    {
        n2 = (n2>>1);
        round++;
    }
    for(int r = 0; r < round; r++) //对比赛轮次循环
    {
        int i = 0;
        int num = 0;
        int member[2] = {0};
        while(i < n)//对比赛队伍循环
        {
            if(result[i] == r)
            {
                member[num++] = i;
            }
            if(num == 2)
            {
                if(*((int*)w + n * member[0] + member[1]) == member[0])
                {
                    result[member[0]]++;
                    stack.push_back(member[1]);
                }
                else
                {
                    result[member[1]]++;
                    stack.push_back(member[0]);
                }
                num = 0;
            }
            i++;
        }
    }

    for(int i = 0; i < n; i++) //冠军进栈
    {
        if(result[i] == round)
        {
            stack.push_back(i);
        }
    }

    for(int i = 0; i < n; i++)
    {
        result[i] = stack.back();
        stack.pop_back();
    }
}

int main()
{
    int w[4][4] = {{0, 0, 2, 0},
                   {0, 1, 1, 3},
                   {2, 1, 2, 3},
                   {0, 3, 3, 3}};
    int order[4] ={1, 2, 3, 4};
    int result[4];
    getRank((int **)w, order, (int *)result, 4);
    return 0;
}

/*----------------------------
Copyright by yuucyf. 2011.08.30
-----------------------------*/
#include "stdafx.h"
#include <iostream>
#include <vector>
#include <assert.h>
using namespace std;



bool CalcPosition(int **ppW, int *pOrder, int *pResult, int nGroup)    //nGroup arg mean n x n
{
    assert(ppW);
    assert(pOrder);
    assert(pResult);
    assert(nGroup > 0);

    int i32I = 0, i32J = 0;
    int nCurPos = nGroup - 1;
    vector<int> vectOrder;
    for (i32I = 0; i32I < nGroup; i32I++)
        vectOrder.push_back(pOrder[i32I]);

    while (vectOrder.size() > 1)
    {
        for (vector<int>::iterator it = vectOrder.begin(); it != vectOrder.end(); ++it)
        {
            if (it + 1 != vectOrder.end())
            {
                if (ppW[*it][*(it+1)] == *it)
                {
                    pResult[nCurPos--] = *(it+1);
                    vectOrder.erase(it+1);
                }
                else
                {
                    pResult[nCurPos--] = *it;
                    it = vectOrder.erase(it);
                }
            }
        }
    }

    if (vectOrder.size() == 1)
        pResult[nCurPos--] = vectOrder[0];

    return true;
}

int _tmain(int argc, _TCHAR* argv[])
{
    int n;
    cout << "请输入队伍数";
    cin >> n;
    cout << endl << "输入实力对比关系" << endl;

    int **ppW = new int* [n];
    assert(*ppW);
    for (int i32I = 0; i32I < n; i32I++)
    {
        ppW[i32I] = new int [n];
        assert(ppW);
        for (int i32J = 0; i32J < n; i32J++)
            cin >> ppW[i32I][i32J];
    }


    int *pOrder = new int[n];
    assert(pOrder);
    cout << "请输入出场顺序" << endl;
    for (int i32I = 0; i32I < n; i32I++)
        cin >> pOrder[i32I];

    int *pRes = new int[n];
    assert(pRes);
    if (CalcPosition(ppW, pOrder, pRes, n))
    {
        cout << endl << "比赛名次为:(大到小)" << endl;
        for (int i32I = 0; i32I < n; i32I++)
            cout << pRes[i32I] << " ";
    }

    //don‘t forget to free memory...

    return 0;
}

#include<iostream>


using namespace std;

#define N 5
void GetResult(int (*w)[N], int* order, int* result)
{
  int i = 0;
  int k = 1;
  int j = N -1;
  while(1)
  {
    i = 0;
    if(i + k > N -1)
    {
      result[j] = order[0];
      break;
    }
    
    while(i + k <= N-1)
    {
      int ii = order[i];

      int jj = order[i+k];
      if(w[ii][jj] == ii)
        result[j--] = jj;
      else
      {
        result[j] = ii;
        order[i]= order[i+k];
        order[i+k] = result[j];
        j --;
      }
      i = i + 2*k;
    }
    k *= 2;
  }
}

int main()
{
  int a[5][5] = {{0,1,2,3,4},{1,1,2,3,4},{2,2,2,3,4},{3,3,3,3,4},{4,4,4,4,4}};
  int order[5] = {4,3,1,2,0};
  int result[5];
  GetResult(a, order, result);

  int i = 0;
  cout << "result order: ";
  while(i < 5)
  {
    cout << result[i++] << ",";
  }
  cout << endl;
}