Sicily.1063. Who's the Boss(排序，确定上下属关系)

/*1063. Who's the Boss  题目大意：给出员工ID，收入，身高。 当a收入大于b且身高大于b，            且a的收入高于b的量是最小时，a为b的直接上司            b的下属也是a的下属。            对查询某个ID，输出其直属上司以及该ID的下属数量  方法：1.先把员工按照收入从大到小降序排列。        2.从底部开始往上遍历数组，依次找出每个员工的直属上司，将自身          fatherID = 上司所处数组下标i           并且初始化下属数量childNum = 0, 同时在相应上司的child          集合中push_back（下属所处数组下标i）          当没有上司时，其fatherID = -1；           3、再从底部往上遍历数组，当发现其下属容器chilid不为空时，           遍历容器，把其下属的childNum+1 加到自身的childNum上        4、针对每一个查询的ID，找出相应的ID后输出结果         */#include <stdlib.h>#include <iostream>#include <algorithm>#include <stdio.h>#include <vector>using namespace std;struct Employee{  int ID, salary, height, fatherID, childNum;   vector<int> child;  };bool compare(Employee a, Employee b){      return a.salary > b.salary ;    }int main (){    int testN;    int employeeNum, queryNum;    scanf("%d", &testN);    //cin >> testN;    Employee employee[30000];    int query[200];    int ID, salary, height;    for(int i=1; i<=testN; i++){       //cin >> employeeNum >> queryNum;       scanf("%d %d", &employeeNum,&queryNum);             for(int k=0; k<employeeNum; k++){          //cin >> employee[k].ID >> employee[k].salary >> employee[k].height;          scanf("%d %d %d", &employee[k].ID, &employee[k].salary , &employee[k].height);          employee[k].childNum = 0;          employee[k].child.clear();          }       sort(employee,employee+employeeNum, compare);                     for(int j=employeeNum-1; j>=0; j--){          int k;          for(k=j-1; k>=0; k--){             if(employee[k].height>= employee[j].height)                break;                        }          if(k>=0){             employee[j].fatherID = k;             employee[k].child.push_back(j);            }          else             employee[j].fatherID  = -1;            }       for(int j=employeeNum-1; j>=0; j--){          if(employee[j].child.size() >0){            for(int k=0; k <employee[j].child.size(); k++ ){                 employee[j].childNum += employee[employee[j].child[k]].childNum + 1;              }                               }         }          for(int k=0; k<queryNum; k++)       {          int queryID ;          //cin >> queryID;          scanf("%d", &queryID);          int j;          for(j=0; j<employeeNum; j++)            if(employee[j].ID == queryID)                break;          if(employee[j].fatherID == -1)               //cout << "0 " << employee[j].childNum << endl;               printf("0 %d\n", employee[j].childNum);              else             //cout << employee[employee[j].fatherID].ID << " " << employee[j].childNum << endl;             printf("%d %d\n", employee[employee[j].fatherID].ID,employee[j].childNum );             }           }        system("pause");    return 0;    }