1080. Graduate Admission (30)

来源：互联网发布：linux饥荒服务器搭建编辑：程序博客网时间：2024/06/03 22:48

题目地址 http://www.patest.cn/contests/pat-a-practise/1080

此题主要是个条件排序，还用到了 map<>映射，还有如何根据题目构造结构和相关的变量，基本上完全按照要求来，下面是一次ac的代码，没写什么注释，不过思路清晰

#include <stdio.h>#include <iostream>  #include <stdlib.h>#include <vector>#include <string>#include <string.h>#include <algorithm>#include <math.h>#include <queue>#include <map>#include <unordered_map>#include <sstream>using namespace std;#define N 40001#define M 101#define K 6struct mydata{  int no;  int ge;// national entrance exam  int gt;  int totalscore;  vector<int> choises;  int rank;  int school;};int n, m, k;int quota[M];vector<mydata> vstu;bool cmp(mydata d1, mydata d2){  if (d1.totalscore > d2.totalscore)    return true;  if (d1.totalscore == d2.totalscore && d1.ge > d2.ge)  {    return true;  }  return false;}vector<int> schools[M]; //第 i个学校 填报的stu的 no map<int, int> um; // no编号的 stu 在 vstu的编号vector<int> luqu[M];bool cmp2(int no1, int no2){  if (vstu[um[no1]].rank > vstu[um[no2]].rank)  {    return true;  }  return false;}bool cmp3(int a, int b){  return a < b;}int main(){  //freopen("in", "r", stdin);  while (scanf("%d%d%d",&n,&m,&k) != EOF)  {    um.clear();    int i,j;    for (i = 0; i < m; i++)    {      scanf("%d", &quota[i]);    }    mydata dt;    int tmp;    vstu.resize(n);    for (i = 0; i < n; i++)    {      dt.no = i;      dt.school = -1;      dt.choises.clear();      scanf("%d%d", &dt.ge, &dt.gt);      dt.totalscore = dt.ge + dt.gt;      for (j = 0; j < k; j++){        scanf("%d", &tmp);        dt.choises.push_back(tmp);        schools[tmp].push_back(dt.no);      }      vstu[i] = dt;    }    sort(vstu.begin(), vstu.end(), cmp);    vstu[0].rank = 1;    um[vstu[0].no] = 0;    for (i = 1; i < n; i++)    {      if (vstu[i].totalscore == vstu[i - 1].totalscore &&        vstu[i].ge == vstu[i - 1].ge)        vstu[i].rank = vstu[i - 1].rank;      else{        vstu[i].rank = i + 1;      }      um[vstu[i].no] = i;    }    int cout[M];    memset(cout, 0, sizeof(cout));    for (i = 0; i < n; i++)    {      while(vstu[i].school == -1)      {        mydata dt = vstu[i];        for (j = 0; j < k; j++)        {          int thChoice = vstu[i].choises[j];          if (cout[thChoice] < quota[thChoice])          {            vstu[i].school = thChoice;            cout[thChoice] ++;            luqu[thChoice].push_back(vstu[i].no);            break;          }          else{            int theLastNo = luqu[thChoice][quota[thChoice] - 1];            if (vstu[i].rank == vstu[um[theLastNo]].rank)            {              cout[thChoice] ++;              luqu[thChoice].push_back(vstu[i].no);              vstu[i].school = thChoice;              break;            }          }        }        if (j == k)          break;      }    }    for (i = 0; i < m; i++)    {      vector<int> vvv = luqu[i];      int lenn = vvv.size();      if (lenn == 0)      {        printf("\n");      }      else{        if (lenn == 1)          printf("%d\n", luqu[i][0]);        else{          sort(vvv.begin(), vvv.end(), cmp3);          printf("%d", vvv[0]);          for (j = 1; j < lenn; j++)          {            printf(" %d", vvv[j]);          }          printf("\n");        }      }    }    //printf("\n");  }  return 0;}

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