奥运排序问题【浙江大学】★

本文主要是介绍奥运排序问题【浙江大学】★，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

牛客网题目链接

注意

1.相同金牌数，相同名次，其他类似。
2.存在人数或者奖牌为0的情况。

#include <iostream>
#include <vector>
#include <string>
#include <cmath>
#include <algorithm>
#include <queue>
#include <cstdio>
#include <cctype>
#include <unordered_map>
#include <map>
#include <cfloat> 
using namespace std;
typedef pair<int, int> PII;
const int N = 245;
struct country{int g, m, pg, pm, ind, rank; //四种排名 
}E[N]; 
bool cmp1(PII a, PII b){return a.first > b.first;
}
bool cmp2(pair<double, int> a, pair<double, int> b){return a.first > b.first;
}
vector<PII> go, me;
vector<pair<double, int>> peg, pem;
int main(){int n, m, Gold, Medal, People;while(cin>>n>>m){go.clear();me.clear();peg.clear();pem.clear();for(int i = 0; i < n; i++){cin>>Gold>>Medal>>People;go.push_back({Gold, i});me.push_back({Medal, i});double t, tt;if(Gold==0 && People==0){ //金牌和人数都为0时 结果为0 t = 0;}else if(People==0 && Gold!=0){ //金牌数不为0 人数为0时 结果为无穷大 t = FLT_MAX;}else t = Gold*1.0/People;if(Medal==0 && People==0){tt = 0;}else if(People==0 && Medal!=0){tt = FLT_MAX;}else tt = Medal*1.0/People;	peg.push_back({t, i});pem.push_back({tt, i});}sort(go.begin(), go.end(), cmp1);sort(me.begin(), me.end(), cmp1);sort(peg.begin(), peg.end(), cmp2);sort(pem.begin(), pem.end(), cmp2);for(int i = 0; i < n; i++){  //具有相同成果的 处于同一排名 //金牌 int ida = go[i].second;if(i == 0 || go[i-1].first != go[i].first){E[ida].g = i + 1;}else E[ida].g = E[go[i-1].second].g; //奖牌int idb = me[i].second;if(i == 0 || me[i-1].first != me[i].first){E[idb].m = i + 1;}else E[idb].m = E[me[i-1].second].m; //人均金牌int idc = peg[i].second;if(i == 0 || peg[i-1].first != peg[i].first){E[idc].pg = i + 1;}else E[idc].pg = E[peg[i-1].second].pg; //人均奖牌 int idd = pem[i].second;if(i == 0 || pem[i-1].first != pem[i].first){E[idd].pm = i + 1;}else E[idd].pm = E[pem[i-1].second].pm; }for(int i = 0; i < n; i++){E[i].rank = 300;if(E[i].g < E[i].rank){E[i].rank = E[i].g;E[i].ind = 1;}if(E[i].m < E[i].rank){E[i].rank = E[i].m;E[i].ind = 2;}if(E[i].pg < E[i].rank){E[i].rank = E[i].pg;E[i].ind = 3;}if(E[i].pm < E[i].rank){E[i].rank = E[i].pm;E[i].ind = 4;}}int x;for(int i = 0; i < m; i++){cin>>x;cout<<E[x].rank<<":"<<E[x].ind<<endl;}	cout<<endl;}return 0;
}