CodeForces 659B Qualifying Contest
来源:互联网 发布:绿霸安全软件 编辑:程序博客网 时间:2024/06/05 04:09
题意:给你n个人,然后有m个地区,每个地区将会选出分数最高的两个人,如果分数最高的不止两个人的话,输出?
思路:每个地区排序然后看看第二第三的分数就好了....
#include<bits/stdc++.h>using namespace std;const int maxn = 1e5+7;vector<pair<int,string> >S[maxn];bool cmp(pair<int,string> A,pair<int,string> B){ return A.first>B.first;}int main(){ int n,m; scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) { string s;int x,y; cin>>s>>x>>y; S[x].push_back(make_pair(y,s)); } for(int i=1;i<=m;i++)sort(S[i].begin(),S[i].end(),cmp); for(int i=1;i<=m;i++) { if(S[i].size()==2)cout<<S[i][0].second<<" "<<S[i][1].second<<endl; else { if(S[i][2].first==S[i][1].first)cout<<"?"<<endl; else cout<<S[i][0].second<<" "<<S[i][1].second<<endl; } }}
0 0
- CodeForces 659B Qualifying Contest
- CodeForces-659B-Qualifying Contest
- Codeforces 659B Qualifying Contest【模拟,读题】
- Codeforces 659B Qualifying Contest 【模拟】
- CodeForces - 659B Qualifying Contest (模拟)水
- CodeForces 659 B. Qualifying Contest(结构体排序的问题)
- B - Qualifying Contest
- CodeForces 659B Qualifying Contest(选2个人参加比赛)
- Codeforces Round #346 (Div. 2)-B. Qualifying Contest(排序)
- Codeforces Round #346 (Div. 2)--B. Qualifying Contest
- Codeforces Round #346 (Div. 2) B. Qualifying Contest
- Codeforces Round #346 (Div. 2) - B Qualifying Contest
- Codeforces Round #346 (Div. 2) B. Qualifying Contest
- Codeforces Round #346 (Div. 2) B - Qualifying Contest 优先队列
- B. Qualifying Contest 结构体模拟
- codeforces contest 785 b题
- codeforces contest 779 B题
- codeforces/contest/797/problem/B
- Android之NDK开发
- 关于JWT
- 对SQL Server的操作
- 两个有序数组元素之积、和的最小K个值
- Matlab中三维直方图的显示方法
- CodeForces 659B Qualifying Contest
- 使用JWT的OAuth2的SSO分析
- android自定义view实现公章效果
- 常用排序算法 Java 实现
- Poj 1151 Atlantis
- LeetCode Algorithms #26 <Remove Duplicates from Sorted Array>
- 对象的创建,访问类中的属性和方法
- 第5周项目1 三角形5(复制构造函数)
- 多态的典型例子(向上转型)