PAT (Advanced Level) 1080. Graduate Admission (30)
简单题。
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<stack>
#include<queue>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std; struct X
{
int id;
int ge,gi;
int ch[];
}s[+];
int n,m,k;
int sch[];
vector<int>ans[];
int le[],li[]; bool cmp(const X&a,const X&b)
{
if(a.ge+a.gi==b.ge+b.gi) return a.ge>b.ge;
return a.ge+a.gi>b.ge+b.gi;
} int main()
{
scanf("%d%d%d",&n,&m,&k);
for(int i=;i<=m;i++) scanf("%d",&sch[i]);
for(int i=;i<=n;i++)
{
s[i].id=i-;
scanf("%d%d",&s[i].ge,&s[i].gi);
for(int j=;j<=k;j++)
{
scanf("%d",&s[i].ch[j]);
s[i].ch[j]++;
}
}
sort(s+,s++n,cmp); for(int i=;i<=n;i++)
{
for(int j=;j<=k;j++)
{
if(sch[s[i].ch[j]]>)
{
sch[s[i].ch[j]]--;
ans[s[i].ch[j]].push_back(s[i].id);
le[s[i].ch[j]]=s[i].ge;
li[s[i].ch[j]]=s[i].gi;
break;
}
if(le[s[i].ch[j]]==s[i].ge&&li[s[i].ch[j]]==s[i].gi)
{
sch[s[i].ch[j]]--;
ans[s[i].ch[j]].push_back(s[i].id);
le[s[i].ch[j]]=s[i].ge;
li[s[i].ch[j]]=s[i].gi;
break;
}
}
} for(int i=;i<=m;i++)
{
sort(ans[i].begin(),ans[i].end());
for(int j=;j<ans[i].size();j++)
{
printf("%d",ans[i][j]);
if(j<ans[i].size()-) printf(" ");
}
printf("\n");
}
return ;
}
PAT (Advanced Level) 1080. Graduate Admission (30)的更多相关文章
- 【PAT甲级】1080 Graduate Admission (30 分)
题意: 输入三个正整数N,M,K(N<=40000,M<=100,K<=5)分别表示学生人数,可供报考学校总数,学生可填志愿总数.接着输入一行M个正整数表示从0到M-1每所学校招生人 ...
- pat 甲级 1080. Graduate Admission (30)
1080. Graduate Admission (30) 时间限制 200 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue It ...
- PAT 甲级 1080 Graduate Admission (30 分) (简单,结构体排序模拟)
1080 Graduate Admission (30 分) It is said that in 2011, there are about 100 graduate schools ready ...
- PAT 1080. Graduate Admission (30)
It is said that in 2013, there were about 100 graduate schools ready to proceed over 40,000 applicat ...
- 1080. Graduate Admission (30)
时间限制 200 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue It is said that in 2013, there w ...
- 1080 Graduate Admission (30)(30 分)
It is said that in 2013, there were about 100 graduate schools ready to proceed over 40,000 applicat ...
- PAT (Advanced Level) 1111. Online Map (30)
预处理出最短路再进行暴力dfs求答案会比较好.直接dfs效率太低. #include<cstdio> #include<cstring> #include<cmath&g ...
- PAT (Advanced Level) 1107. Social Clusters (30)
简单并查集. #include<cstdio> #include<cstring> #include<cmath> #include<vector> # ...
- PAT (Advanced Level) 1103. Integer Factorization (30)
暴力搜索. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #i ...
随机推荐
- Shell脚本,自动化发布tomcat项目【转】
Shell脚本,自动化发布tomcat项目脚本. 1. vko2c_auto_build_by_scp.sh 文件内容: #---------------------start------------ ...
- bind启动时提示953端口被使用
部署DNS的时候遇到个奇葩的问题,总是提示 couldn't add command channel 0.0.0.0#953: address in use 实际上系统上并没有进程使用953端口.查询 ...
- phpStorm 2016.1.2 最新版激活方法【亲测可用】
测试日期:2016-07-29 下载地址:https://yunpan.cn/c6mWAGbExcyjf 访问密码 00fb 1.windows版本 菜单help >>>> ...
- libevent linux安装
wget http://monkey.org/~provos/libevent-1.4.13-stable.tar.gzwget http://downloads.sourceforge.net/le ...
- struts2获取请求参数的三种方式及传递给JSP参数的方式
接上一篇文章 package test; import com.opensymphony.xwork2.ActionSupport; import javax.servlet.http.*; impo ...
- 怎样让pl sql developer 界面视图复位
tools->preferences->user interface->appearance->reset docking工具-首选项-用户界面-外观-复位停放
- ubuntu 安装LNMP
How To Install Linux, nginx, MySQL, PHP (LEMP) stack on Ubuntu 12.04 PostedJune 13, 2012 802.8kviews ...
- 【多重背包模板】poj 1014
#include <iostream> #include <stdio.h> #include <cstring> #define INF 100000000 us ...
- 【bfs】 poj 3984 maze 队列存储
#include <iostream> #include <stdio.h> #include <cstring> #define Max 0x7f7f7f7f u ...
- 集合工具类CollectionUtils、ListUtils、SetUtils、MapUtils探究
之前一直以为集合工具类只有CollectionUtils,主要用它的isEmpty(final Collection<?> coll)静态方法来判断一个给定的集合是否为null或者是否长度 ...