Sequence
http://poj.org/problem?id=2442
用STL写的时间为:5657MS
#include<cstdio>
#include<algorithm>
#include<queue>
#define MAXN 2005
using namespace std;
int main()
{
int t,n,m,c;
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&m);
int a[MAXN];
priority_queue< int,vector<int>,greater<int> >q;
priority_queue< int,vector<int>,less<int> >p;
for(int i=; i<m; i++)
{
scanf("%d",&c);
q.push(c);
}
for(int j=; j<n; j++)
{
for(int k=; k<m; k++)
{
scanf("%d",&a[k]);
}
while(!q.empty())
{
int mm=q.top();
q.pop();
for(int h=; h<m; h++)
{
if(p.size()==m&&p.top()>mm+a[h])
{
p.pop();
p.push(mm+a[h]);
}
else if(p.size()<m)
{
p.push(mm+a[h]);
}
}
}
while(!p.empty())
{
q.push(p.top());
p.pop();
}
}
int mark=;
for(int i=;i<m;i++)
{
if(mark)
{
printf("%d",q.top());
mark=;
}
else printf(" %d",q.top());
q.pop();
}
printf("\n");
}
return ;
}
我用堆写的时间:3969MS
#include<cstdio>
#include<algorithm>
#include<queue>
#define MAXN 2005
long long a[],b[];
using namespace std;
int len=,len1=;
void up1(int n)
{
a[++len]=n;
int p=len;
int q=p/;
long long m=a[p];
while(q>=&&m<a[q])
{
a[p]=a[q];
p=q;
q=p/;
}
a[p]=m;
}
void up2(int n)
{
b[++len1]=n;
int p=len1;
int q=p/;
long long m=b[p];
while(q>=&&m>b[q])
{
b[p]=b[q];
p=q;
q=p/;
}
b[p]=m;
}
void down1(int p)
{
a[]=a[len--];
int q=p*;
long long m=a[p];
while(q<=len)
{
if(q<len&&a[q]>a[q+])
q++;
if(a[q]>=m) break;
else
{
a[p]=a[q];
p=q;
q=p*;
}
}
a[p]=m;
}
int pop1()
{
long long r=a[];
return r;
}
int pop2()
{
long long r=b[];
return r;
}
void down2(int p)
{
b[]=b[len1--];
int q=p*;
long long m=b[p];
while(q<=len1)
{
if(q<len1&&b[q]<b[q+])
q++;
if(b[q]<=m) break;
else
{
b[p]=b[q];
p=q;
q=p*;
}
}
b[p]=m;
} int main()
{
int t,n,m,c;
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&m);
int aa[MAXN];
for(int i=; i<m; i++)
{
scanf("%d",&c);
up1(c);
}
for(int j=; j<n; j++)
{
for(int k=; k<m; k++)
{
scanf("%d",&aa[k]);
}
while(len!=)
{
int mm=pop1();
down1();
for(int h=; h<m; h++)
{
if(len1>=m&&b[]>mm+aa[h])
{
down2();
up2(mm+aa[h]);
}
else if(len1<m)
{
up2(mm+aa[h]);
}
}
}
while(len1!=)
{
up1(b[]);
down2();
}
}
int mark=;
for(int i=;i<m;i++)
{
if(mark)
{
printf("%lld",a[]);
mark=;
}
else printf(" %lld",a[]);
down1();
}
printf("\n");
}
return ;
}
Sequence的更多相关文章
- oracle SEQUENCE 创建, 修改,删除
oracle创建序列化: CREATE SEQUENCE seq_itv_collection INCREMENT BY 1 -- 每次加几个 STA ...
- Oracle数据库自动备份SQL文本:Procedure存储过程,View视图,Function函数,Trigger触发器,Sequence序列号等
功能:备份存储过程,视图,函数触发器,Sequence序列号等准备工作:--1.创建文件夹 :'E:/OracleBackUp/ProcBack';--文本存放的路径--2.执行:create or ...
- DG gap sequence修复一例
环境:Oracle 11.2.0.4 DG 故障现象: 客户在备库告警日志中发现GAP sequence提示信息: Mon Nov 21 09:53:29 2016 Media Recovery Wa ...
- Permutation Sequence
The set [1,2,3,-,n] contains a total of n! unique permutations. By listing and labeling all of the p ...
- [LeetCode] Sequence Reconstruction 序列重建
Check whether the original sequence org can be uniquely reconstructed from the sequences in seqs. Th ...
- [LeetCode] Binary Tree Longest Consecutive Sequence 二叉树最长连续序列
Given a binary tree, find the length of the longest consecutive sequence path. The path refers to an ...
- [LeetCode] Verify Preorder Sequence in Binary Search Tree 验证二叉搜索树的先序序列
Given an array of numbers, verify whether it is the correct preorder traversal sequence of a binary ...
- [LeetCode] Longest Consecutive Sequence 求最长连续序列
Given an unsorted array of integers, find the length of the longest consecutive elements sequence. F ...
- [LeetCode] Permutation Sequence 序列排序
The set [1,2,3,…,n] contains a total of n! unique permutations. By listing and labeling all of the p ...
- Leetcode 60. Permutation Sequence
The set [1,2,3,-,n] contains a total of n! unique permutations. By listing and labeling all of the p ...
随机推荐
- js提交前弹出提示框
<form target="_blank" name="f1" method="post" action="sub2.php ...
- Struts2接收参数的几种方式
一.用Action属性 在action里定义要接收的参数,并提供相应的set和get方法. 如: public class LoginAction extends ActionSupport { pr ...
- ASP.NET DropDownList1_SelectedIndexChanged使用
DropDownList1.AutoPostBack 属性 今天写代码给DropDownList1添加DropDownList1_SelectedIndexChanged事件,在运行测试时发现Drop ...
- [Angular 2] Using Pipes to Filter Data
Pipes allow you to change data inside of templates without having to worry about changing it in the ...
- Entity Framework CodeFirst------数据迁移(二)
众所周知当我们的项目涉及到数据库时,随着需求或大或小的 变更后,我们之前设计好的数据模型会发生部分的更改,导致数据表.或者数据字段的增加.修改等,这个时候我们就需要对数据库结构进行修改,如果我们之前采 ...
- Linux 基本命令(持续更新ing)
cd -> 变换路径 //文件一般存在/var/路径下,var为可修改存储盘 ls -> 列出所有隐藏文件与相关文件的属性 #ls -al ...
- Day6 - Python基础6 面向对象编程
Python之路,Day6 - 面向对象学习 本节内容: 面向对象编程介绍 为什么要用面向对象进行开发? 面向对象的特性:封装.继承.多态 类.方法. 引子 你现在是一家游戏公司的开发 ...
- AlertDialog dismiss 和 cancel方法的区别
AlertDialog使用很方便,但是有一个问题就是:dismiss方法和cancel方法到底有什么不同? AlertDialog继承与Dialog,现在各位看看结构图: 然后在Dialog类中找到了 ...
- HDU5289
题意:求解存在最大差值小于给定K值的所有区间段. 输入: T(测试数据) n(数组个数)K(给定区间值的范围) ai...(数组值) 输出: ss(所有满足符合条件的区间段) 思路:二分+ST算法,首 ...
- jq之简单表单验证
<body> <form method="post" action=""> <div class="int"& ...