ZOJ 2702 Unrhymable Rhymes 贪心
贪心。能凑成一组就算一组
Unrhymable Rhymes
Time Limit: 10 Seconds Memory Limit: 32768 KB Special Judge
An amateur poet Willy is going to write his first abstract poem. Since abstract art does not give much care to the meaning of the poem, Willy is planning to impress listeners with unusual
combinations of words. He prepared n lines of the future poem, but suddenly noticed that not all of them rhyme well.
Though abstractionist, Willy strongly respects canons of classic poetry. He is going to write the poem that would consist of quatrains. Each quatrain consists of two pairs of rhymed lines.
Therefore there can be four types of quatrains, if we denote rhymed lines with the same letter, these types are “AABB”, “ABAB”, “ABBA” and “AAAA”.
Willy divided the lines he composed into groups, such that in each group any line rhymes with any other one. He assigned a unique integer number to each group and wrote the number of
the group it belongs next to each line. Now he wants to drop some lines from the poem, so that it consisted of correctly rhymed quatrains. Of course, he does not want to change the order of the lines.
Help Willy to create the longest poem from his material.
Input
There are mutilple cases in the input file.
The first line of each case contains n --- the number of lines Willy has composed (1 <= n <= 4000 ). It is followed by n integer numbers denoting the rhyme
groups that lines of the poem belong to. All numbers are positive and do not exceed 109 .
There is an empty line after each case.
Output
On the first line of the output file print k --- the maximal number of quatrains Willy can make. After that print 4k numbers --- the lines that should form the poem.
There should be an empty line after each case.
Sample Input
15
1 2 3 1 2 1 2 3 3 2 1 1 3 2 2 3
1 2 3
Sample Output
3
1 2 4 5
7 8 9 10
11 12 14 15 0
Source: Andrew Stankevich's Contest #9
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector> using namespace std; const int maxn=4400; int n;
vector<int> v[maxn];
vector<int> ans; void init()
{
for(int i=0; i<=n+10; i++) v[i].clear();
ans.clear();
} int a[maxn],b[maxn]; int main()
{
bool first=true;
while(scanf("%d",&n)!=EOF)
{
if(first) first=false;
else putchar(10); init();
for(int i=0; i<n; i++)
{
scanf("%d",a+i);
b[i]=a[i];
}
sort(b,b+n);
int m=unique(b,b+n)-b;
int two=-1;
for(int i=0;i<n;i++)
{
a[i]=lower_bound(b,b+m,a[i])-b;
v[a[i]].push_back(i);
int vas=v[a[i]].size();
if(vas>=2)
{
if(two!=-1)
{
if(two!=a[i])
{
ans.push_back(v[two][0]);
ans.push_back(v[two][1]);
ans.push_back(v[a[i]][0]);
ans.push_back(v[a[i]][1]);
/// clear
for(int j=0; j<=n+10; j++) v[j].clear();
two=-1;
}
else if(two==a[i]&&vas>=4)
{
for(int j=0;j<4;j++)
{
ans.push_back(v[a[i]][j]);
}
for(int j=0; j<=n+10; j++) v[j].clear();
two=-1;
}
}
else
{
two=a[i];
}
}
}
printf("%d\n",(int)ans.size()/4);
sort(ans.begin(),ans.end());
for(int i=0,sz=ans.size();i<sz;i++)
{
printf("%d",ans[i]+1);
if((i+1)%4==0) putchar(10);
else putchar(32);
}
}
return 0;
}
ZOJ 2702 Unrhymable Rhymes 贪心的更多相关文章
- ZOJ 2702 Unrhymable Rhymes
Unrhymable Rhymes Time Limit:10000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu De ...
- ZOJ 2702 Unrhymable Rhymes(DP)
题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1702 题目大意:给定有很多数字组成的诗,譬如 “AABB”, “AB ...
- POJ 1862 & ZOJ 1543 Stripies(贪心 | 优先队列)
题目链接: PKU:http://poj.org/problem?id=1862 ZJU:http://acm.zju.edu.cn/onlinejudge/showProblem.do?proble ...
- 2017CCPC秦皇岛G ZOJ 3987Numbers(大数+贪心)
Numbers Time Limit: 2 Seconds Memory Limit: 65536 KB DreamGrid has a nonnegative integer n . He ...
- ZOJ 3946 Highway Project 贪心+最短路
题目链接: http://www.icpc.moe/onlinejudge/showProblem.do?problemCode=3946 题解: 用dijkstra跑单元最短路径,如果对于顶点v,存 ...
- zoj 2921 Stock(贪心)
Optiver sponsored problem. After years of hard work Optiver has developed a mathematical model that ...
- zoj 1375||poj 1230(贪心)
Pass-Muraille Time Limit: 2 Seconds Memory Limit: 65536 KB In modern day magic shows, passing t ...
- POJ - 2349 ZOJ - 1914 Arctic Network 贪心+Kru
Arctic Network The Department of National Defence (DND) wishes to connect several northern outposts ...
- ZOJ 3870 Team Formation 贪心二进制
B - Team Formation Description For an upcoming progr ...
随机推荐
- [python]文件操作read&readline&readlines
(1)read是将整个文件读入内存,将整个文件的内容当作一个字符串 (2)readline是一行一行的读如内存,每一次读的一行为一个字符串 (3)readlines是一次将整个文件读入内存,但是将整个 ...
- 20155202 2016-2017-2 《Java程序设计》第5周学习总结
20155202 2016-2017-2 <Java程序设计>第5周学习总结 教材学习内容总结 第八章:异常处理 java中所有错误会包装成对象,可以尝试(try)执行程序并捕捉(catc ...
- Sortable.js参数
所有的事件回调函数都有两个参数:event和ui,浏览器自有event对象,和经过封装的ui对象ui.helper - 表示sortable元素的JQuery对象,通常是当前元素的克隆对象ui.pos ...
- [转]Centos 安装Sublime text 3
本文简单介绍在CentOS上安装Sublime text 3, 转自:Centos 安装Sublime text 3 Step 1 :建立软件安装目录 # mkdir /opt # cd /opt S ...
- Linux内存管理 【转】
转自:http://blog.chinaunix.net/uid-25909619-id-4491368.html Linux内存管理 摘要:本章首先以应用程序开发者的角度审视Linux的进程内存管理 ...
- 【mac】7z 终端命令行
链接:http://www.2cto.com/os/201410/341079.html 7z指令 7z是7zip压缩工具的常用压缩文件格式.7zip是一个开源的压缩工具,软件本身十分小巧,功能强大, ...
- OpenLayers 3 之 地图图层数据来源(ol.source)详解
原文地址 source 是 Layer 的重要组成部分,表示图层的来源,也就是服务地址.除了在构造函数中制定外,可以使用 layer.setSource(source) 稍后指定.一.包含的类型 ol ...
- 设置adb server的端口号
在操作系统的系统环境里面,加一个环境变量: ANDROID_ADB_SERVER_PORT,值为9999,看自己喜欢.
- Linux学习笔记:sed删除、插入数据
一.sed删除文件第一行 sed -i '1d' file.txt -- 删除第一行 sed -i 'nd' file.txt -- 删除第n行 sed -i '$d' file.txt -- 删除最 ...
- django orm按天统计发布单数量
夜深了,先上代码和数据,明天再实现可视化图表. from datetime import datetime, timedelta from django.http import JsonRespons ...