题目:Problem J. Terminal
Input file: standard input
Output file: standard input
Time limit: 2 seconds
Memory limit: 256 mebibytes
N programmers from M teams are waiting at the terminal of airport. There are two shuttles at the exit
of terminal, each shuttle can carry no more than K passengers.
Now employees of the airport service need to choose one of the shuttles for each programmer. Note that:
• programmers already formed a queue before assignment to the shuttles;
• each second next programmer in the queue goes to the shuttle he or she is assigned for;
• when all programmers, assigned to the shuttle, are in, shuttle immediately closes door and leaves
the terminal;
• no two programmers from the same team may be assigned to the different shuttles;
• each programmer must be assigned to one of shuttles.
Check if its possible to find such as assignment; if the answer is positive, find minimum sum of waiting
times for each programmer. Waiting time for a person is defined as time when shuttle with this person
left the terminal; it takes one second to programmer to leave the queue and enter the assigned shuttle.
At moment 0 the first programmer begins moving to his shuttle.
Input
First line of the input contains three positive integers N, M and K (M ≤ 2000, M ≤ N ≤ 105, K ≤ 105).
Second line contains description of the queue — N space-separated integers Ai — ids of team of each
programmer in order they are placed in the queue (1 ≤ Ai ≤ M).
Output
If it is impossible to assign programmers to she shuttles following the rules above, print -1. Otherwise
print one integer — minimum sum of waiting times for all programmers.
Examples

standard input standard input
7 3 5
2 2 1 1 1 3 1
39
12 3 9
1 1 1 2 3 2 2 2 2 2 2 2
116
2 1 2
1 1
4

思路:

   如果存在可行解,那么最后一个人一定会上车,不如直接选定上第二辆车,所以第二辆车是第n秒开的。

  然后枚举上第一辆车的最后一个队的最后一个人是什么时候上车的。

  怎么判断可行呢?01背包即可。

 #include <bits/stdc++.h>

 using namespace std;

 #define MP make_pair
#define PB push_back
typedef long long LL;
typedef pair<int,int> PII;
const double eps=1e-;
const double pi=acos(-1.0);
const int K=1e6+;
const int mod=1e9+; int n,m,k;
LL ls[],rd[],cnt[];
LL ans=2e18;
bool dp[][];
bool cmp(int a,int b)
{
return ls[a]<ls[b];
}
int main(void)
{
//freopen("in.txt","r",stdin);
scanf("%d%d%d",&n,&m,&k);
for(int i=;i<=m;i++) rd[i]=i;
for(int i=,x;i<=n;i++)
scanf("%d",&x),ls[x]=i,cnt[x]++;
sort(rd+,rd+m+,cmp);
dp[][]=;
for(int i=,now=,pre=;i<=m;i++)
{
//printf("%d\n",rd[i]);
for(int j=;j<=k;j++)
if(dp[pre][j]&&j+cnt[rd[i]]<=k&&n-j-cnt[rd[i]]<=k)
ans=min(ans,1LL*(j+cnt[rd[i]])*ls[rd[i]]+1LL*(n-j-cnt[rd[i]])*n);
for(int j=;j<=k;j++)
{
dp[now][j]|=dp[pre][j];
if(j+cnt[rd[i]]<=k)
dp[now][j+cnt[rd[i]]]|=dp[pre][j];
dp[pre][j]=;
}
swap(now,pre);
}
if(ans==2e18) printf("-1\n");
else printf("%lld\n",ans);
return ;
}

XVII Open Cup named after E.V. Pankratiev Stage 14, Grand Prix of Tatarstan, Sunday, April 2, 2017 Problem J. Terminal的更多相关文章

  1. XVII Open Cup named after E.V. Pankratiev Stage 14, Grand Prix of Tatarstan, Sunday, April 2, 2017 Problem F. Matrix Game

    题目: Problem F. Matrix GameInput file: standard inputOutput file: standard inputTime limit: 1 secondM ...

  2. XVII Open Cup named after E.V. Pankratiev Stage 14, Grand Prix of Tatarstan, Sunday, April 2, 2017 Problem L. Canonical duel

    题目:Problem L. Canonical duelInput file: standard inputOutput file: standard outputTime limit: 2 seco ...

  3. XVII Open Cup named after E.V. Pankratiev Stage 14, Grand Prix of Tatarstan, Sunday, April 2, 2017 Problem A. Arithmetic Derivative

    题目:Problem A. Arithmetic DerivativeInput file: standard inputOutput file: standard inputTime limit: ...

  4. 【二分】【字符串哈希】【二分图最大匹配】【最大流】XVII Open Cup named after E.V. Pankratiev Stage 14, Grand Prix of Tatarstan, Sunday, April 2, 2017 Problem I. Minimum Prefix

    给你n个字符串,问你最小的长度的前缀,使得每个字符串任意循环滑动之后,这些前缀都两两不同. 二分答案mid之后,将每个字符串长度为mid的循环子串都哈希出来,相当于对每个字符串,找一个与其他字符串所选 ...

  5. XVII Open Cup named after E.V. Pankratiev Stage 14, Grand Prix of Tatarstan, Sunday, April 2, 2017 Problem D. Clones and Treasures

    题目:Problem D. Clones and TreasuresInput file: standard inputOutput file: standard outputTime limit: ...

  6. 【二分图】【并查集】XVII Open Cup named after E.V. Pankratiev Stage 14, Grand Prix of Tatarstan, Sunday, April 2, 2017 Problem L. Canonical duel

    给你一个网格(n<=2000,m<=2000),有一些炸弹,你可以选择一个空的位置,再放一个炸弹并将其引爆,一个炸弹爆炸后,其所在行和列的所有炸弹都会爆炸,连锁反应. 问你所能引爆的最多炸 ...

  7. 【动态规划】【滚动数组】【bitset】XVII Open Cup named after E.V. Pankratiev Stage 14, Grand Prix of Tatarstan, Sunday, April 2, 2017 Problem J. Terminal

    有两辆车,容量都为K,有n(10w)个人被划分成m(2k)组,依次上车,每个人上车花一秒.每一组的人都要上同一辆车,一辆车的等待时间是其停留时间*其载的人数,问最小的两辆车的总等待时间. 是f(i,j ...

  8. 【枚举】【最小表示法】XVII Open Cup named after E.V. Pankratiev Stage 14, Grand Prix of Tatarstan, Sunday, April 2, 2017 Problem F. Matrix Game

    给你一个n*m的字符矩阵,将横向(或纵向)全部裂开,然后以任意顺序首尾相接,然后再从中间任意位置切开,问你能构成的字典序最大的字符串. 以横向切开为例,纵向类似. 将所有横排从大到小排序,枚举最后切开 ...

  9. 【推导】【构造】XVII Open Cup named after E.V. Pankratiev Stage 14, Grand Prix of Tatarstan, Sunday, April 2, 2017 Problem E. Space Tourists

    给你n,K,问你要选出最少几个长度为2的K进制数,才能让所有的n位K进制数删除n-2个元素后,所剩余的长度为2的子序列至少有一个是你所选定的. 如果n>K,那么根据抽屉原理,对于所有n位K进制数 ...

随机推荐

  1. ES5与ES6的继承

    JavaScript本身是一种神马语言: 提到继承,我们常常会联想到C#.java等面向对象的高级语言(当然还有C++),因为存在类的概念使得这些语言在实际的使用中抽象成为一个对象,即面向对象.Jav ...

  2. Nginx配置里的fastcgi_index和index

    在配置nginx时有时会遇到, 所以记录一下 location ^~ /wechat/ { index index.php; fastcgi_pass 127.0.0.1:9000; fastcgi_ ...

  3. iOS开发之--获取验证码倒计时及闪烁问题解决方案

    大家在做验证码的时候一般都会用到倒计时,基本上大家实现的方式都差不多,先贴出一些代码来.. -(void)startTime{ __block ; //倒计时时间 dispatch_queue_t q ...

  4. SQL TRIM()函数去除字符串头尾空格

    SQL TRIM()函数去除字符串头尾空格 SQL 中的 TRIM 函数是用来移除掉一个字串中的字头或字尾.最常见的用途是移除字首或字尾的空白.这个函数在不同的资料库中有不同的名称: MySQL: T ...

  5. JZOJ.5306【NOIP2017模拟8.18】棋盘游戏

    Description 这个游戏上在一个无限大的棋盘上, 棋盘上只有一颗棋子在位置(x,y)(x,y>=0)棋盘的左下角是(0,0)Amphetamine每次都是第一个移动棋子,然后Amphet ...

  6. 分布式项目中 linux 服务器 部署jar 应用脚本 deploy.sh

    在实际项目的部署中,尤其是分布式项目,有很多服务的jar包需要 部署,这里抽取出公用的 deploy的脚本 下面是不含jdk配置的 #!/bin/bash JAVA_OPTIONS_INITIAL=- ...

  7. warning: Now you can provide attr "wx:key" for a "wx:for" to improve performance.

    小程序开发过程中在写for循环的时候会出现如下报错 warning: Now you can provide attr "wx:key" for a "wx:for&qu ...

  8. js数组和字符串去重复几种方法

    js数组去重复几种方法 第一种:也是最笨的吧. Array.prototype.unique1 = function () { var r = new Array(); label:for(var i ...

  9. 170412、Spring Boot Quartz介绍

    (1)什么是Quartz? (2)Quartz的特点: (3)Quartz专用词汇说明: (4)Quartz任务调度基本实现原理: 接下来看下具体的内容: (1)什么是Quartz? Quartz是一 ...

  10. 深入学习AngularJS中数据的双向绑定机制

    来自:http://www.jb51.net/article/80454.htm Angular JS (Angular.JS) 是一组用来开发Web页面的框架.模板以及数据绑定和丰富UI组件.它支持 ...