Arranging Your Team

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1218    Accepted Submission(s): 360

Problem Description
Your country has qualified for the FIFA 2010 South Africa World Cup. As the coach, you have made up the 23 men squad. Now you must select 11 of them as the starters. As is well known, there are four positions in soccer: goalkeeper, defender, midfielder and striker. Your favorite formation is 4-4-2, that is, you should choose 4 defenders, 4 midfielders, 2 strikers, and of course, 1 goalkeeper. As a retired ACMer, you want to write a program to help you make decision. Each person's ability has been evaluated as a positive integer. And what's more, for some special pairs of persons, if the two people are both on the field, there will be an additional effect (positive or negative). Now you should choose the 11 persons to make the total value maximum.
 
Input
There are multiple test cases, separated by an empty line. The first 23 lines of each test case indicate each person's name Si, ability value Vi, and position. The length of each name is no more than 30, and there are no whitespaces in the names. All the names are different. The ability values are positive integers and no more than 100. The position is one of "goalkeeper", "defender", "midfielder" and "striker".

Then an integer M indicates that there are M special pairs. Each of the following M lines contains Si, Sj and Cij, means that if Si and Sj are both on the field, the additional profit is Cij. (-100 ≤ Cij ≤ 100). Si and Sj are different strings, and must be in the previous 23 names. All the (Si, Sj) pairs
are different.

 
Output
Output one line for each test case, indicating the maximum total ability values, that is, the total ability values of the 11 persons plus the additional effects. If you cannot choose a 4-4-2 formation, output "impossible" instead.
 
Sample Input
Buffon 90 goalkeeper
De_Sanctis 80 goalkeeper
Marchetti 80 goalkeeper
Zambrotta 90 defender
Cannavaro 90 defender
Chiellini 90 defender
Maggio 90 defender
Bonucci 80 defender
Criscito 80 defender
Bocchetti 80 defender
Pirlo 90 midfielder
Gattuso 90 midfielder
De_Rossi 90 midfielder
Montolivo 90 midfielder
Camoranesi 80 midfielder
Palombo 80 midfielder
Marchisio 80 midfielder
Pepe 80 midfielder
Iaquinta 90 striker
Di_Natale 90 striker
Gilardino 80 striker
Quagliarella 80 striker
Pazzini 80 striker
1
Pirlo Quagliarella 50

ZhangSan01 50 goalkeeper
ZhangSan02 50 defender
ZhangSan03 50 defender
ZhangSan04 50 defender
ZhangSan05 50 defender
ZhangSan06 50 defender
ZhangSan07 50 defender
ZhangSan08 50 defender
ZhangSan09 50 defender
ZhangSan10 50 defender
ZhangSan11 50 defender
ZhangSan12 50 defender
ZhangSan13 50 defender
ZhangSan14 50 defender
ZhangSan15 50 defender
ZhangSan16 50 midfielder
ZhangSan17 50 midfielder
ZhangSan18 50 midfielder
ZhangSan19 50 midfielder
ZhangSan20 50 midfielder
ZhangSan21 50 midfielder
ZhangSan22 50 midfielder
ZhangSan23 50 midfielder
0

 
Sample Output
1030
impossible
 
Source
 
Recommend
zhouzeyong
 
题意:。。懒得打,身为体育迷一看就懂把。
思路:搜索。适当剪枝即可。
/*
* Author: Joshua
* Created Time: 2014年08月29日 星期五 18时52分48秒
* File Name: hdu3720.cpp
*/
#include<cstdio>
#include<cstring>
#include<map>
#include<string>
using namespace std;
#define inf 100000000
char s[35],ps[35];
map<string,int> mp;
const int limit[5]={0,1,4,4,2};
int v[25],g[25][25],id[25],cnt[5],t[12],ans; int gao(char x[])
{
if (x[0]=='g') return 1;
if (x[0]=='d') return 2;
if (x[0]=='m') return 3;
return 4;
} void init()
{
int m,x;
mp.clear();
memset(g,0,sizeof(g));
for (int i=1;i<=23;++i)
{
mp[s]=i;
id[i]=gao(ps);
if (i<23) scanf("%s%d%s",s,&v[i+1],ps);
}
scanf("%d",&m);
for (int i=1;i<=m;++i)
{
scanf("%s%s%d",s,ps,&x);
g[mp[s]][mp[ps]]=x;
g[mp[ps]][mp[s]]=x;
}
} void updata()
{
int temp=0;
for (int i=1;i<=11;++i)
temp+=v[t[i]];
for (int i=1;i<=11;++i)
for (int j=i+1;j<=11;++j)
temp+=g[t[i]][t[j]];
if (temp>ans) ans=temp;
} void dfs(int x,int y)
{
if (y==11)
{
updata();
return;
}
if (x>23) return;
for (int i=0;i<=1;++i)
{
cnt[id[x]]+=i;
if (cnt[id[x]]<=limit[id[x]])
{ if (i) t[y+1]=x;
dfs(x+1,y+i);
}
cnt[id[x]]-=i;
}
} void solve()
{
memset(cnt,0,sizeof(cnt));
ans=-inf;
dfs(1,0);
if (ans==-inf) printf("impossible\n");
else printf("%d\n",ans);
} int main()
{
while (scanf("%s%d%s",s,&v[1],ps)!=EOF)
{
init();
solve();
}
return 0;
}

  

hdu3720 Arranging Your Team的更多相关文章

  1. HDU 3720 Arranging Your Team(DFS)

    题目链接 队内赛里,匆匆忙忙写的. #include <cstdio> #include <cstring> #include <iostream> #includ ...

  2. hdu 3720 Arranging Your Team 枚举

    不可能解可以直接判断. 搭配产生的附加分可以用一个二维数组保存. 枚举1442,4种类型的人,因为总人数只有23个,所以可以搜索暴力枚举,然后保存最优解. 注意trick,答案可能为负数,所以初始化a ...

  3. HDU 3720 Arranging Your Team

    先分组,然后暴力:注意  初始化时不要为0 会有负数:我直接二进制枚举: dfs是正解:呵呵 #include <iostream> #include <cstdio> #in ...

  4. Arranging Your Team

    http://acm.hust.edu.cn/vjudge/contest/view.action?cid=35800#problem/D #include <iostream> #inc ...

  5. Arranging Your Team HDU - 3720 【DFS】

    思路 题意:此题大意是指首先给你23个队员的信息,包括他们的名字,能力值,在赛场上的职位.然后给出几个若能满足某两个队员同时在球场上就额外加上一定的值.最后让你从23个队员中选出11个人,使得最终的v ...

  6. Configure a VLAN on top of a team with NetworkManager (nmcli) in RHEL7

    SOLUTION VERIFIED September 13 2016 KB1248793 Environment Red Hat Enterprise Linux 7 NetworkManager ...

  7. Create a Team in RHEL7

    SOLUTION VERIFIED September 13 2016 KB2620131 Environment Red Hat Enterprise Linux 7 NetworkManager ...

  8. Team Leader 你不再只是编码, 来炖一锅石头汤吧

    h3{ color: #000; padding: 5px; margin-bottom: 10px; font-weight: bolder; background-color: #ccc; } h ...

  9. Configure bridge on a team interface using NetworkManager in RHEL 7

    SOLUTION IN PROGRESS February 29 2016 KB2181361 environment Red Hat Enterprise Linux 7 Teaming,Bridg ...

随机推荐

  1. Serv-u Mysql数据库用户

    Serv-u 关联Mysql数据库用户需要用到ODBC数据源,windows不自带支持MySQL.所以要网上下载自己安装 官网下载地址:http://dev.mysql.com/downloads/c ...

  2. 开源个.NetCore写的 - 并发请求工具PressureTool

    本篇和大家分享的是一个 并发请求工具,并发往往代表的就是压力,对于一些订单量比较多的公司这种情况很普遍,也因此出现了很多应对并发的解决方案如:分布式,队列,数据库锁等: 对于没有遇到过或者不可能线上来 ...

  3. CocoaPods私有库管理

    简介: 前一篇文章已经介绍过如果安装使用CocoaPods,下面将要介绍如果通过CocoaPods和git来维护我们私有的库. 个人或公司在开发过程中,会积累很多可以复用的代码包,有些我们不想开源,又 ...

  4. phpMyAdmin安装部署

    phpMyAdmin 是一个用PHP编写的软件工具,可以通过web方式控制和操作MySQL数据库.通过phpMyAdmin 可以完全对数据库进行操作,例如建立.复制和删除数据等等.如果使用合适的工具, ...

  5. Eclipse中安装MemoryAnalyzer插件及使用

    Eclipse中安装MemoryAnalyzer插件 一.简介 Eclipse作为JAVA非常好用的一款IDE,其自带的可扩展插件非常有利于JAVA程序员的工作效率提升. MemoryAnalyzer ...

  6. poj 3253 Fence Repair 优先队列

    poj 3253 Fence Repair 优先队列 Description Farmer John wants to repair a small length of the fence aroun ...

  7. Monotonicity 2[POI2010]

    题目描述 给出N个正整数a[1..N],再给出K个关系符号(>.<或=)s[1..k].选出一个长度为L的子序列(不要求连续),要求这个子序列的第i项和第i+1项的的大小关系为s[(i-1 ...

  8. mysql控制台出现“unknown column 'password' in 'field list'问题

    今天在windows系统上使用MySQL命令时,出现下面的"unknown column 'password' in 'field list'问题 解决办法如下,使用authenticati ...

  9. Dubbo实战快速入门 (转)

    Dubbo是什么? Dubbo[]是一个分布式服务框架,致力于提供高性能和透明化的RPC远程服务调用方案,以及SOA服务治理方案. 其核心部分包含: 远程通讯: 提供对多种基于长连接的NIO框架抽象封 ...

  10. Android 原生 Intent 分享支持的那些事

    版权声明: 本账号发布文章均来自公众号,承香墨影(cxmyDev),版权归承香墨影所有. 每周会统一更新到这里,如果喜欢,可关注公众号获取最新文章. 未经允许,不得转载. 一.前言 对于一个 App ...