B. Shower Line
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Many students live in a dormitory. A dormitory is a whole new world of funny amusements and possibilities but it does have its drawbacks.

There is only one shower and there are multiple students who wish to have a shower in the morning. That's why every morning there is a line of five people in front of the dormitory shower door. As soon as the shower opens, the first person from the line
enters the shower. After a while the first person leaves the shower and the next person enters the shower. The process continues until everybody in the line has a shower.

Having a shower takes some time, so the students in the line talk as they wait. At each moment of time the students talk in pairs: the
(2i - 1)-th man in the line (for the current moment) talks with the
(2i)-th one.

Let's look at this process in more detail. Let's number the people from 1 to 5. Let's assume that the line initially looks as 23154 (person number 2 stands at the beginning of the line). Then, before the shower opens, 2 talks with 3, 1 talks with 5, 4 doesn't
talk with anyone. Then 2 enters the shower. While 2 has a shower, 3 and 1 talk, 5 and 4 talk too. Then, 3 enters the shower. While 3 has a shower, 1 and 5 talk, 4 doesn't talk to anyone. Then 1 enters the shower and while he is there, 5 and 4 talk. Then 5
enters the shower, and then 4 enters the shower.

We know that if students i and
j talk, then the i-th student's happiness increases by
gij and the
j-th student's happiness increases by
gji. Your task is to find such initial order of students in the line that the total happiness of all students will be maximum in the end. Please note that some pair of students may have a talk several
times. In the example above students 1 and 5 talk while they wait for the shower to open and while 3 has a shower.

Input

The input consists of five lines, each line contains five space-separated integers: the
j-th number in the
i-th line shows gij (0 ≤ gij ≤ 105). It is guaranteed
that gii = 0 for all
i.

Assume that the students are numbered from 1 to 5.

Output

Print a single integer — the maximum possible total happiness of the students.

Sample test(s)
Input
0 0 0 0 9
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
7 0 0 0 0
Output
32
Input
0 43 21 18 2
3 0 21 11 65
5 2 0 1 4
54 62 12 0 99
87 64 81 33 0
Output
620

题目大意:输出最大的欢乐值。

思路:要输出最大的欢乐值,应该枚举全部的情况,找出最大的欢乐值,数据较小能够枚举。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<string>
#include<cmath>
#include<algorithm>
#define LL int
#define inf 0x3f3f3f3f
using namespace std;
int a[6][6];
int f[]={0,1,2,3,4};
bool bj;
int main()
{
LL n,m,i,j,k,l;
LL a1,a2,a3,a4;
for(i=0;i<5;i++)
{
for(j=0;j<5;j++)
{
scanf("%d",&a[i][j]);
}
}
for(i=0;i<4;i++)
{
for(j=i+1;j<5;j++)
{
a[i][j]=a[j][i]=a[i][j]+a[j][i];//枚举出全部可能的欢乐值
}
}
LL ans=0;
do//注意运行后推断
{
ans=max(ans,a[f[0]][f[1]]*2+a[f[1]][f[2]]*2+a[f[2]][f[3]]+a[f[3]][f[4]]);<span id="transmark"></span>
}
while(next_permutation(f,f+5))//注意排序问题。由于数组从0下标開始,若为1则不能够这么写
printf("%d\n",ans);
return 0;
}

Codeforces Round #247 (Div. 2) B的更多相关文章

  1. Codeforces Round #247 (Div. 2) ABC

    Codeforces Round #247 (Div. 2) http://codeforces.com/contest/431  代码均已投放:https://github.com/illuz/Wa ...

  2. Codeforces Round #247 (Div. 2) B - Shower Line

    模拟即可 #include <iostream> #include <vector> #include <algorithm> using namespace st ...

  3. Codeforces Round #247 (Div. 2)

    A.水题. 遍历字符串对所给的对应数字求和即可. B.简单题. 对5个编号全排列,然后计算每种情况的高兴度,取最大值. C.dp. 设dp[n][is]表示对于k-trees边和等于n时,如果is== ...

  4. Codeforces Round #247 (Div. 2) C题

    赛后想了想,然后就过了.. 赛后....... 我真的很弱啊!想那么多干嘛? 明明知道这题的原型就是求求排列数,这不就是 (F[N]-B[N]+100000007)%100000007: F[N]是1 ...

  5. Codeforces Round #247 (Div. 2) C. k-Tree (dp)

    题目链接 自己的dp, 不是很好,这道dp题是 完全自己做出来的,完全没看题解,还是有点进步,虽然这个dp题比较简单. 题意:一个k叉树, 每一个对应权值1-k, 问最后相加权值为n, 且最大值至少为 ...

  6. [Codeforces Round #247 (Div. 2)] A. Black Square

    A. Black Square time limit per test 1 second memory limit per test 256 megabytes input standard inpu ...

  7. Codeforces Round #247 (Div. 2) D. Random Task

    D. Random Task time limit per test 1 second memory limit per test 256 megabytes input standard input ...

  8. Codeforces Round #247 (Div. 2) C D

    这题是一个背包问题 这样的 在一个k子树上 每个节点都有自己的k个孩子 然后 从原点走 走到 某个点的 和为 N 且每条的 长度不小于D 就暂停问这样的 路有多少条,  呵呵 想到了 这样做没有把他敲 ...

  9. 「专题训练」k-Tree(CodeForces Round #247 Div.2 C)

    题意与分析(Codeforces-431C) 题意是这样的:给出K-Tree--一个无限增长的树,它的每个结点都恰有\(K\)个孩子,每个节点到它\(K\)个孩子的\(K\)条边的权重各为\(1,2, ...

随机推荐

  1. 中国版 Office 365 (X-Tenant / Tango) 功能验证报告 - 1 简介

    花了点时间做了一次Office 365 X-Tenant的 POC,对过程做了记录和总结,在这里会陆续分享: (一) 简介 这次POC的系统环境是模拟一个公司的生产环境: 1. 公司总部在国外,拥有 ...

  2. 迅为iTOP-4418嵌入式开发板初体验

    iTOP-4418开发板预装 Android4.4.4 系统, 支持9.7 寸.7 寸.4.3 寸屏幕. 参数:核心板参数 尺寸 50mm*60mm高度 核心板连接器为1.5mmCPU ARM Cor ...

  3. 高阶函数与接口混入和java匿名类

    高阶函数与接口混入和java匿名类. 高阶函数中的组件(参量)函数相当于面向对象中的混入(接口)类. public abstract class Bird { private String name; ...

  4. Windos无法验证文件数组签名

    参考链接:https://jingyan.baidu.com/article/09ea3ede6982c4c0aede39e6.html Windows无法验证文件数字签名而无法启动,照以下去做,可以 ...

  5. go protobuf 编码与解码

    package main import ( "encoding/hex" "fmt" "github.com/golang/protobuf/prot ...

  6. 循环冗余校验(CRC)算法入门

    http://blog.csdn.net/liyuanbhu/article/details/7882789 前言 CRC校验(循环冗余校验)是数据通讯中最常采用的校验方式.在嵌入式软件开发中,经常要 ...

  7. 循环实现数组 map 方法

    //循环实现数组 map 方法 const selfMap = function (fn, context) { let arr = Array.prototype.slice.call(this) ...

  8. 利用jquery制作滚动到指定位置触发动画

    <!DOCTYPE html><html><head> <meta charset="utf-8"> <title>利用 ...

  9. Spring Data Redis入门示例:数据序列化 (四)

    概述 RedisTemplate默认使用的是基于JDK的序列化器,所以存储在Redis的数据如果不经过相应的反序列化,看到的结果是这个样子的: 可以看到,出现了乱码,在程序层面上,不会影响程序的运行, ...

  10. 小程序调用支付报错:jsapi缺少参数: total_fee

    这种情况通常是因为在调用的时候,package参数有问题导致: wx.requestPayment中package参数必须是package:"prepay_id=wx21********** ...