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.

Examples
Input

Copy
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

Copy
32
Input

Copy
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

Copy
620
Note

In the first sample, the optimal arrangement of the line is 23154. In this case, the total happiness equals:

(g23 + g32 + g15 + g51) + (g13 + g31 + g54 + g45) + (g15 + g51) + (g54 + g45) = 32.

 
next_permutation即可;
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream> //#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 1000005
#define inf 0x3f3f3f3f
#define INF 9999999999
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++) inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
} ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; } /*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/ ll qpow(ll a, ll b, ll c) {
ll ans = 1;
a = a % c;
while (b) {
if (b % 2)ans = ans * a%c;
b /= 2; a = a * a%c;
}
return ans;
}
/*
int n, m;
int st, ed;
struct node {
int u, v, nxt, w;
}edge[maxn<<1]; int head[maxn], cnt; void addedge(int u, int v, int w) {
edge[cnt].u = u; edge[cnt].v = v; edge[cnt].w = w;
edge[cnt].nxt = head[u]; head[u] = cnt++;
} int rk[maxn]; int bfs() {
queue<int>q;
ms(rk);
rk[st] = 1; q.push(st);
while (!q.empty()) {
int tmp = q.front(); q.pop();
for (int i = head[tmp]; i != -1; i = edge[i].nxt) {
int to = edge[i].v;
if (rk[to] || edge[i].w <= 0)continue;
rk[to] = rk[tmp] + 1; q.push(to);
}
}
return rk[ed];
}
int dfs(int u, int flow) {
if (u == ed)return flow;
int add = 0;
for (int i = head[u]; i != -1 && add < flow; i = edge[i].nxt) {
int v = edge[i].v;
if (rk[v] != rk[u] + 1 || !edge[i].w)continue;
int tmpadd = dfs(v, min(edge[i].w, flow - add));
if (!tmpadd) { rk[v] = -1; continue; }
edge[i].w -= tmpadd; edge[i ^ 1].w += tmpadd; add += tmpadd;
}
return add;
}
ll ans;
void dinic() {
while (bfs())ans += dfs(st, inf);
}
*/ int g[10][10];
int a[6];
int main()
{
//ios::sync_with_stdio(0);
//memset(head, -1, sizeof(head));
for (int i = 1; i <= 5; i++)
for (int j = 1; j <= 5; j++)cin >> g[i][j];
for (int i = 1; i <= 5; i++) {
a[i] = i;
}
ll maxx = -inf;
do {
ll ans = 0;
ans += (g[a[1]][a[2]] + g[a[2]][a[1]] + g[a[3]][a[4]] + g[a[4]][a[3]]);
ans += (g[a[2]][a[3]] + g[a[3]][a[2]] + g[a[4]][a[5]]+g[a[5]][a[4]]);
ans += (g[a[3]][a[4]] + g[a[4]][a[3]]); ans += (g[a[4]][a[5]] + g[a[5]][a[4]]);
maxx = max(maxx, ans);
} while (next_permutation(a + 1, a + 1 + 5));
cout << maxx << endl;
return 0;
}

CF431B Shower Line的更多相关文章

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

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

  2. codeforces 431 B Shower Line【暴力】

    题意:给出五个人的编号,分别为 1 2 3 4 5,他们在排队, 最开始的时候,1和2可以交谈,3和4可以交谈 然后1走了之后,2和3交谈,4和5可以交谈 2走了之后,3和4可以交谈, 3走了之后,4 ...

  3. codeforces B. Shower Line 解题报告

    题目链接:http://codeforces.com/contest/431/problem/B 题目意思:给出5 * 5 的矩阵.从这个矩阵中选出合理的安排次序,使得happiness之和最大.当第 ...

  4. codeforces431B

    Shower Line CodeForces - 431B Many students live in a dormitory. A dormitory is a whole new world of ...

  5. 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 ...

  6. Codeforces Round #247 (Div. 2) B

    B. Shower Line time limit per test 1 second memory limit per test 256 megabytes input standard input ...

  7. POJ 3669 Meteor Shower(流星雨)

    POJ 3669 Meteor Shower(流星雨) Time Limit: 1000MS    Memory Limit: 65536K Description 题目描述 Bessie hears ...

  8. poj 3669 Meteor Shower

                                                                                                      Me ...

  9. POJ3669(Meteor Shower)(bfs求最短路)

    Meteor Shower Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 12642   Accepted: 3414 De ...

随机推荐

  1. PD中设置外键约束名称生成规则

    选择Database—>Edit Current DBMS选择Scripts->Objects->Reference->ConstName可以发现右侧的Value为: FK_% ...

  2. oracle系统函数(日期函数)

    oracle系统函数(日期函数) 调用日期函数时,首先要明确两个概念,时间戳和日期是不同的,日期中包括年月日但不包括小时分钟秒,时间戳包括年月日小时分钟秒.在oracle中,一般情况下,函数中包含da ...

  3. 《Android应用性能优化》 第8章 图形

    1.例子中 30个部件的xml setContentView 几乎占用了从onCreate() 到 onResume() 结束之前所有时间的99% 因为展开布局的开销很大.要尽量用不同的布局方式.比如 ...

  4. oracle DML-(insert、select、update、delete)

    一.插入记录INSERT INTO table_name (column1,column2,...) values ( value1,value2, ...); 示例:insert into emp ...

  5. linux设置自动获取IP地址

    右键单击,选择设置 勾选桥接模式

  6. 详解CSS盒模型(转)

    详解CSS盒模型   阅读目录 一些基本概念 盒模型 原文地址:http://luopq.com/2015/10/26/CSS-Box-Model/ 本文主要是学习CSS盒模型的笔记,总结了一些基本概 ...

  7. 【总结整理】行内标签span设置position:absolute/float属性可以设置宽度与高度

    postion:absolute 跳出文本流,不是行内元素,设置宽高有效,我的理解. 引用下曹刘阳写的<编写高质量代码-web前端开发修炼之道>一书中看到的一句话:position:abs ...

  8. Struts2框架01【如果使用struts框架】【利用struts框架写一个 hello world】

    1 什么是Struts2框架 基于MVC设计模式的web应用框架 Struts2框架是一个轻量级的MVC流程框架 轻量级是指程序的代码不是很多,运行时占用的资源不是很多,MVC流程框架就是说它是支持分 ...

  9. Java3D读取3DMax模型并实现鼠标拖拽、旋转、滚轮缩放等功能

    /**-------------------------------------------------代码区--------------------------------------------- ...

  10. Luogu 3168 [CQOI2015]任务查询系统

    区间修改单点查询,又观察到是一个k小,考虑主席树上做差分 一开始样例疯狂挂,后来发现主席树在一个历史版本上只能修改一次,所以要开2*n个根结点,记录一下每个时间对应的根结点编号 然后80分,考虑到当一 ...