Codeforces Round #247 (Div. 2) B. Shower Line(简单模拟)
来源:互联网 发布:桐乡淘宝美工 编辑:程序博客网 时间:2024/05/17 08:21
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.
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.
Print a single integer — the maximum possible total happiness of the students.
0 0 0 0 90 0 0 0 00 0 0 0 00 0 0 0 07 0 0 0 0
32
0 43 21 18 23 0 21 11 6554 62 12 0 995 2 0 1 487 64 81 33 0
620
In the first sample, the optimal arrangement of the line is 23154. In this case, the total happiness equals:
#include <queue>#include <functional>#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <stack>#include <vector>#include <set>#include <map>#include <string>#include <cmath>#include <cstdlib>#include <ctime>#include <assert.h>#define REP(i,k,n) for(int i=k;i<n;i++)#define REPP(i,k,n) for(int i=k;i<=n;i++)#define scan(d) scanf("%d",&d)#define scann(n,m) scanf("%d%d",&n,&m)#define mst(a,k) memset(a,k,sizeof(a));#define LL long long#define eps 1e-8#define INF 0x6f6f6f6f#define mod 1000000007#define PI acos(-1.)using namespace std;#define N 15#define M 1005int num[N][N];int f(int a,int b,int c,int d,int e) //暴力模拟{ return num[a][b]+num[b][a]+num[c][d]+num[d][c]+num[b][c]+num[c][b]+num[d][e]+num[e][d]+num[c][d]+num[d][c]+num[d][e]+num[e][d];}int main(){ for(int i=0;i<5;i++){ for(int j=0;j<5;j++){ scanf("%d",&num[i][j]); } } int ans = 0; for(int i=0;i<5;i++){ // 5层for 暴力枚举 for(int j=0;j<5;j++){ if(i==j) continue; for(int k=0;k<5;k++){ if(i==k||j==k) continue; for(int l=0;l<5;l++){ if(i==l||j==l||k==l) continue; for(int t=0;t<5;t++){ if(i==t||j==t||k==t||l==t) continue; ans = max(ans,f(i,j,k,l,t)); } } } } } cout<<ans<<endl; return 0;}
- Codeforces Round #247 (Div. 2) B. Shower Line(简单模拟)
- Codeforces Round #247(Div. 2) B. Shower Line 暴力
- Codeforces #247 (Div. 2) B. Shower Line
- Codeforces Round #247 (Div. 2)a.Black Square,b.Shower Line,c.k-Tree前三题
- Codeforces Round #107 (Div. 2) B (简单模拟)
- Codeforces Round #323 (Div. 2) B 模拟
- Codeforces Round #362 (Div. 2) B 模拟
- Codeforces Round #315 (Div. 2) B 模拟
- Codeforces Round #105 (Div. 2) B 模拟
- Codeforces Round #392 (Div. 2)B (模拟)
- Codeforces Round #199 (Div. 2) -- B. Xenia and Spies (简单模拟)
- Codeforces Round #387 (Div. 2)B. Mammoth's Genome Decoding(简单模拟)
- B. Tell Your World Codeforces Round #431 (Div. 2)(简单模拟)
- codeforces 431B - Shower Line
- Codeforces 431B Shower Line
- codeforces 431B Shower Line
- codeforces 431B Shower Line
- Codeforces Round #256 (Div. 2) B. Suffix Structures(模拟)
- 生成ssh key及配置多个ssh key
- go新建文件权限与设置不符
- RedHat相关问题解决
- 在vs2013中编译运行调试Notepad++源码
- POJ 2828 - Buy Tickets(线段树)
- Codeforces Round #247 (Div. 2) B. Shower Line(简单模拟)
- Qt for Android 环境配置
- Java学习笔记--并行API
- Linux 系统创建自动运行脚本
- 重载与重写的区别
- HDU4352 lis+状压+数位dp
- Spring(三)Spring整合Struts2、Hibernate
- 打印菱形星号组合
- [c语言]通过指针获取函数内部变量值