codeforce 247 div.2 B

题目如下：

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 g_ij and the j-th student's happiness increases by g_ji. 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 g_ij (0 ≤ g_ij ≤ 10⁵). It is guaranteed that g_ii = 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

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

这道题我没想到用next_permutation这个函数，所以自己实现了全排列，并且用循环实现的，但鉴于这道题比较特殊，只有五个人，所以可以直接枚举其实要简单许多，所以我的代码复杂一些。

我的代码：

#include

int temp=0;
int g[5][5];
void swap(int *a,int *b)
{
    int temp=*a;
    *a=*b;
    *b=temp;
}

void perm(int list[],int k,int m)
{    //产生list[k:m]的所有排列
    int happy=0;
    if(k==m)
    {  //只剩下一个元素

        for(int i=1;i<=5;i++)
        {

                for(int j=1;j<=6-i;j++)
                {
                    if(j%2!=0&&j+1<=6-i)
                        happy+=g[list[j-2+i]-1][list[j+i-1]-1]+g[list[j+i-1]-1][list[j-2+i]-1];


                }
        }
        if(happy>temp)
            temp=happy;
    }
    else    //还有多个元素待排列，递归产生排列
    {
        for(int i=k;i<=m;i++)
        {
            swap(&list[i],&list[k]);//将第i个元素作为第一个
            perm(list,k+1,m);
            swap(&list[i],&list[k]);//将第i个元素换回原来的位置，准备用第i+1个做第一个
        }
    }
}

int main()
{
    for(int i=0;i<=4;i++)
for(int j=0;j<=4;j++)
    scanf("%d",&g[i][j]);

    int list[5];
    for(int i=0;i<=4;i++)
    list[i]=i+1;
perm(list,0,4);
printf("%d\n",temp);
    return 0;
}

更简单的代码：

#include 
#include 

using namespace std;

int A[5][5];
int P[5] = {0, 1, 2, 3, 4};

int val(int i, int j)
{
	return A[P[i]][P[j]] + A[P[j]][P[i]];
}

int main()
{
	for(int i = 0; i < 5; i++) for(int j = 0; j < 5; j++) cin >> A[i][j];
	int best = 0;
	do
	{
		best = max(best, val(0, 1) + val(2, 3) + val(1, 2) + val(3, 4) + val(2, 3) + val(3, 4));
	}
	while(next_permutation(P, P + 5));
	cout << best;
}