Matrix

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Given an n*n matrix A, whose entries Ai,j are integer numbers ( 0 <= i < n, 0 <= j < n ). An operation SHIFT at row i ( 0 <= i < n ) will move the integers in the row one position
right, and the rightmost integer will wrap around to the leftmost column.

You can do the SHIFT operation at arbitrary row, and as many times as you like. Your task is to minimize

Input

The input consists of several test cases. The first line of each test case contains an integer n. Each of the following n lines contains n integers, indicating the matrix A. The input
is terminated by a single line with an integer -1. You may assume that 1 <= n <= 7 and |A_i,j| < 10⁴.

Output

For each test case, print a line containing the minimum value of the maximum of column sums.

Sample Input

2

4 6

3 7

3

1 2 3

4 5 6

7 8 9

-1

Sample Output

11

15

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Source: Asia 2004, Shanghai (Mainland China), Preliminary

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题目的意思是每一行可以任意移动求每列和的最大值的最小值

暴力DFS枚举行与行之间关系，可以开一个数组A保存每一行第一个数位置效率n^(n-1)，求最小n^2 ，总共n^(n+1)

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cmath>

#include <algorithm>

#include <queue>

#include <set>

#include <stack>

#include <map>

#include <functional>

#include <bitset>

#include <string>

using namespace std;

#define LL long long

#define INF 0x3f3f3f3f

int a[10][10];

int ans[10];

int n,mn;

void dfs(int x)

{

    if(x==n)

    {

        int mx=-1;

        for(int i=0; i<n; i++)

        {

            int ans1=0;

            for(int j=0; j<n; j++)

            {

                int pos=ans[j]+i;

                if(pos>=n) pos-=n;

                ans1+=a[j][pos];

            }

            mx=max(ans1,mx);

        }

        mn=min(mn,mx);

        return;

    }

    for(int i=0; i<n; i++)

    {

        ans[x]=i;

        dfs(x+1);

    }

}

int main()

{

    while(~scanf("%d",&n))

    {

        if(n==-1)

            break;

        for(int i=0; i<n; i++)

            for(int j=0; j<n; j++)

                scanf("%d",&a[i][j]);

        mn=INF;

        ans[0]=0;

        dfs(1);

        printf("%d\n",mn);

    }

    return 0;

}