ZOJ2418-Matrix

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

题意：给你一个矩阵，每个矩阵的每一行可以依次往右移动一个位置，最右边移出来的可以补到最左边，也就是循环移动啦，移动次数不限。问你移动后每列元素的和的最大值最小是多少。

解题思路：暴力枚举每行移了几次，第一行可以假设它不动

#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 0x3f3f3f3fint a[10][10];int ans[10];int n,mi;void dfs(int x){    if(x==n)    {        int ma=-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];            }            ma=max(ans1,ma);        }        mi=min(mi,ma);        return;    }    for(int i=0; i<n; i++)    {        ans[x]=i;        dfs(x+1);    }}int main(){    while(~scanf("%d",&n)&&n!=-1)    {        for(int i=0; i<n; i++)            for(int j=0; j<n; j++)                scanf("%d",&a[i][j]);        mi=INF;        ans[0]=0;        dfs(1);        printf("%d\n",mi);    }    return 0;}